KC Sinha Mathematics Solution Class 10 Chapter 7 Quadratic Equations Exercise 7.1

Exercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5

Exercise 7.1

Question 1 A

Which of the following is a quadratic polynomial?

(i) 2 - $\frac{1}{3}$ x²

(ii) x+ $\frac{1}{\sqrt{\mathrm{x}}}$

(iii) x+ $\frac{1}{x}$

(iv) x² +3 $\sqrt{x}$ +2

Sol :
(i) On solving the equations,

$2-\frac{1}{3} x^{2}$
Re-writing in the format of ax² + bx + c = 0

$\left(-\frac{1}{3}\right) x^{2}+x(0)+2=0$
∵

$a=-\frac{1}{3}$ , b = 0 & c = 2
So, following the ideal pattern of a quadratic polynomial

$2-\frac{1}{3} x^{2}$ is a quadratic polynomial.

(ii) On solving the equations,
x+

$\frac{1}{\sqrt{\mathrm{x}}}$
∵ it can’t be re-written in the format of ax² + bx + c = 0
So, following the ideal pattern of a quadratic polynomial, x+

$\frac{1}{\sqrt{\mathrm{x}}}$ is not a quadratic polynomial.

(iii) On solving the equations,
x+

$\frac{1}{x}$
∵ it can’t be re-written in the format of ax² + bx + c = 0
So, following the ideal pattern of a quadratic polynomial, x+

$\frac{1}{x}$ is not a quadratic polynomial.

(iv) On solving the equations,
x² + 3√x + 2
∵ it can’t be re-written in the format of ax² + bx + c = 0
So, following the ideal pattern of a quadratic polynomial, x² + 3√x + 2 is not a quadratic polynomial.

Question 1 B

Which of the following is a quadratic polynomial?
(i) 2 x²+1

(ii) $x^{2}+\frac{1}{\sqrt{x}}$

(iii) $\sqrt{x^{2}+1}+\frac{1}{\sqrt{x}}$

(iv) $3 \sqrt{x^{2}+1}+x$

Sol :
(i) On solving the equations,
2x² + 1 = 0
Re-writing in the format of ax² + bx + c = 0

(2)x² + (0)x + 1 = 0
∵ a = 2 b = 0 & c = 1.
So, following the ideal pattern of a quadratic polynomial 2x² + 1 is a quadratic polynomial.

(ii) On solving the equations,

$\mathrm{x}^{2}+\frac{1}{\sqrt{\mathrm{x}}}$
∵ it can’t be re-written in the format of ax² + bx + c = 0
So, following the ideal pattern of a quadratic polynomial,

$x^{2}+\frac{1}{\sqrt{x}}$ is not a quadratic polynomial.

(iii) On solving the equations,

$\sqrt{x^{2}+1}+\frac{1}{\sqrt{x}}$
∵ it can’t be re-written in the format of ax² + bx + c = 0
So, following the ideal pattern of a quadratic polynomial,

$\sqrt{x^{2}+1}+\frac{1}{\sqrt{x}}$ is not a quadratic polynomial.

(iv) On solving the equations,
3√(x²+ 1) + x = 0
∵ it can’t be re-written in the format of ax² + bx + c = 0
So, following the ideal pattern of a quadratic polynomial, 3√(x²+ 1) + x = 0 is not a quadratic polynomial.

Question 2

Which of the following is a polynomial?

(i) $2 x+\frac{1}{3 x^{2}}$

(ii) $\frac{\sqrt{3}}{2}+x^{2}$

(iii) y² + y^-3

(iv) $3 \sqrt{x}+7$

Sol :
(i) On solving the equations,

$2 x+\frac{1}{3 x^{2}}=0$

$2 x+3 x^{-2}=0$
∵ After simplifying the equation, one of the term has a negative (-2) exponent.
So, following the ideal pattern of a polynomial,

$2 x+\frac{1}{3 x^{2}}=0$ is not a polynomial.

(ii) On solving the equations,

$\frac{\sqrt{3}}{2}+x^{2}=0$
∵ After simplifying the equation, as it has a positive (2) exponent.
So, following the ideal pattern of a polynomial,

$\frac{\sqrt{3}}{2}+x^{2}=0$ is a polynomial.

(iii) On solving the equations,

$\mathrm{y}^{2}+\frac{1}{\mathrm{y}^{3}}=0$

$\mathrm{y}^{2}+\mathrm{y}^{-3}=0$
∵ After simplifying the equation, as the one of the term has a negative (-3) exponent.
So, following the ideal pattern of a polynomial,

$y^{2}+\frac{1}{y^{3}}=0$ is not a polynomial.

(iv) On solving the equations,

$3 \sqrt{x}+7=0$

$3 x^{\frac{1}{2}}+7=0$
∵ After simplifying the equation, as the expression has a degree of

$\frac{1}{2}$ .
So, following the ideal pattern of a polynomial,

$3 x^{\frac{1}{2}}+7=0$ is not a polynomial.

