| Exercise 2.1 Exercise 2.2 Exercise 2.3 |
Exercise 2.2
Question 1 A
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
x2 – 3Sol :
Let f (x) = x2 – 3
Now, if we recall the identity
(a2 – b2) = (a – b)(a + b)
Using this identity, we can write
x2 – 3 = (x – √3) (x + √3)
So, the value of x2 – 3 is zero when x = √3 or x = – √3
Therefore, the zeroes of x2 – 3 are √3 and – √3.
Verification
Now,
Sum of zeroes = α + β = √3 + ( – √3) = 0 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{0}{1}=0$
Product of zeroes = αβ = (√3)( – √3) = – 3 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{1}=-3$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 2x2 – 8x + 6
By splitting the middle term, we get
f(x) = 2 x2 – (2 + 6)x + 6 [∵ – 8 = – (2 + 6) and 2×6 = 12]
= 2 x2 – 2x – 6x + 6
= 2x(x – 1) – 6(x – 1)
= (2x – 6) (x – 1)
On putting f(x) = 0, we get
(2x – 6) (x – 1) = 0
⇒2x – 6 = 0 or x – 1 = 0
⇒x = 3 or x = 1
Thus, the zeroes of the given polynomial 2 x2 – 8x + 6 are 1 and 3
Verification
Sum of zeroes = α + β = 3 + 1 = 4 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{2}=4$
So, the relationship between the zeroes and the coefficients is verified.
Question 1 B
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
2x2 – 8x + 6Sol :
Let f(x) = 2x2 – 8x + 6
By splitting the middle term, we get
f(x) = 2 x2 – (2 + 6)x + 6 [∵ – 8 = – (2 + 6) and 2×6 = 12]
= 2 x2 – 2x – 6x + 6
= 2x(x – 1) – 6(x – 1)
= (2x – 6) (x – 1)
On putting f(x) = 0, we get
(2x – 6) (x – 1) = 0
⇒2x – 6 = 0 or x – 1 = 0
⇒x = 3 or x = 1
Thus, the zeroes of the given polynomial 2 x2 – 8x + 6 are 1 and 3
Verification
Sum of zeroes = α + β = 3 + 1 = 4 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{2}=4$
Product of zeroes = αβ = (3)(1) = 3 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{6}{2}=3$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = x2 – 2x – 8
By splitting the middle term, we get
f(x) = x2 – 4x + 2x – 8 [∵ – 2 = 2 – 4 and 2×4 = 8]
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
On putting f(x) = 0, we get
(x + 2) (x – 4) = 0
⇒ x + 2 = 0 or x – 4 = 0
⇒x = – 2 or x = 4
Thus, the zeroes of the given polynomial x2 – 2x – 8 are – 2 and 4
Verification
Sum of zeroes = α + β = – 2 + 4 = 2 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-2)}{1}=2$
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{6}{2}=3$
So, the relationship between the zeroes and the coefficients is verified.
Question 1 C
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
x2 – 2x – 8Sol :
Let f(x) = x2 – 2x – 8
By splitting the middle term, we get
f(x) = x2 – 4x + 2x – 8 [∵ – 2 = 2 – 4 and 2×4 = 8]
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
On putting f(x) = 0, we get
(x + 2) (x – 4) = 0
⇒ x + 2 = 0 or x – 4 = 0
⇒x = – 2 or x = 4
Thus, the zeroes of the given polynomial x2 – 2x – 8 are – 2 and 4
Verification
Sum of zeroes = α + β = – 2 + 4 = 2 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-2)}{1}=2$
Product of zeroes = αβ = ( – 2)(4) = – 8 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-8}{1}=-8$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 3x2 + 5x – 2
By splitting the middle term, we get
f(x) = 3x2 + (6 – 1)x – 2 [∵ 5 = 6 – 1 and 2×3 = 6]
= 3x2 + 6x – x – 2
= 3x(x + 2) – 1(x + 2)
= (3x – 1) (x + 2)
On putting f(x) = 0 , we get
(3x – 1) (x + 2) = 0
⇒ 3x – 1 = 0 or x + 2 = 0
$\Rightarrow \mathrm{x}=\frac{1}{3}$ or x = – 2
Thus, the zeroes of the given polynomial 3x2 + 5x – 2 are – 2 and
Verification
So, the relationship between the zeroes and the coefficients is verified.
Question 1 D
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
3x2 + 5x – 2Let f(x) = 3x2 + 5x – 2
By splitting the middle term, we get
f(x) = 3x2 + (6 – 1)x – 2 [∵ 5 = 6 – 1 and 2×3 = 6]
= 3x2 + 6x – x – 2
= 3x(x + 2) – 1(x + 2)
= (3x – 1) (x + 2)
On putting f(x) = 0 , we get
(3x – 1) (x + 2) = 0
⇒ 3x – 1 = 0 or x + 2 = 0
$\Rightarrow \mathrm{x}=\frac{1}{3}$ or x = – 2
Thus, the zeroes of the given polynomial 3x2 + 5x – 2 are – 2 and
Verification
Sum of zeroes
$=\alpha+\beta=\frac{1}{3}+(-2)$
$=\frac{1-6}{3}=-\frac{5}{3}$
or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{5}{3}$
$=\alpha \beta=\left(\frac{1}{3}\right)(-2)=\frac{-2}{3}$
Product of zeroes or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-2}{3}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 3x2 – x – 4
By splitting the middle term, we get
f(x) = 3x2 – (4 – 3)x – 4 [∵ – 1 = 3 – 4 and 4×3 = 12]
= 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (3x – 4) (x + 1)
On putting f(x) = 0, we get
(3x – 4) (x + 1) = 0
⇒ 3x – 4 = 0 or x + 1 = 0
$\Rightarrow x=\frac{4}{3}$ or x = – 1
Thus, the zeroes of the given polynomial 3x2 – x – 4 are – 1 and $\frac{4}{3}$
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-2}{3}$
So, the relationship between the zeroes and the coefficients is verified.
