Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 - 2x - 8$
(ii) $4s^2 - 4s + 1$
(iii) $6x^2 - 3 - 7x$
(iv) $4u^2 + 8u$
(v) $t^2 -15$
(vi) $3x^2 - x - 4$.

AcademicMathematicsNCERTClass 10

To find:

Here, we have to find the zeros of the given quadratic polynomials and verify the relationship between the zeroes and the coefficients. 

Solution:

(i) Let $f(x)=x^2 - 2x - 8$

To find the zeros of f(x), we have to put $f(x)=0$.

This implies,

$x^2 - 2x - 8 = 0$

$x^2 - 4x + 2x - 8 = 0$

$x(x - 4) + 2(x - 4) = 0$

$(x - 4)(x + 2) = 0$

$x-4=0$ and $x+2=0$

$x = 4$ and $x = -2$

Therefore, the zeros of the quadratic equation $f(x) = x^2 - 2x - 8$ are $4$ and $-2$.

Verification:

We know that, 

Sum of zeros $= -\frac{coefficient\ of\ x}{coefficient\ of\ x^2}$

$= -\frac{(-2)}{1}$

$=2$

Sum of the zeros of $f(x)=4+(-2)=4-2=2$ 

Product of roots $= \frac{constant}{coefficient\ of\ x^2}$

$= \frac{(-8)}{1}$

$= -8$

Product of the roots of $f(x)=4\times(-2) =-8$

Hence, the relationship between the zeros and their coefficients is verified.

(ii) Let $g(s)=4s^2 - 4s + 1$

To find the zeros of g(s), we have to put $g(s)=0$.

This implies,

$4s^2 – 4s +1 = 0$

$4s^2 – 2s -2s +1 = 0$

$2s(s – 1) -1(2s – 1) = 0$

$(2s – 1)(2s- 1) = 0$

$2s-1=0$ and $2s-1=0$

$2s= 1$ and $2s= 1$

$s=\frac{1}{2}$ and $s=\frac{1}{2}$

Therefore, the zeros of the quadratic equation $g(s) = 4s^2 – 4s +1$ are $\frac{1}{2}$ and $\frac{1}{2}$.

Verification:

We know that, 

Sum of zeros $= -\frac{coefficient\ of\ s}{coefficient\ of\ s^2}$

$= –\frac{(-4)}{4}$

$=1$

Sum of the zeros of $g(s)=\frac{1}{2}+\frac{1}{2}=1$

Product of roots $= \frac{constant}{coefficient\ of\ s^2}$

$= \frac{1}{4}$

Product of the roots of $g(s)=\frac{1}{2}\times\frac{1}{2} =\frac{1}{4}$

Hence, the relationship between the zeros and their coefficients is verified.

(iii) Let $f(x) =6x^2-3-7x$.

To find the zeros of $f(x)$, we have to put $f(x)=0$

This implies,

$6x^2-9x+2x-3=0$

$3x( 2x-3)+1( 2x-3)=0$

$( 2x-3)( 3x+1)=0$

$( 2x-3)=0$ and $( 3x+1)=0$

$x=\frac{3}{2}$  and $ x=-\frac{1}{3}$

Therefore, the zeros of the quadratic equation $6x^2-3-7x$ are $\frac{3}{2}$ and $\frac{1}{3}$

Verification:

We know that,

Sum of zeros $= -\frac{coefficient\ of\ x}{coefficient\ of\ x^2}$

$=-(\frac{-7}{6})$

$=\frac{7}{6}$

Sum of zeros of $f(x)=\frac{3}{2}+( -\frac{1}{3})$

$=\frac{9-2}{6}$

$=\frac{7}{6}$ 

Product of roots$=\frac{constant}{coefficient\ of\ x^2}$

$=\frac{-3}{6}$

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

product of the roots of $f(x) =\frac{3}{2}\times( -\frac{1}{3})$

$=-\frac{3}{6}$

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

Hence, the relationship between the zeros and their coefficients is verified.

(iv) Let $f(u)=4u^2 + 8u$

To find the zeros of f(u), we have to put $f(u)=0$.

This implies,

$4u^2 + 8u=0$

$4u(u+2) = 0$

$4u=0$ and $u+2=0$

$u = 0$ and $u = -2$

Therefore, the zeros of the quadratic equation $f(u) = 4u^2 + 8u=0$ are $0$ and $-2$.

Verification:

We know that, 

Sum of zeros $= -\frac{coefficient\ of\ x}{coefficient\ of\ x^2}$

$= -\frac{(8)}{4}$

$=-2$

Sum of  zeros of $f(u)=0+(-2)=0-2=-2$ 

Product of roots $= \frac{constant}{coefficient\ of\ x^2}$

$= \frac{0}{4}$

$= 0$

Product of the roots of $f(u)=0\times(-2) =0$

Hence, the relationship between the zeros and their coefficients is verified.

(v) Let $f(t)=t^2 - 15$

To find the zeros of f(t), we have to put $f(t)=0$.

This implies,

$t^2 - 15=0$

$t^2-(\sqrt{15})^2=0$

$(t-\sqrt{15})(t+\sqrt{15}) = 0$

$t-\sqrt{15}=0$ and $t+\sqrt{15}=0$

$t=\sqrt{15}$ and $t=-\sqrt{15}$

Therefore, the zeros of the given quadratic equation are $\sqrt{15}$ and $-\sqrt{15}$.

Verification:

We know that, 

Sum of zeros $= -\frac{coefficient\ of\ x}{coefficient\ of\ x^2}$

$= -\frac{0}{1}$

$=0$

Sum of zeros of $f(t)=\sqrt{15}-\sqrt{15}=0$ 

Product of roots $= \frac{constant}{coefficient\ of\ x^2}$

$= \frac{-15}{1}$

$=-15$

Product of the roots of $f(t)=(\sqrt{15})\times(-\sqrt{15}) =-15$

Hence, the relationship between the zeros and their coefficients is verified.

(vi) Let $f(x)=3x^2 - x - 4$

To find the zeros of f(x), we have to put $f(x)=0$.

This implies,

$3x^2 - x - 4 = 0$

$3x^2 - 4x + 3x - 4 = 0$

$3x(x +1 ) -4 (x + 1) = 0$

$(3x - 4)(x + 1) = 0$

$3x-4=0$ and $x+1=0$

$x = \frac{4}{3}$ and $x = -1$

Therefore, the zeros of the quadratic equation $f(x) = 3x^2 - x - 4$ are $\frac{4}{3}$ and $-1$.

Verification:

We know that, 

Sum of zeros $= -\frac{coefficient\ of\ x}{coefficient\ of\ x^2}$

$= -\frac{(-1)}{3}$

$=\frac{1}{3}$

Sum of zeros of $f(x)=\frac{4}{3}+(-1)=\frac{4-3}{3}=\frac{1}{3}$ 

Product of roots $= \frac{constant}{coefficient\ of\ x^2}$

$= \frac{(-4)}{3}$

$= -\frac{4}{3}$

Product of the roots of $f(x)=\frac{4}{3}\times(-1) =-\frac{4}{3}$

Hence, the relationship between the zeros and their coefficients is verified.

raja
Updated on 10-Oct-2022 13:19:34

Advertisements