Review question

Ref: R7972

## Solution

If $x^2 = ax + b$ has solutions $x = -2$ and $x = 4$, find the values of $a$ and $b$.

By substituting in the values $x = -2$ and $x = 4$, we know that \begin{align*} 4 &= -2a + b, \\ 16 &= 4a + b. \end{align*}

We can solve these equations simultaneously: subtracting the first from the second to eliminate $b$ gives $12=6a,$ so $a=2$. Substituting this into the first equation then gives $b=4+2a=8$.

Finally, we can check our working using the second equation: $4a+b=4\times2+8=16$, as we required.

Alternatively, we can say $x^2-ax-b=(x+2)(x-4)$ for all $x$.

Multiplying out and comparing coefficients gives $a = 2, b = 8$, as before.