Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$p(y)\ =\ y^{2} \ +\ \left(\frac{3\sqrt{5}}{2}\right) y\ –\ 5$

Given:


$p(y)=y^2+(\frac{3\sqrt5}{2})y-5$

To find:

Here, we have to find the zeros of p(y) . 

Solution:

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

This implies,

$y^2+(\frac{3\sqrt5}{2})y-5= 0$

$y^2-(\frac{\sqrt{5}}{2})y+2\sqrt5y-5= 0$

$y(y-\frac{\sqrt{5}}{2})+2\sqrt{5}(y-\frac{\sqrt{5}}{2})= 0$

$(y- \frac{\sqrt{5}}{2})(y+2\sqrt5)= 0$

$y-\frac{\sqrt{5}}{2}=0$ and $y+2\sqrt5=0$

$y=\frac{\sqrt{5}}{2}$ and $y=-2\sqrt{5}$

Therefore, the zeros of the quadratic equation $p(y)=y^2+(\frac{3\sqrt5}{2})y-5$ are $\frac{\sqrt{5}}{2}$ and $-2\sqrt5$.

Verification:

We know that, 
Sum of zeros $= -\frac{coefficient of y}{coefficient of y^2}$

                       $= –(\frac{\frac{3\sqrt5}{2}}{1})$

                       $=-\frac{3\sqrt{5}}{2}$

Sum of the zeros of $p(y) =\frac{\sqrt{5}}{2}+(-2\sqrt5) =\frac{-\sqrt{5}+(-4\sqrt{5}}{2})=-\frac{3\sqrt{5}}{2}$

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

                            $=\frac{-5}{1}$

                            $= -5$

Product of the roots of $p(y)==\frac{\sqrt{5}}{2}\times(-2\sqrt5) =-5$

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

Updated on: 10-Oct-2022

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