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: $a^2=2b+15$ and $b^2=2a+15$
Subtracting, $a^2-b^2 = -2(a-b)$. As $a \ne b$,
$a+b = - 2$
Also adding both equations, $a^2+b^2 = 2(a+b)+30 = 26$
$(a+b)^2 = a^2+b^2+2ab \implies 4 = 26 + 2ab$
$ab = -11$
A: 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
$$ ab=-11.$$
A: 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$.
A: This is not a straightforward way to answer the question, but illustrates a matter that the problem touches on.
We can treat the given equations as "solutions" to the problem:  if $ \ y^2 \ = \ 2x + 15 \ $ describes a transformation such that $ \ T(a) \ = \ b \ $ and $ \ T(b) \ = \ a \ \ , $ and so $ \ T( \ T(a)  \ ) \ = \ a \ $ and $ \ T( \ T(b)  \ ) \ = \ b \ \ , $ with $ \ a \neq b \ \ , $ find the product $ \ ab \ \ . $
Of course the original equation $ \ y^2 \ = \ 2x + 15 \ $ is not a function of $ \ x \ \ . $  Further, using either of its associated square-root functions alone do not suffice:  $ \ f(x) \ = \ \sqrt{2x + 15} \ $ has the fixed point $ \ x \ = \ 5 \ $ and $ \ g(x) \ = \ -\sqrt{2x + 15} \ \ , $ the fixed point $ \ x \ = \ -3 \ \ , $ as is easily computed and is seen in the graph below.
But the specification in the equation can be satisfied if we "link" the two functions so as to determine $ \ f(a) \ = \ b \ , \ g(b) \ = \ a \ $ and thus $ \ g( \ f(a) \ ) \ = \ a \ $ and $ \ f( \ g(b) \ ) \ = \ b \ \ . $  This produces a pair of transformations
$$  f( \ g(x) \ ) \  =    \ \sqrt{ \ 2·(-\sqrt{2x + 15}) \ + \ 15} \ \ \ \text{and}  \ \ \  g( \ f(x) \ ) \  = \    -\sqrt{ \ 2·(\sqrt{2x + 15}) \ + \ 15} \ \ . $$
So we now seek to solve $ \ g( \ f(a) \ ) \ = \  -\sqrt{ \ 2·(\sqrt{ \ 2a + 15 \ }) \ + \ 15} \ = \ a \ \ $ [solving $ \ f( \ g(b) \ ) \ = \ b \ \ $ will take us to the same place].  This looks like it will be spectacularly unhelpful; let us proceed:
$$ - \ \sqrt{ \ 2·(\sqrt{ \ 2a + 15 \ }) \ + \ 15} \ \ = \ \ a \ \ \Rightarrow \ \ 2· \sqrt{ \ 2a + 15 \ } \ \ = \ \ a^2 \ - \ 15 $$
$$ \Rightarrow \ \ 4·( 2a + 15 ) \ \ = \ \ (a^2 \ - \ 15)^2 \ \ \Rightarrow \ \ a^4 \ - \ 30a^2 \ - \ 8a \ + \ 165 \ \ = \ \ 0 \ \ . $$
Certainly we didn't want to have to deal with a quartic equation.  But it turns out (not coincidentally) that the fixed points of $ \ f(x) \ $ and $ \ g(x) \ $ that we found earlier are roots of this equation.  We thus find that the polynomial can be factored as
$$  a^4 \ - \ 30a^2 \ - \ 8a \ + \ 165 \ \ = \ \ (a + 3)·(a - 5)·(a^2 + 2a - 11) \ \ = \ \  0 \ \ , $$
for which the zeroes of the quadratic factor are the sought-after $ \ a \ = \ -1 + 2\sqrt3 \ $ and $ \ b \ = \ -1 - 2\sqrt3 \ \ , $   the product of these being the constant term of that factor, $ \ \mathbf{ab \ = \ -11} \ \ . $  [These zeroes are also seen in the graph.]