Question 1 E
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
3x2 – x – 4Let f(x) = 3x2 – x – 4
By splitting the middle term, we get
f(x) = 3x2 – (4 – 3)x – 4 [∵ – 1 = 3 – 4 and 4×3 = 12]
= 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (3x – 4) (x + 1)
On putting f(x) = 0, we get
(3x – 4) (x + 1) = 0
⇒ 3x – 4 = 0 or x + 1 = 0
$\Rightarrow x=\frac{4}{3}$ or x = – 1
Thus, the zeroes of the given polynomial 3x2 – x – 4 are – 1 and $\frac{4}{3}$
Verification
Sum of zeroes
$=\alpha+\beta=\frac{4}{3}+(-1)$
$=\frac{4-3}{3}=\frac{1}{3}$
or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-1)}{3}=\frac{1}{3}$
Product of zeroes
$=\alpha \beta=\left(\frac{4}{3}\right)(-1)=\frac{-4}{3}$
or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-4}{3}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = x2 + 7x + 10
By splitting the middle term, we get
f(x) = x2 + 5x + 2x + 10 [∵ 7 = 2 + 5 and 2×5 = 10]
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
On putting f(x) = 0 , we get
(x + 2) (x + 5) = 0
⇒ x + 2 = 0 or x + 5 = 0
⇒x = – 2 or x = – 5
Thus, the zeroes of the given polynomial x2 + 7x + 10 are – 2 and – 5
Verification
Sum of zeroes = α + β = – 2 + ( – 5) = – 7 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{7}{1}=-7$
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-4}{3}$
So, the relationship between the zeroes and the coefficients is verified.
Question 1 F
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
x2 + 7x + 10Sol :
Let f(x) = x2 + 7x + 10
By splitting the middle term, we get
f(x) = x2 + 5x + 2x + 10 [∵ 7 = 2 + 5 and 2×5 = 10]
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
On putting f(x) = 0 , we get
(x + 2) (x + 5) = 0
⇒ x + 2 = 0 or x + 5 = 0
⇒x = – 2 or x = – 5
Thus, the zeroes of the given polynomial x2 + 7x + 10 are – 2 and – 5
Verification
Sum of zeroes = α + β = – 2 + ( – 5) = – 7 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{7}{1}=-7$
Product of zeroes = αβ = ( – 2)( – 5) = 10 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{10}{1}=10$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = t2 – 15
Now, if we recall the identity
(a2 – b2) = (a – b)(a + b)
Using this identity, we can write
t2 – 15 = (t – √15) (x + √15)
So, the value of t2– 15 is zero when t = √15 or t = – √15
Therefore, the zeroes of t2– 15 are √15 and – √15.
Verification
Now,
Sum of zeroes = α + β = √15 + ( – √15) = 0 or
$=-\frac{\text { Coefficient of } \mathrm{t}}{\text { Coefficient of } \mathrm{t}^{2}}=-\frac{0}{1}=0$
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{10}{1}=10$
So, the relationship between the zeroes and the coefficients is verified.
Question 1 G
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
t2 – 15Let f(x) = t2 – 15
Now, if we recall the identity
(a2 – b2) = (a – b)(a + b)
Using this identity, we can write
t2 – 15 = (t – √15) (x + √15)
So, the value of t2– 15 is zero when t = √15 or t = – √15
Therefore, the zeroes of t2– 15 are √15 and – √15.
Verification
Now,
Sum of zeroes = α + β = √15 + ( – √15) = 0 or
$=-\frac{\text { Coefficient of } \mathrm{t}}{\text { Coefficient of } \mathrm{t}^{2}}=-\frac{0}{1}=0$
Product of zeroes = αβ = (√15)( – √15) = – 15 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{t}^{2}}=\frac{-15}{1}=-15$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 4 s2 – 4s + 1
By splitting the middle term, we get
f(x) = 4 s2 – (2 – 2)s + 1 [∵ – 4 = – (2 + 2) and 2×2 = 4]
= 4 s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
On putting f(x) = 0 , we get
(2s – 1) (2s – 1) = 0
⇒ 2s – 1 = 0 or 2s – 1 = 0
$\Rightarrow s=\frac{1}{2}$ or $s=\frac{1}{2}$
Thus, the zeroes of the given polynomial 4 s2 – 4s + 1 are $\frac{1}{2}$ and $\frac{1}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{1}{2}+\frac{1}{2}$=1
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{t}^{2}}=\frac{-15}{1}=-15$
So, the relationship between the zeroes and the coefficients is verified.
Question 1 H
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
4 s2 – 4s + 1Sol :
Let f(x) = 4 s2 – 4s + 1
By splitting the middle term, we get
f(x) = 4 s2 – (2 – 2)s + 1 [∵ – 4 = – (2 + 2) and 2×2 = 4]
= 4 s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
On putting f(x) = 0 , we get
(2s – 1) (2s – 1) = 0
⇒ 2s – 1 = 0 or 2s – 1 = 0
$\Rightarrow s=\frac{1}{2}$ or $s=\frac{1}{2}$
Thus, the zeroes of the given polynomial 4 s2 – 4s + 1 are $\frac{1}{2}$ and $\frac{1}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{1}{2}+\frac{1}{2}$=1
or
$=-\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}=-\frac{(-4)}{4}=1$
$=-\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}=-\frac{(-4)}{4}=1$
Product of zeroes $=\alpha \beta=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}=\frac{1}{4}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 8x2 – 22x – 21
By splitting the middle term, we get
f(x) = 8x2 – 28x + 6x – 21
= 4x(2x – 7) + 3(2x – 7)
= (4x + 3) (2x – 7)
On putting f(x) = 0 , we get
(4x + 3) (2x – 7) = 0
⇒ 4x + 3 = 0 or 2x – 7 = 0
Question 2 A
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
8x2– 22x – 21Sol :
Let f(x) = 8x2 – 22x – 21
By splitting the middle term, we get
f(x) = 8x2 – 28x + 6x – 21
= 4x(2x – 7) + 3(2x – 7)
= (4x + 3) (2x – 7)
On putting f(x) = 0 , we get
(4x + 3) (2x – 7) = 0
⇒ 4x + 3 = 0 or 2x – 7 = 0
⇒ x = $\frac{-3}{4}$ or x = $\frac{7}{2}$
Thus, the zeroes of the given polynomial 8x2 – 22x – 21 are $\frac{-3}{4}$ and $\frac{7}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-3}{4}+\frac{7}{2}=\frac{-3+14}{4}=\frac{11}{4}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-22)}{8}=\frac{11}{4}$
The product of zeroes $=\alpha \beta=\frac{-3}{4} \times \frac{7}{2}=\frac{-21}{8}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-21}{8}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 2 x2 – 7x
In this the constant term is zero.
f(x) = 2 x2 – 7x
= x(2x – 7)
On putting f(x) = 0 , we get
x(2x – 7) = 0
⇒ 2x – 7 = 0 or x = 0
⇒ X = $\frac{7}{2}$ or x = 0
Thus, the zeroes of the given polynomial 8x2 – 22x – 21 are $\frac{-3}{4}$ and $\frac{7}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-3}{4}+\frac{7}{2}=\frac{-3+14}{4}=\frac{11}{4}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-22)}{8}=\frac{11}{4}$
The product of zeroes $=\alpha \beta=\frac{-3}{4} \times \frac{7}{2}=\frac{-21}{8}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-21}{8}$
So, the relationship between the zeroes and the coefficients is verified.
