If sum of the square of zeroes of the polynomial $f(x)=x^2−8x+k$ is $40$, find the value of $k$.

Given: Sum of the square of zeroes of the polynomial $f(x)=x^2−8x+k$ is $40$

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

Solution: 

We are given, sum of the square of zeros of polynomial $f(x)=x^2−8x+k$ is $40$.

Let take $\alpha$ & $\beta$ are the roots of $f(x)=x^2−8x+k$

here $\alpha^2+\beta^2=40$

$(\alpha+\beta)^2−2\alpha\beta=40$

Now, sum of zeros $\alpha+\beta=a−b​$

$\alpha+\beta=−(−8)=8$

product of zeros $\alpha\beta=ac​$

$\alpha\beta=1k​=k$

so, $(\alpha+\beta)^2−2\alpha\beta=40$

$\Rightarrow (8)^2−2(k)=40$

$\Rightarrow 64−40=2k$

$\Rightarrow 2k=24$

$\Rightarrow k=12$

 

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Updated on: 10-Oct-2022

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