Prove that if $a = 8 - b$ and $c^2 = ab - 16$ , then $a = b$ .

Question

Prove that if $a = 8 - b$ and $c^2 = ab - 16$ , then $a = b$ .

My attempt :- If $a = b$ then definitely this works as we get $a = b = 4$ . Tried to think of this by means of contradiction but couldn't any useful info . After working a lot of ways in the end I got $c^2 + ab = a(b + 2) + b(a - 2)$ , but can't see how it will help.

Can anyone give me a hint to this problem?

Answer 1

If you substitute $a$ into $c^2$, you will get:

$c^2= (8-b)b-16$ $\qquad \Rightarrow$ $c^2 = 8b - b^2 -16 \qquad \Rightarrow c^2 = -(b-4)^2$

Since $c^2$ must be non-negative, then $-(b-4)^2$ must be equal 0. So, $b-4=0 \Rightarrow b=4.$

Done?

Answer 2

Just substitute the value of $a$ in the second equation and use the fact that $c^2 \geq 0$.