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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