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Question:
If $x=1$ and $y=6$ is a solution of the equation $8 x-a y+a^{2}=0$, find the values of $a$.
Solution:
We are given, $8 x$-ay $+a^{2}=0(1,6)$ is a solution of equation $8 x-a y+a^{2}=0$
Substituting $x=1$ and $y=6$ in $8 x-a y+a^{2}=0$, we get $8 \times 1-a \times 6+a^{2}=0$
$\Rightarrow a^{2}-6 a+8=0$
Using quadratic factorization $a^{2}-4 a-2 a+8=0 a(a-4)-2(a-4)=0(a-2)(a-4)=0$
a = 2, 4
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