If the sum of the zeroes of the polynomial $P(x)=( k^{2}-14)x^{2}-2x-12$ is $1$. Then find the value of $k$.
Given: The sum of the zeroes of the polynomial $P(x)=( k^{2}-14)x^{2}-2x-12$ is $1$.
To do: To find the value of $k$.
Solution:
The given polynomial is $P(x)=( k^{2}-14)x^{2}-2x-12$
The sum of zeros of a quadratic equation is $-\frac{b}{a}$.
Hence, the sum of the zeros of the given quadratic equation$=-\frac{-2}{k^2-14}$
$=\frac{2}{k^2-14}$
As given, the sum of the zeroes is $1$.
$\Rightarrow \frac{2}{k^2-14}=1$
$\Rightarrow k^2-14=2$
$\Rightarrow k^2=14+2=16$
$\Rightarrow k=\pm\sqrt{16}$
$\Rightarrow k=\pm4$
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