If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?

If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?

I tried to solve it as follows:

I state that $p=ab$

$p^2=(2b+15)(2a+15)$

$p^2=4ab+30(a+b)+225$

$p^2=4p+30(a+b)+225$

and this is where I got stuck. I don't know how to get over this hurdle. could you please explain to me how to solve the question?

$a^2=2b+15$ and $b^2=2a+15$

Subtracting, $a^2-b^2 = -2(a-b)$. As $a \ne b$,

$a+b = - 2$

Also adding both equations, $a^2+b^2 = 2(a+b)+30 = 26$

$(a+b)^2 = a^2+b^2+2ab \implies 4 = 26 + 2ab$

$ab = -11$

Alternatively,

If $a≠b$, then

• $a^2-b^2=2(b-a)$

$$ a+b =-2$$

• $(-2-b)^2-2b-15=0$

$$b^2+2b-11=0$$

• $a^2-2(-2-a)-15=0$

$$a^2+2a-11=0$$

Then, by Vieta

$$ ab=-11.$$

Let $ a + b = c$.

Then, the quadratic $x^2 - 2 (c-x) -15 = x^2 +2x + (-15-2c) = 0 $ has distinct roots $a , b$.
Hence, these are the roots to the quadratic.
Thus, by Vieta's formula, $ a+b = - 2 \Rightarrow c = -2$ and $ ab = -15 - 2c = -11$.