Question 3

Fill in the blanks:

(i) x² + x + 3 is a ....... polynomial.
(ii) axⁿ + bx + c is a quadratic polynomial if n = ......
(iii) The value of the quadratic polynomial x² — 5x + 4 for x = — 1 is ......
(iv) The degree of the polynomial 2x² + 4x — x³is ..........
(v) A real number a will be called the zero of the quadratic polynomial ax² + bx + c if ........ is equal to zero.
Sol :
(i) Quadratic, because it is in the form of ax² + bx² + c = 0
(ii) n = 2, and also a≠0, as it will make the polynomial 0.
(iii) Putting the value of x = -1, in x² — 5x + 4
(-1)² - 5(-1) + 4
1 + 5 + 4
10

The value is 10.
(iv) ∵ The degree is the highest power of the term in the expression, so it is 3.
(v) ∵ The zeroes of a polynomial are α & β.
∴ to be zero, αx²+b α+c=0 & βx²+b β+c=0

Question 4 A

Find the zeroes of the quadratic polynomial 9 — x².

Sol :
-x² + 9 = 0
-x² = -9
x² = 9
∴ x = ±3
∴ The zeroes of the given polynomial are 3 & -3.

Question 4 B

Find the zeroes of the quadratic polynomial 4x²-1.

Sol :
4x² - 1 = 0
4x² = 1

$x^{2}=\frac{1}{4}$
∴

$x=\pm \frac{1}{2}$
∴ The zeroes of the given polynomial are

$\frac{1}{2}$ &

$-\frac{1}{2}$ .

Question 4 C

Which of the following are the zeroes of the quadratic polynomial 9 — 4 x² ?

(a) 4

(b) 9
(c) $\frac{3}{2}$

(d) $\frac{2}{3}$
Sol :
9 - 4x² = 0
4x² = 9

$x^{2}=\frac{9}{4}$
∴

$x=\pm \frac{3}{2}$
∴ The zeroes of the given polynomial are

$\frac{3}{2}$ &

$-\frac{3}{2}$ and the option (c) is correct.

Question 4 D

Find the zeroes of the polynomial $4-\frac{1}{2} x^{2}$

(a) 2 (b) $2 \sqrt{2}$ (c) 0 (d) 4
Sol :

$4-\frac{1}{2} x^{2}=0$

$\frac{1}{2} x^{2}=4$
x² =8
∴ x = ±2√2
∴ The zeroes of the given polynomial 2√2 and the option (b) is correct.

Question 5 A

Is — 2 a zero of the quadratic polynomial 3x² + x — 10?

Sol :
Putting the value of -2, in the given polynomial,
3(-2)² + (-2) – 10
3(4) – 2 – 10
12 – 12
0
∵ the value comes out to be 0.
∴ -2 is one of the zeroes and, yes 3x² + x — 10 is a quadratic polynomial.

Question 5 B

Is — 1 a zero of the quadratic polynomial x² + 2x — 3?

Sol :
Putting the value of -1, in the given polynomial,
(-1)² + 2(-1) – 3
1 – 2 – 3
3 – 3
0
∵ the value comes out to be 0.
∴ -1 is one of the zeroes of the given polynomial.

Question 6 A

Which of the following is a polynomial? Find its degree and the zeroes.
$2-\frac{1}{2} x^{2}$

Sol :
∵ The highest power is 2, so the degree is also 2.
Equating the expression with 0,

$2-\frac{1}{2} x^{2}=0$

$2=\frac{1}{2} x^{2}$
x² = 4
∴ x = ±2
Yes, the above expression is a polynomial, as it has no negative powers in any of the terms and its zeroes are 2 & -2.

Question 6 B

Which of the following is a polynomial? Find its degree and the zeroes.
$x+\frac{1}{\sqrt{x}}$

Sol :
∵ the power of a term is in negative

$\left(-\frac{1}{2}\right)$ .
∴ The above given expression is not a polynomial.

Question 7 A

Which of the following is a polynomial '? Find its zeroes.

(i) $x^{2}+\sqrt{x}+2$

(ii) $x+\frac{1}{x}$

(iii)

$4-\frac{1}{4} x^{2}$
Sol :
In the above expressions, only the thirdone has the positive power unlike others.
∴ It is the only polynomial.
Equating the expression with 0,

$4-\frac{1}{4} x^{2}=0$

$4=\frac{1}{4} x^{2}$
x² = 16
∴ x = ±4
The zeroes of the polynomial

$4-\frac{1}{4} x^{2}$ are 4 & -4.

Question 7 B

Which of the following expressions is a polynomial? Find the degree and zeroes of the polynomial.

(i) $\frac{x}{2}+\frac{2}{x}$

(ii) x² +2x

Sol :
In the above expressions, only the secondone has a positive power, unlike others.
∴ It is the only polynomial.
Equating the expression with 0,
x² + 2x = 0
x ( x+2) = 0
∴ x = 0 or x + 2 = 0
X = 0 Or x = -2
The zeroes of the polynomial x² +2x are 2 & -2, having a degree of 2, being the highest power of the terms in the same expression.