Question 2 B
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
2x2 – 7xSol :
Let f(x) = 2 x2 – 7x
In this the constant term is zero.
f(x) = 2 x2 – 7x
= x(2x – 7)
On putting f(x) = 0 , we get
x(2x – 7) = 0
⇒ 2x – 7 = 0 or x = 0
⇒ X = $\frac{7}{2}$ or x = 0
Thus, the zeroes of the given polynomial 2x2 – 7x are 0 and $\frac{7}{2}$
Verification
Sum of zeroes $=\alpha+\beta=0+\frac{7}{2}=\frac{7}{2}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-7)}{2}=\frac{7}{2}$
Product of zeroes $=\alpha \beta=0 \times \frac{7}{2}=0$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{0}{2}=0$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = 10x2 + 3x – 1
By splitting the middle term, we get
f(x) = 10x2 – 2x + 5x – 1
= 2x(5x – 1) + 1(5x – 1)
= (2x + 1) (5x – 1)
On putting f(x) = 0, we get
(2x + 1) (5x – 1) = 0
⇒ 2x + 1 = 0 or 5x – 1 = 0
Verification
Sum of zeroes $=\alpha+\beta=0+\frac{7}{2}=\frac{7}{2}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-7)}{2}=\frac{7}{2}$
Product of zeroes $=\alpha \beta=0 \times \frac{7}{2}=0$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{0}{2}=0$
So, the relationship between the zeroes and the coefficients is verified.
Question 2 C
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
10x2 + 3x – 1Sol :
Let f(x) = 10x2 + 3x – 1
By splitting the middle term, we get
f(x) = 10x2 – 2x + 5x – 1
= 2x(5x – 1) + 1(5x – 1)
= (2x + 1) (5x – 1)
On putting f(x) = 0, we get
(2x + 1) (5x – 1) = 0
⇒ 2x + 1 = 0 or 5x – 1 = 0
⇒ X = $\frac{-1}{2}$ or X =$\frac{1}{5}$
Thus, the zeroes of the given polynomial
10x2 + 3x – 1 are $\frac{-1}{2}$ and $\frac{1}{5}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-1}{2}+\frac{1}{5}=\frac{-5+2}{10}=\frac{-3}{10}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{3}{10}=-\frac{3}{10}$
Product of zeroes $=\alpha \beta=\frac{-1}{2} \times \frac{1}{5}=\frac{-1}{10}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-1}{10}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = px2 + (2q – p2)x – 2pq
f(x) = px2 + 2qx – p2 x – 2pq
= x(px + 2q) – p(px + 2q)
= (x – p) (px + 2q)
On putting f(x) = 0, we get
(x – p) (px + 2q) = 0
⇒ x – p = 0 or px + 2q = 0
⇒x = p or X = $\frac{-2 q}{p}$
Thus, the zeroes of the given polynomial px2 + (2q – p2)x – 2pq are p and $\frac{-2 q}{p}$
Verification
Sum of zeroes $=\alpha+\beta=p+\frac{(-2 q)}{p}=\frac{p^{2}-2 q}{p}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{\left(-\mathrm{p}^{2}+2 \mathrm{q}\right)}{\mathrm{p}}=\frac{\mathrm{p}^{2}-2 \mathrm{q}}{\mathrm{p}}$
Product of zeroes $=\alpha \beta=p \times \frac{-2 q}{p}=-2 q$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-2 \mathrm{pq}}{\mathrm{p}}=-2 \mathrm{q}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = x2 – (2a + b)x + 2ab
f(x) = x2 – 2ax – bx + 2ab
= x(x – 2a) – b(x – 2a)
= (x – 2a) (x – b)
On putting f(x) = 0 , we get
(x – 2a) (x – b) = 0
⇒ x – 2a = 0 or x – b = 0
⇒x = 2a or x = b
Thus, the zeroes of the given polynomial x2 – (2a + b)x + 2ab are 2a and b
Verification
Sum of zeroes = α + β = 2a + b or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-2 \mathrm{a}-\mathrm{b})}{1}=2 \mathrm{a}+\mathrm{b}$
Product of zeroes = αβ = 2a × b = 2ab or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{2 \mathrm{ab}}{1}=2 \mathrm{ab}$
So, the relationship between the zeroes and the coefficients is verified.
Sol :
Let f(x) = r2s2x2 + 6rstx + 9t2
Now, if we recall the identity
(a + b)2 = a2 + b2 + 2ab
Using this identity, we can write
r2s2x2 + 6rstx + 9t2 = (rsx + 3t)2
On putting f(x) = 0 , we get
(rsx + 3t)2 = 0
Thus, the zeroes of the given polynomial
10x2 + 3x – 1 are $\frac{-1}{2}$ and $\frac{1}{5}$Verification
Sum of zeroes $=\alpha+\beta=\frac{-1}{2}+\frac{1}{5}=\frac{-5+2}{10}=\frac{-3}{10}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{3}{10}=-\frac{3}{10}$
Product of zeroes $=\alpha \beta=\frac{-1}{2} \times \frac{1}{5}=\frac{-1}{10}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-1}{10}$
So, the relationship between the zeroes and the coefficients is verified.
Question 2 D
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
px2 + (2q – p2)x – 2pq, p≠0Sol :
Let f(x) = px2 + (2q – p2)x – 2pq
f(x) = px2 + 2qx – p2 x – 2pq
= x(px + 2q) – p(px + 2q)
= (x – p) (px + 2q)
On putting f(x) = 0, we get
(x – p) (px + 2q) = 0
⇒ x – p = 0 or px + 2q = 0
⇒x = p or X = $\frac{-2 q}{p}$
Thus, the zeroes of the given polynomial px2 + (2q – p2)x – 2pq are p and $\frac{-2 q}{p}$
Verification
Sum of zeroes $=\alpha+\beta=p+\frac{(-2 q)}{p}=\frac{p^{2}-2 q}{p}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{\left(-\mathrm{p}^{2}+2 \mathrm{q}\right)}{\mathrm{p}}=\frac{\mathrm{p}^{2}-2 \mathrm{q}}{\mathrm{p}}$
Product of zeroes $=\alpha \beta=p \times \frac{-2 q}{p}=-2 q$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-2 \mathrm{pq}}{\mathrm{p}}=-2 \mathrm{q}$
So, the relationship between the zeroes and the coefficients is verified.
