Question

If $x=1$and $y=6$ is a solution of $8x-ky+k^2=0$, find the values of $k$.

Answer 1

Finding the value of $k$:

Substituting the value of the given solution in the given equation we will get the value of $k$.

Since $x=1$and $y=6$ is a solution of $8x-ky+k^2=0$,

By putting the values of $x$ and $y$ in the given equation, we get;

∴ $8\times 1-k\times 6+k^2=0$

$\Rightarrow k^2-6k+8=0$

$\Rightarrow k^2-4k-2k+8=0$

$\Rightarrow k(k-4)-2(k-4)=0$

$\Rightarrow (k-4)(k-2)=0$

$\Rightarrow k-4=0\&$ $k-2=0$

$\Rightarrow k=4,2$
Hence, the value of $k$ is $4\&2$.