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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$h(s)\ =\ 2s^2\ –\ (1\ +\ 2\sqrt{2})s\ +\ \sqrt{2}$
Given:
$h(s) = 2s^2– (1+2\sqrt{2})s+\sqrt{2}$
To find:
Here, we have to find the zeros of h(s) .
Solution:
To find the zeros of h(s), we have to put $h(s)=0$.
This implies,
$h(s) = 2s^2– (1+2\sqrt{2})s+\sqrt{2}= 0$
$2s^2-s-2\sqrt{2}s+\sqrt{2}= 0$
$s(2s-1)-\sqrt{2}(2s-1)= 0$
$(2s-1) (s-\sqrt{2})= 0$
$2s-1=0$ and $s-\sqrt{2}=0$
$2s = 1$ and $s= \sqrt{2}$
$s=\frac{1}{2}$ and $s=\sqrt{2}$
Therefore, the zeros of the quadratic equation $h(s) = 2s^2– (1+2\sqrt{2})s+\sqrt{2}$ are $\frac{1}{2}$ and $\sqrt{2}$.
Verification:
We know that,
Sum of zeros $= -\frac{coefficient of s}{coefficient of s^2}$
$= –(\frac{-(1+2\sqrt{2})}{2})$
$=\frac{1}{2}+\sqrt{2}$
Sum of the zeros of $h(s)=\frac{1}{2}+\sqrt{2}$
Product of roots $= \frac{constant}{coefficient of x^2}$
$=\frac{\sqrt{2}}{2}$
Product of the roots of $h(s)=\frac{1}{2}\times \sqrt{2)}=\frac{\sqrt{2}}{2}$
Hence, the relationship between the zeros and their coefficients is verified.
Updated on: 10-Oct-2022
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