Question 2 E
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
x2 – (2a + b)x + 2abSol :
Let f(x) = x2 – (2a + b)x + 2ab
f(x) = x2 – 2ax – bx + 2ab
= x(x – 2a) – b(x – 2a)
= (x – 2a) (x – b)
On putting f(x) = 0 , we get
(x – 2a) (x – b) = 0
⇒ x – 2a = 0 or x – b = 0
⇒x = 2a or x = b
Thus, the zeroes of the given polynomial x2 – (2a + b)x + 2ab are 2a and b
Verification
Sum of zeroes = α + β = 2a + b or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-2 \mathrm{a}-\mathrm{b})}{1}=2 \mathrm{a}+\mathrm{b}$
Product of zeroes = αβ = 2a × b = 2ab or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{2 \mathrm{ab}}{1}=2 \mathrm{ab}$
So, the relationship between the zeroes and the coefficients is verified.
Question 2 F
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
r2s2x2 + 6rstx + 9t2Sol :
Let f(x) = r2s2x2 + 6rstx + 9t2
Now, if we recall the identity
(a + b)2 = a2 + b2 + 2ab
Using this identity, we can write
r2s2x2 + 6rstx + 9t2 = (rsx + 3t)2
On putting f(x) = 0 , we get
(rsx + 3t)2 = 0
$\Rightarrow X=\frac{-3 t}{r s}, \frac{-3 t}{r s}$
Thus, the zeroes of the given polynomial r2s2x2 + 6rstx + 9t2 are $\frac{-3 t}{r s}$ and$\frac{-3 t}{r s}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-3 t}{r s}+\frac{-3 t}{r s}=-\frac{6 t}{r s}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{6 \mathrm{rst}}{\mathrm{r}^{2} \mathrm{s}^{2}}=-\frac{6 \mathrm{t}}{\mathrm{rs}}$
Product of zeroes $=\alpha \beta=\frac{-3 \mathrm{t}}{\mathrm{rs}} \times \frac{-3 \mathrm{t}}{\mathrm{rs}}=\frac{9 \mathrm{t}^{2}}{\mathrm{r}^{2} \mathrm{s}^{2}}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{9 \mathrm{t}^{2}}{\mathrm{r}^{2} \mathrm{s}^{2}}$
So, the relationship between the zeroes and the coefficients is verified.
Question 3 A
Find the zeroes of the quadratic polynomial 5x2 – 8x – 4 and verify the relationship between the zeroes and the coefficients of the polynomial.
Sol :Let f(x) = 5x2 – 8x – 4
By splitting the middle term, we get
f(x) = 5x2 – 10x + 2x – 4
= 5x(x – 2) + 2(x – 2)
= (5x + 2) (x – 2)
On putting f(x) = 0 we get
(5x + 2) (x – 2) = 0
⇒ 5x + 2 = 0 or x – 2 = 0
⇒ X = $\frac{-2}{5}$ or x = 2
Thus, the zeroes of the given polynomial
5x2 – 8x – 4 are $\frac{-2}{5}$ and 2
Verification
Sum of zeroes $=\alpha+\beta=\frac{-2}{5}+2=\frac{-2+10}{5}=\frac{8}{5}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{5}=\frac{8}{5}$
Product of zeroes $=\alpha \beta=\frac{-2}{5} \times 2=\frac{-4}{5}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-4}{5}$
So, the relationship between the zeroes and the coefficients is verified.
Let f(x) = 4 x2 – 4x – 3
By splitting the middle term, we get
f(x) = 4 x2 – 6x + 2x – 3
= 2x(2x – 3) + 1(2x – 3)
= (2x + 1) (2x – 3)
On putting f(x) = 0, we get
(2x + 1) (2x – 3) = 0
⇒ 2x + 1 = 0 or 2x – 3 = 0
⇒ X = $\frac{-1}{2}$ or X = $\frac{3}{2}$
Thus, the zeroes of the given polynomial
4x2 – 4x – 3 are $\frac{-1}{2}$ and $\frac{3}{2}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{-1}{2}+\frac{3}{2}=\frac{-1+3}{2}=1$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-4)}{4}=1$
Product of zeroes $=\alpha \beta=\frac{-1}{2} \times \frac{3}{2}=\frac{-3}{4}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{4}$
So, the relationship between the zeroes and the coefficients is verified.
Let f(x) = √3 x2 – 8x + 4√3
By splitting the middle term, we get
(x) = √3 x2 – 6x – 2x + 4√3
= √3 x(x – 2√3) – 2(x – 2√3)
= (√3 x – 2) (x – 2√3)
On putting f (x) = 0, we get
(√3 x – 2) (x – 2√3) = 0
⇒ √3 x – 2 = 0 or x – 2√(3 = 0)
$\Rightarrow \mathrm{X}=\frac{2}{\sqrt{3}}$ or X = $2 \sqrt{3}$
Thus, the zeroes of the given polynomial
√3 x2 – 8x + 4√3 are $\frac{2}{\sqrt{3}}$ and $2 \sqrt{3}$
Verification
Sum of zeroes $=\alpha+\beta=\frac{2}{\sqrt{3}}+2 \sqrt{3}=\frac{2+6}{\sqrt{3}}=\frac{8}{\sqrt{3}}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{\sqrt{3}}=\frac{8}{\sqrt{3}}$
Product of zeroes $=\alpha \beta=\frac{2}{\sqrt{3}} \times 2 \sqrt{3}=4$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{4 \sqrt{3}}{\sqrt{3}}=4$
So, the relationship between the zeroes and the coefficients is verified.
(iii) α2β + αβ2 (iv)$\frac{1}{\alpha}+\frac{1}{\beta}$
(v)$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ (vi) α – β
(vii) α3 + β3 (viii)$\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}$
Sol :
Let the quadratic polynomial be 2 x2 + 3x – 6, and its zeroes are α and β.
We have
$\alpha+\beta=\frac{-b}{2}$ and $\alpha \beta=\frac{c}{a}$
Here, a = 2 , b = 3 and c = – 6
$\alpha+\beta=\frac{-b}{a}=\frac{-3}{2}$ ….(1)
$\alpha \beta=\frac{c}{a}=\frac{-6}{2}=-3$ ….(2)
(i) α2 + β2
We have to find the value of α2 + β2
Now, if we recall the identity
(a + b)2 = a2 + b2 + 2ab
Using the identity, we get
(α + β)2 = α2 + β2 + 2αβ
$\Rightarrow\left(\frac{-3}{2}\right)^{2}=\alpha^{2}+\beta^{2}+2(-3)$ {from eqn (1) &(2)} $\Rightarrow \frac{9}{4}=\alpha^{2}+\beta^{2}-6$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{9}{4}+6$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{9+24}{4}=\frac{33}{4}$
Thus, the zeroes of the given polynomial
5x2 – 8x – 4 are $\frac{-2}{5}$ and 2Verification
Sum of zeroes $=\alpha+\beta=\frac{-2}{5}+2=\frac{-2+10}{5}=\frac{8}{5}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{5}=\frac{8}{5}$
Product of zeroes $=\alpha \beta=\frac{-2}{5} \times 2=\frac{-4}{5}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-4}{5}$
So, the relationship between the zeroes and the coefficients is verified.
