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.