# If sum of the squares of zeroes of the quadratic polynomial $6x^2+x+k$ is $\frac{25}{36}$, then find the value of $k$.

Given: Sum of the squares of zeroes of the quadratic polynomial $6x^2+x+k$ is $\frac{25}{36}$.

To do: To find the value of $k$.

Solution:

On comparing given expression with $ax^2+bx+c$

$a=6,\ b=1,\ c=k$

let zeroes of polynomials are $\alpha$ and $\beta$

Sum of the zeroes $=\alpha+\beta=-\frac{b}{a}=-\frac{1}{6}\ ----( i)$

Product of the roots $=\alpha\beta=\frac{c}{a}=\frac{k}{6}\ ----( ii)$

As given sum of the squares of zeroes $=\frac{25}{36}$

$\alpha^2+\beta^2=\frac{25}{36}\ ---( iii)$

On squaring both sides of $( i)$

$(\alpha+\beta)^2=(-\frac{1}{6})^2$

$\Rightarrow \alpha^2+\beta^2+2\alpha\beta=\frac{1}{36}$

$\Rightarrow \frac{25}{36}+2\alpha\beta=\frac{1}{36}$                                      [from $( iii)$]

$\Rightarrow 2\alpha\beta=\frac{1}{36}-\frac{25}{36}$

$\Rightarrow 2\alpha\beta=\frac{( 1-25)}{36}$

$\Rightarrow 2\alpha\beta=-\frac{24}{36}$

$\Rightarrow \alpha\beta= -\frac{24}{2\times 36}$

$\Rightarrow \alpha\beta=-\frac{1}{3}\ ---( iv)$

but $( ii) =( iv)$

$\Rightarrow \frac{k}{6}=-\frac{1}{3}$

$\Rightarrow k=-\frac{1}{3}\times 6$

$\Rightarrow k=-2$

Thus, the value of $k$ is $-2$.

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Updated on: 10-Oct-2022

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