Question 3 B
Find the zeroes of the quadratic polynomial 4 x2 – 4x – 3 and verify the relationship between the zeroes and the coefficients of the polynomial.
Sol :Let f(x) = 4 x2 – 4x – 3
By splitting the middle term, we get
f(x) = 4 x2 – 6x + 2x – 3
= 2x(2x – 3) + 1(2x – 3)
= (2x + 1) (2x – 3)
On putting f(x) = 0, we get
(2x + 1) (2x – 3) = 0
⇒ 2x + 1 = 0 or 2x – 3 = 0
⇒ X = $\frac{-1}{2}$ or X = $\frac{3}{2}$
Thus, the zeroes of the given polynomial
4x2 – 4x – 3 are $\frac{-1}{2}$ and $\frac{3}{2}$Verification
Sum of zeroes $=\alpha+\beta=\frac{-1}{2}+\frac{3}{2}=\frac{-1+3}{2}=1$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-4)}{4}=1$
Product of zeroes $=\alpha \beta=\frac{-1}{2} \times \frac{3}{2}=\frac{-3}{4}$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{4}$
So, the relationship between the zeroes and the coefficients is verified.
Question 3 C
Find the zeroes of the quadratic polynomial √3 x2 – 8x + 4√3 .
Sol :Let f(x) = √3 x2 – 8x + 4√3
By splitting the middle term, we get
(x) = √3 x2 – 6x – 2x + 4√3= √3 x(x – 2√3) – 2(x – 2√3)
= (√3 x – 2) (x – 2√3)
On putting f (x) = 0, we get
(√3 x – 2) (x – 2√3) = 0
⇒ √3 x – 2 = 0 or x – 2√(3 = 0)
$\Rightarrow \mathrm{X}=\frac{2}{\sqrt{3}}$ or X = $2 \sqrt{3}$
Thus, the zeroes of the given polynomial
√3 x2 – 8x + 4√3 are $\frac{2}{\sqrt{3}}$ and $2 \sqrt{3}$Verification
Sum of zeroes $=\alpha+\beta=\frac{2}{\sqrt{3}}+2 \sqrt{3}=\frac{2+6}{\sqrt{3}}=\frac{8}{\sqrt{3}}$ or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(-8)}{\sqrt{3}}=\frac{8}{\sqrt{3}}$
Product of zeroes $=\alpha \beta=\frac{2}{\sqrt{3}} \times 2 \sqrt{3}=4$ or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{4 \sqrt{3}}{\sqrt{3}}=4$
So, the relationship between the zeroes and the coefficients is verified.
Question 4
If α and β be the zeroes of the polynomial 2x2 + 3x – 6, find the values of
(i) α2 + β2 (ii) α2 + β2 + αβ(iii) α2β + αβ2 (iv)$\frac{1}{\alpha}+\frac{1}{\beta}$
(v)$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ (vi) α – β
(vii) α3 + β3 (viii)$\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}$
Sol :
Let the quadratic polynomial be 2 x2 + 3x – 6, and its zeroes are α and β.
We have
$\alpha+\beta=\frac{-b}{2}$ and $\alpha \beta=\frac{c}{a}$
Here, a = 2 , b = 3 and c = – 6
$\alpha+\beta=\frac{-b}{a}=\frac{-3}{2}$ ….(1)
$\alpha \beta=\frac{c}{a}=\frac{-6}{2}=-3$ ….(2)
(i) α2 + β2
We have to find the value of α2 + β2
Now, if we recall the identity
(a + b)2 = a2 + b2 + 2ab
Using the identity, we get
(α + β)2 = α2 + β2 + 2αβ
$\Rightarrow\left(\frac{-3}{2}\right)^{2}=\alpha^{2}+\beta^{2}+2(-3)$ {from eqn (1) &(2)} $\Rightarrow \frac{9}{4}=\alpha^{2}+\beta^{2}-6$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{9}{4}+6$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{9+24}{4}=\frac{33}{4}$
(ii) α2 + β2 + αβ
$\alpha^{2}+\beta^{2}=\frac{33}{4}$ { from part (i)}
and, we have αβ = – 3
So, $\alpha^{2}+\beta^{2}+\alpha \beta=\frac{33}{4}+(-3)$
= $\frac{33-12}{4}$
= $=\frac{21}{4}$
(iii) α2β + α β2
Firstly, take $\alpha \beta$ common, we get
αβ(α + β)
and we already know the value of $\alpha \beta$ and $\alpha+\beta$.
So, α2β + α β2 = αβ(α + β)
$=(-3)\left(\frac{-3}{2}\right)$ {from eqn (1) and (2)}
$=\frac{9}{2}$
(iv) $\frac{1}{\alpha}+\frac{1}{\beta}$
Let’s take the LCM first then we get,
$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}$
$=\frac{\left(\frac{-3}{2}\right)}{-3}$
$=\frac{1}{2}$
$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}$
$=\frac{\left(\frac{-3}{2}\right)}{-3}$
$=\frac{1}{2}$
(v) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
Let’s take the LCM first then we get,
$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\beta \alpha}$
$=\frac{\left(\frac{33}{4}\right)}{-3}$ {from part(i) and eqn (2)}
$=\frac{-11}{4}$
Let’s take the LCM first then we get,
$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\beta \alpha}$
$=\frac{\left(\frac{33}{4}\right)}{-3}$ {from part(i) and eqn (2)}
$=\frac{-11}{4}$
(vi) $\alpha-\beta$
Now, recall the identity
(a – b)2 = a2 + b2 – 2ab
Using the identity , we get
(α – β)2 = α2 + β2 – 2 αβ
$\Rightarrow(\alpha-\beta)^{2}=\left(\frac{33}{4}\right)-2(-3)$ {from part(i) and eqn (2)}
$\Rightarrow(\alpha-\beta)^{2}=\frac{33}{4}+6$
$\Rightarrow(\alpha-\beta)^{2}=\frac{33+24}{4}=\frac{57}{4}$
$\Rightarrow(\alpha-\beta)=\pm \frac{\sqrt{57}}{2}$
(vii) $\alpha^{3}+\beta^{3}$
Now, recall the identity
(a + b)3 = a3 + b3 + 3a2 b + 3ab2
Using the identity, we get
⇒(α + β)3 = α3 + β3 + 3α2 β + 3αβ2
$\Rightarrow\left(\frac{-3}{2}\right)^{3}=\alpha^{3}+\beta^{3}+3\left(\alpha^{2} \beta+\alpha \beta^{2}\right)$
$\Rightarrow \frac{-27}{8}=\alpha^{3}+\beta^{3}+3 \times \frac{9}{2}$
$\Rightarrow \alpha^{3}+\beta^{3}=\frac{-27-108}{8}$
$\Rightarrow \alpha^{3}+\beta^{3}=-\frac{135}{4}$
(viii) $\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}$
Let’s take the LCM first then we get,
$\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\beta \alpha}$
$=\frac{\left(\frac{-135}{8}\right)}{-3}$ {from part(vii) and eqn (2)}
$=\frac{45}{8}$
Question 5
If α and β be the zeroes of the polynomial ax2 + bx + c, find the values of
(i) $\mathrm{a}^{2}+\beta^{2}$(ii) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
(ii) $\alpha^{3}+\beta^{3}$
Sol :
Let the quadratic poynomial be ax2 + bx + c , and its zeroes be α and β.
