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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.
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