Question 7 C

Which among the expressions 1- $\frac{1}{16}$ z² and z² + z + 1 is a polynomial in z? Find its zeroes and degree.

Sol :
∵ The highest power is 2, so the degree is also 2, in both the expressions.
Equating the expression with 0,

$1-\frac{1}{16} z^{2}=0$

$1=\frac{1}{16} z^{2}$
x² =16
∴ x = ±4
Yes, the above expression (

$1-\frac{1}{16} z^{2}$ ⁾ is a polynomial, and its zeroes are 4 & -4.
Equating the expression with 0,
z² + z + 1 =0
Using Sreedharacharya formula,

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
ax²+bx+c = 0

$\mathrm{x}=\frac{(-(1)) \pm \sqrt{(1)^{2}-4(1)(1)}}{2(1)}$

$x=\frac{-1 \pm \sqrt{1-4}}{2}$

$x=\frac{-1 \pm \sqrt{-3}}{2}$
∵ it does not have real values.
∴ The zeroes of z² + z + 1 are complex numbers, though it is a polynomial having the degree 2.

Question 8

Find the zeroes of the quadratic polynomial x² — 6x + 8.

Sol :
Equating the expression with 0,
x² — 6x + 8 = 0

On factorising it further,
x² - 4x - 2x + 8 = 0
x(x - 4) – 2(x - 4) = 0
(x - 4) (x - 2) = 0

∴ x = 4 or x = 2

∴ The zeroes of x² — 6x + 8 are 4 & 2.

Question 9 A

Find the zeroes of the quadratic polynomial:
2x² + x — 1

Sol :
Equating the expression with 0,
2x² +x -1 = 0

On factorising it further,
2x² -x + 2x - 1 = 0
x(2x - 1) +1(2x - 1) = 0
(2x - 1) (x + 1) = 0
∴ x =

$\frac{1}{2}$ or x = -1
∴ The zeroes of 2x² + x — 1 are

$\frac{1}{2}$ and -1.

Question 9 B

Find the zeroes of the quadratic polynomial:

2x²— 5x + 2
Sol :
Equating the expression with 0,
2x² - 5x + 2 = 0

On factorising it further,
2x² - 4x- x + 2 = 0
2x(x - 2) -1(x - 2) = 0
(2x - 1) (x - 2) = 0
∴ x =

$\frac{1}{2}$ or x = 2
∴ The zeroes of 2x²— 5x + 2 are

and 2.

Question 9 C

Find the zeroes of the quadratic polynomial:

5x² - 4x — 1
Sol :
Equating the expression with 0,
5x² - 4x - 1 = 0

On factorising it further,
5x² - 5x + x - 1 = 0
5x(x - 1) +1(x - 1)=0
(5x + 1) (x - 1) = 0
∴ x = -

$\frac{1}{5}$ or x = 1
∴ The zeroes of 5x² - 4x — 1 are -

and 1.

Question 9 D

Find the zeroes of the quadratic polynomial:

x² — 2x + 3
Sol :
Equating the expression with 0,
x² — 2x + 3= 0

Using Sreedharacharya formula,

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
ax²+bx+c = 0

$x=\frac{(-(-2)) \pm \sqrt{(-2)^{2}-4(1)(3)}}{2(1)}$
x

$=\frac{2 \pm \sqrt{4-12}}{2}$

$x=\frac{2 \pm \sqrt{-8}}{2}$
∵ it does not have real values.
∴ The zeroes of x² — 2x + 3 are complex numbers.

Question 9 E

Find the zeroes of the quadratic polynomial:

3x² — 10x + 3
Sol :
Equating the expression with 0,
3x² - 10x + 3 = 0
On factorising it further,
3x² - 9x - x + 3 = 0
3x(x - 3) - 1(x - 3) = 0
(3x - 1) (x - 3) = 0
∴ x =

$\frac{1}{3}$ or x = 3
∴ The zeroes of 3x² — 10x + 3 are

$\frac{1}{3}$ and 3.

Question 9 F

Find the zeroes of the quadratic polynomial:
3x² + 5x + 2

Sol :
Equating the expression with 0,
3x² + 5x + 2 = 0
On factorising it further,
3x² + 3x + 2x + 2 = 0
3x(x + 1) + 2(x + 1) = 0
(3x + 2) (x + 1) = 0
∴ x = -

$\frac{2}{3}$ or x = -1
∴ The zeroes of 3x² + 5x + 2 are –

$\frac{2}{3}$ and -1.

Question 9 G

Find the zeroes of the quadratic polynomial:

4x² — x — 5
Sol :
Equating the expression with 0,
4x² — x — 5 = 0
On factorising it further,
4x² + 4x - 5x - 5 = 0
4x(x + 1) -5(x + 1)=0
(4x - 5) (x + 1) = 0
∴ x =

$\frac{5}{4}$ or x = -1
∴ The zeroes of 4x² — x — 5 are

$\frac{5}{4}$ and -1.

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