We have
α + β = $\frac{-b}{a}$ and α β = $\frac{c}{a}$
(i) $\mathrm{a}^{2}+\beta^{2}$
We have to find the value of $\alpha^{2}+\beta^{2}$
Now, if we recall the identity
(a + b)2 = a2 + b2 + 2ab
Using the identity, we get $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$
$\Rightarrow\left(\frac{-b}{a}\right)^{2}=\alpha^{2}+\beta^{2}+2 \times \frac{c}{a}$ {from eqn (1) & (2)}
$\Rightarrow \frac{b^{2}}{a^{2}}=\alpha^{2}+\beta^{2}+\frac{2 c}{a}$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a}$
$\Rightarrow \alpha^{2}+\beta^{2}=\frac{b^{2}-2 c a}{a^{2}}$
(ii) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$
Let’s take the LCM first then we get,
$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\beta \alpha}$
$=\frac{\left(\frac{b^{2}-2 c a}{a^{2}}\right)}{\frac{c}{a}}$ {$\because \alpha \beta=\frac{c}{a}$}
$=\frac{b^{2}-2 c a}{c a}$
(iii) $\alpha^{3}+\beta^{3}$
Now, recall the identity
(a + b)3 = a3 + b3 + 3a2 b + 3ab2
Using the identity, we get
⇒(α + β)3 = α3 + β3 + 3α2 β + 3αβ2
$\Rightarrow\left(\frac{-b}{a}\right)^{3}=\alpha^{3}+\beta^{3}+3\left(\alpha^{2} \beta+\alpha \beta^{2}\right)$
$\Rightarrow\left(\frac{-b^{3}}{a^{3}}\right)=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta)$
$\Rightarrow \alpha^{3}+\beta^{3}=\left(\frac{-b^{3}}{a^{3}}\right)+3 \times \frac{c}{a} \times\left(\frac{-b}{a}\right)$
$\Rightarrow \alpha^{3}+\beta^{3}=\frac{3 a b c-b^{3}}{a^{3}}$
Question 6
If α, β are the zeroes of the quadratic polynomial x2 + kx = 12, such that α – β = 1, find the value of k.
Sol :The given quadratic polynomial is x2 + kx = 12 and $\alpha{-} \beta$ = 1
If we rearrange the polynomial then we get
p(x) = x2 + kx – 12
We have,
$\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta=\frac{c}{a}$
So, $\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}=\frac{-\mathrm{k}}{1}=-\mathrm{k}$ …(1)
$\alpha \beta=\frac{c}{a}=\frac{-12}{1}=-12$ …(2)
Now, if we recall the identities
(a + b)2 = a2 + b2 + 2ab
Using the identity, we get
(α + β)2 = α2 + β2 + 2 αβ
( – k)2 = α2 + β2 + 2( – 12)
⇒ α2 + β2 = k2 + 24 …(3)
Again, using the identity
(a – b)2 = a2 + b2 – 2ab
Using the identity, we get
(α – β)2 = α2 + β2 – 2 αβ
(1)2 = α2 + β2 – 2( – 12) {∵ (α – β) = 1}
⇒ α2 + β2 = 1 – 24
⇒ α2 + β2 = – 23 …(4)
From eqn (3) and (4), we get
k2 + 24 = – 23
⇒ k2 = – 23 – 24
⇒ k2 = – 47
Now the square can never be negative, so the value of k is imaginary.
Question 7
If the sum of squares of the zeroes of the quadratic polynomial x2 – 8x + k is 40, find k.
Sol :Given : p(x) = x2 – 8x + k
α2 + β2 = 40
We have,
$\alpha+\beta=\frac{-b}{a}$
and $\alpha \beta=\frac{c}{a}$
So, $\alpha+\beta=\frac{-b}{a}=\frac{-(-8)}{1}=8$ …(1)
$\alpha \beta=\frac{c}{a}=\frac{k}{1}=k$ …(2)
Now, if we recall the identities
(a + b)2 = a2 + b2 + 2ab
Using the identity, we get
(α + β)2 = α2 + β2 + 2 αβ
(8)2 = 40+ 2(k)
⇒2k = 64 – 40
$\Rightarrow \mathrm{k}=\frac{24}{2}$= 12
Question 8 A
If one zero of the polynomial (α2 + 9) x2 + 13x + 6α is reciprocal of the other, find the value of a.
Sol :Let one zero of the given polynomial is $\alpha$
According to the given condition,
The other zero of the polynomial is $=\frac{1}{a}$
We have,
Product of zeroes, $\alpha \beta=\frac{c}{a}=\frac{6 \alpha}{\alpha^{2}+9}$
$\Rightarrow \alpha \times \frac{1}{\alpha}=\frac{6 \alpha}{\alpha^{2}+9}$
⇒ α2 – 6α + 9 = 0
⇒ α2 – 3α – 3α + 9 = 0
⇒ α(α – 3) – 3(α – 3) = 0
⇒(α – 3)(α – 3) = 0
⇒(α – 3) = 0 & (α – 3) = 0
⇒ α = 3, 3
Question 8 B
If the product of zeroes of the polynomial α2 – 6x – 6 is 4, find the value of a.
Sol :Given Product of zeroes, α β = 4
p(x) = α x2 – 6x – 6
to find: value of α
We know,
Product of zeroes , $\alpha \beta=\frac{c}{a}=\frac{-6}{\alpha}$
$\Rightarrow 4=\frac{-6}{\alpha}$
$\Rightarrow \alpha=\frac{-6}{4}=\frac{-3}{2}$
Question 8 C
If (x + a) is a factor 2x2 + 2ax + 5x + 10, find a.
