If $\alpha$ and $\beta$ are zeroes of $p( x)=kx^2+4x+4$, such that $\alpha^2+\beta^2=24$, find $k$.


Given: $\alpha$ and $\beta$ are zeroes of $p( x)=kx^2+4x+4$, such that $\alpha^2+\beta^2=24$.

To do: To find $k$.


Solution:


As given  $\alpha$ and $\beta$ are zeroes of $p( x)=kx^2+4x+4$

$\Rightarrow \alpha+\beta=-\frac{4}{k}$

And $\alpha\beta=\frac{4}{k}$

Given, $\alpha^2+\beta^2=24$

$\Rightarrow ( \alpha+\beta)^2-2\alpha\beta=24$

$\Rightarrow (-\frac{4}{k})^2-2\times\frac{4}{k}$=24$

$\Rightarrow \frac{16}{k^2}-\frac{8}{k}=24$

$\Rightarrow \frac{16-8k}{k^2}=24$

$\Rightarrow 16-8k=24k^2$

$\Rightarrow 24k^2+8k-16=0$

$\Rightarrow 24k^2+24k-16k-16=0$

$\Rightarrow 24k( k+1)-16( k+1)=0$

$\Rightarrow ( 24k-16)( k+1)=0$

If  $24k-16=0\ \Rightarrow k=-\frac{16}{24}$

If $k+1=0\ \Rightarrow k=-1$

Thus, $k=-\frac{16}{24},\ -1$.

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Updated on: 10-Oct-2022

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