Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:

$g(s)\ =\ 4s^2\ –\ 4s\ +\ 1$

 Given:


$g(s) = 4s^2 – 4s+1$

To find:

Here, we have to find the zeros of g(s). 

Solution:

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.

Updated on: 10-Oct-2022

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