Sol :Given x + a is a factor of $2 x^{2}+2 a x+5 x+10$
So,
g(x) = x + ax + a = 0
⇒x = – a
Putting the value x = – a in the given polynomial, we get
2( – a)2 + 2a( – a) + 5( – a) + 10 = 0
2a2 – 2a2 – 5a + 10 = 0
– 5a + 10 = 0
$a=\frac{-10}{-5}$
a = 2
Question 9 A
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
1,1Sol :
Given: Sum of zeroes = α + β = 1
Product of zeroes = αβ = 1
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (1)x + 1
= x2 – x + 1
Question 9 B
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
0,3Sol :
Given: Sum of zeroes = α + β = 0
Product of zeroes = αβ = 3
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (0)x + 3
= x2 + 3
Question 9 C
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
$\frac{1}{4},-1$Sol :
Given: Sum of zeroes = α + β = 1/4
Product of zeroes = αβ = – 1
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= $X^{2}-\left(\frac{1}{4}\right) x+(-1)$
= $x^{2}-\frac{x}{4}-1$
= $\frac{4 x^{2}-x-4}{4}$
We can consider 4x2 – x – 4 as required quadratic polynomial because it will also satisfy the given conditions.
Question 9 D
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
4,1Sol :
Given: Sum of zeroes = $\alpha+\beta$ = 4
Product of zeroes = $\alpha \beta$ = 1
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (4)x + 1
= x2 – 4x + 1
Question 9 E
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
$\frac{10}{3},-1$Sol :
Given: Sum of zeroes $=\alpha+\beta=\frac{10}{3}$
Product of zeroes = αβ = – 1
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= $x^{2}-\left(\frac{10}{3}\right) X+(-1)$
= $X^{2}-\frac{10 x}{3}-1$
= $\frac{3 x^{2}-10 x-3}{3}$
We can consider 3x2 – 10x – 3 as required quadratic polynomial because it will also satisfy the given conditions.
Question 9 F
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
$-\frac{1}{2},-\frac{1}{2}$Sol :
Given: Sum of zeroes $=\alpha+\beta=-\frac{1}{2}$
Product of zeroes $=\alpha \beta=-\frac{1}{2}$
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= $X^{2}-\left(-\frac{1}{2}\right) X+\left(-\frac{1}{2}\right)$
= $X^{2}+\frac{x}{2}-\frac{1}{2}$
= $\frac{2 x^{2}+x-1}{2}$
We can consider 2x2 + x – 1 as required quadratic polynomial because it will also satisfy the given conditions.
Question 10 A
Find quadratic polynomial whose zeroes are :
3, – 3Sol :
Let the zeroes of the quadratic polynomial be
α = 3 , β = – 3
Then, α + β = 3 + ( – 3) = 0
αβ = 3 × ( – 3) = – 9
Sum of zeroes = α + β = 0
Product of zeroes = αβ = – 9
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (0)x + ( – 9)
= x2 – 9
Question 10 B
Find quadratic polynomial whose zeroes are :
$\frac{2+\sqrt{5}}{2}, \frac{2-\sqrt{5}}{2}$Let the zeroes of the quadratic polynomial be
$\alpha=\frac{2+\sqrt{5}}{2}, \beta=\frac{2-\sqrt{5}}{2}$
Then, $\alpha+\beta=\frac{2+\sqrt{5}}{2}+\frac{2-\sqrt{5}}{2}=\frac{2+\sqrt{5}+2-\sqrt{5}}{2}=2$
$\alpha \beta=\frac{2+\sqrt{5}}{2} \times \frac{2-\sqrt{5}}{2}$
$=\frac{(2+\sqrt{5})(2-\sqrt{5})}{4}=\frac{4-5}{4}$
$=\frac{-1}{4}$
Sum of zeroes = α + β = 2
Product of zeroes = $\alpha \beta$ = $-\frac{1}{4}$
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= $X^{2}-(2) X+\left(-\frac{1}{4}\right)$
= $X^{2}-2 X-\frac{1}{4}$
= $\frac{4 x^{2}-8 x-1}{2}$
We can consider 4x2 – 8x – 1 as required quadratic polynomial because it will also satisfy the given conditions.
Question 10 C
Find quadratic polynomial whose zeroes are :
$3+\sqrt{7}, 3-\sqrt{7}$Sol :
Let the zeroes of the quadratic polynomial be
α = 3 + √7, β = 3 – √7
Then, α + β = 3 + √7 + 3 – √7 = 6
αβ = (3 + √(7)) × (3 – √7) = 9 – 7 = 2
Sum of zeroes = α + β = 6
Product of zeroes = αβ = 2
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (6)x + 2
= x2 – 6x + 2
Question 10 D
Find quadratic polynomial whose zeroes are :
$1+2 \sqrt{3}, 1-2 \sqrt{3}$Let the zeroes of the quadratic polynomial be
α = 1 + 2√3, β = 1 – 2√3
Then, α + β = 1 + 2√3 + 1 – 2√3 = 2
αβ = (1 + 2√3) × (1 – 2√3) = 1 – 12 = – 11
Sum of zeroes = α + β = 2
Product of zeroes = αβ = – 11
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (2)x + ( – 11)
= x2 – 2x – 11
Question 10 E
Find quadratic polynomial whose zeroes are :
$\frac{2-\sqrt{3}}{3}, \frac{2+\sqrt{3}}{3}$Let the zeroes of the quadratic polynomial be
$\alpha=\frac{2-\sqrt{3}}{3}, \beta=\frac{2+\sqrt{3}}{3}$
Then, $\alpha+\beta=\frac{2-\sqrt{3}}{3}+\frac{2+\sqrt{3}}{3}=\frac{2-\sqrt{3}+2+\sqrt{3}}{3}=\frac{4}{3}$
$\alpha \beta=\frac{2-\sqrt{3}}{3} \times \frac{2+\sqrt{3}}{3}=\frac{(2-\sqrt{3})(2+\sqrt{3})}{9}=\frac{4-3}{9}=\frac{1}{9}$
Sum of zeroes $=\alpha+\beta=\frac{4}{3}$
Product of zeroes $=\alpha \beta=\frac{1}{9}$
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= $X^{2}-\left(\frac{4}{3}\right) X+\left(\frac{1}{9}\right)$
= $X^{2}-\frac{4}{3} X+\frac{1}{9}$
= $\frac{9 x^{2}-12 x+1}{2}$
We can consider 9x2 – 12x + 1 as required quadratic polynomial because it will also satisfy the given conditions.
Question 10 F
Find quadratic polynomial whose zeroes are :
$\sqrt{2}, 2 \sqrt{2}$Sol :
Let the zeroes of the quadratic polynomial be
α = √2, β = 2√2
Then, α + β = √2 + 2√2 = √2 (1 + 2) = 3√2
αβ = √2× 2√2 = 4
Sum of zeroes = α + β = 3√2
Product of zeroes = αβ = 4
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (3√2)x + 4
= x2 – 3√2 x + 4
Question 11
Find the quadratic polynomial whose zeroes are square of the zeroes of the polynomial x2 – x – 1.
Sol :Let the zeroes of the polynomial x2 – x – 1 be α and β
We have,
$\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}$
and $\alpha \beta=\frac{c}{a}$
So,
$\alpha+\beta=\frac{-b}{a}=-\frac{-1}{1}=1$
$\alpha \beta=\frac{c}{a}=\frac{-1}{1}=-1$
Now, according to the given condition,
α2 β2 = ( – 1)2 = 1
& (α + β)2 = α2 + β2 + 2 α β
α2 + β2 = (α + β)2 – 2αβ
α2 + β2 = (1)2 – 2( – 1)
α2 + β2 = 3So, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (3)x + 1
= x2 – 3x + 1
Question 12 A
If α and β be the zeroes of the polynomial x2 + 10x + 30, then find the quadratic polynomial whose zeroes are α + 2β and 2α + β.
Sol :Given : p(x) = x2 + 10x + 30
So, Sum of zeroes $=\alpha+\beta=\frac{-b}{a}=\frac{-10}{1}=-10$ …(1)
Product of zeroes $=\alpha \beta=\frac{c}{a}=\frac{30}{1}=30$ …(2)
Now,
Let the zeroes of the quadratic polynomial be
α^' = α + 2β , β' = 2α + β
Then, α’ + β’ = α + 2β + 2α + β = 3α + 3β = 3(α + β)
α'β’ = (α + 2β) ×(2α + β) = 2α2 + 2β2 + 5αβ
Sum of zeroes = 3(α + β)
Product of zeroes = 2α2 + 2β2 + 5αβ
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (3(α + β))x + 2α2 + 2β2 + 5αβ
= x2 – 3( – 10)x + 2 (α2 + β2) + 5(30) {from eqn (1) & (2)}
= x2 + 30x + 2(α2 + β2 + 2αβ – 2αβ) + 150
= x2 + 30x + 2 (α + β)2 – 4αβ + 150
= x2 + 30x + 2( – 10)2 – 4(30) + 150
= x2 + 30x + 200 – 120 + 150
= x2 + 30x + 230
So, the required quadratic polynomial is x2 + 30x + 230
Question 12 B
If α and β be the zeroes of the polynomial x2 + 4x + 3, find the quadratic polynomial whose zeroes are $1+\frac{\alpha}{\beta}$ and $1+\frac{\beta}{\alpha}$.
Sol :Given : p(x) = x2 + 4x + 3
So, Sum of zeroes $=\alpha+\beta=\frac{-b}{a}=\frac{-4}{1}=-4$ …(1)
Product of zeroes $=\alpha \beta=\frac{c}{a}=\frac{3}{1}=3$ …(2)
Now,
Let the zeroes of the quadratic polynomial be
$\alpha^{\prime}=1+\frac{\alpha}{\beta}, \beta^{\prime}=1+\frac{\beta}{\alpha}$
Then,
$\alpha^{\prime}+\beta^{\prime}=1+\frac{\alpha}{\beta}+1+\frac{\beta}{\alpha}$
$=\frac{2 \alpha \beta+\alpha^{2}+\beta^{2}}{\alpha \beta}=\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
$\alpha^{\prime} \beta^{\prime}=\left(1+\frac{\alpha}{\beta}\right) \times\left(1+\frac{\beta}{\alpha}\right)$
$=1+\frac{\beta}{\alpha}+\frac{\alpha}{\beta}+1=\frac{2 \alpha \beta+\alpha^{2}+\beta^{2}}{\alpha \beta}$
$=\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
Sum of zeroes $=\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
Product of zeroes $=\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
$=\mathrm{x}^{2}-\frac{(\alpha+\beta)^{2}}{\alpha \beta} \mathrm{x}+\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
$=x^{2}-\frac{(4)^{2}}{3} x+\frac{(4)^{2}}{3}$ {from eqn (1) & (2)}
$=x^{2}-\frac{16}{3} x+\frac{16}{3}$
$=\frac{3 x^{2}-16 x+16}{3}$
So, the required quadratic polynomial is 3x2 – 16x + 16
Let the zeroes of the quadratic polynomial be
α = 1 , β = – 3
Then, α + β = 1 + ( – 3) = – 2
αβ = 1 × ( – 3) = – 3
Sum of zeroes = α + β = – 2
Product of zeroes = αβ = – 3
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – ( – 2)x + ( – 3)
= x2 + 2x – 3
Verification
Sum of zeroes = α + β = 1 + ( – 3) = – 2 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(2)}{1}=-2$
= x2 – (sum of zeroes)x + product of zeroes
$=\mathrm{x}^{2}-\frac{(\alpha+\beta)^{2}}{\alpha \beta} \mathrm{x}+\frac{(\alpha+\beta)^{2}}{\alpha \beta}$
$=x^{2}-\frac{(4)^{2}}{3} x+\frac{(4)^{2}}{3}$ {from eqn (1) & (2)}
$=x^{2}-\frac{16}{3} x+\frac{16}{3}$
$=\frac{3 x^{2}-16 x+16}{3}$
So, the required quadratic polynomial is 3x2 – 16x + 16
Question 13 A
Find a quadratic polynomial whose zeroes are 1 and – 3. Verify the relation between the coefficients and zeroes of the polynomial.
Sol :Let the zeroes of the quadratic polynomial be
α = 1 , β = – 3
Then, α + β = 1 + ( – 3) = – 2
αβ = 1 × ( – 3) = – 3
Sum of zeroes = α + β = – 2
Product of zeroes = αβ = – 3
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – ( – 2)x + ( – 3)
= x2 + 2x – 3
Verification
Sum of zeroes = α + β = 1 + ( – 3) = – 2 or
$=-\frac{\text { Coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{2}}=-\frac{(2)}{1}=-2$
Product of zeroes = αβ = (1)( – 3) = – 3 or
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{1}=-3$
$=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{x}^{2}}=\frac{-3}{1}=-3$
So, the relationship between the zeroes and the coefficients is verified.
Question 13 B
Find the quadratic polynomial sum of whose zeroes in 8 and their product is 12. Hence find the zeroes of the polynomial.
Sol :Given: Sum of zeroes = α + β = 8
Product of zeroes = αβ = 12
Then, the quadratic polynomial
= x2 – (sum of zeroes)x + product of zeroes
= x2 – (8)x + 12
= x2 – 8x + 12
Now, we have
$\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta=\frac{c}{a}$
So,
$\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}}=-\frac{-8}{1}=8$
$\alpha \beta=\frac{c}{a}=\frac{12}{1}=12$
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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