$S$ , $P$ are sum and product of real roots of $x^4-7x^2-5=0$ what is the value of $2P^2-3SP+2S$? 
If $S$ and $P$ are sum and product of real roots of $x^4-7x^2-5=0$
what is the value of $2P^2-3SP+2S$ ?
$1)59-7\sqrt{69}\qquad\qquad2)7+\sqrt{69}\qquad\qquad3)50\qquad\qquad4)59+7\sqrt{69}$

I solved this problem as follow:
Since $f(x)=x^4-7x^2-5$ is an even function, sum of the roots is equal to zero Hence we are looking for the value of $2P^2$.
$$x^2=\dfrac{7\pm\sqrt{69}}{2}\quad\text{Since $x^2\ge0$, $\quad x^2=\dfrac{7+\sqrt{69}}{2}$ is acceptable}$$
$$x_1=+\sqrt{\dfrac{7+\sqrt{69}}{2}}\quad\text{,}\quad x_2=-\sqrt{\dfrac{7+\sqrt{69}}{2}}$$
$$2P^2=2\times \left(\dfrac{7+\sqrt{69}}{2}\right)^2=\dfrac{49+69+14\sqrt{69}}{2}=59+7\sqrt{69}$$
Is my answer right? Also I'm very eager to see if it is possible to solve this with other approaches.
 A: Observe that $f$ is a function of the quadratic in $z = x^2$:  $$f(x) = z^2 - 7z - 5.$$  This has roots $$x^2 = z = \frac{7 \pm \sqrt{(-7)^2 - 4(-5)}}{2} = \frac{7 \pm \sqrt{69}}{2}.$$  Since the negative root for $z$ yields non-real roots for $x$, we can discard these, and find that the two real roots of $f$ are $$x \in \pm \sqrt{\frac{7 + \sqrt{69}}{2}}.$$  Their sum $S$ is indeed zero, and their product is simply $$P = - \frac{7 + \sqrt{69}}{2}.$$  Thus $$2P^2 = 2 \left(\frac{7 + \sqrt{69}}{2}\right)^2 = \frac{1}{2} (49 + 14\sqrt{69} + 69) = 59 + 7 \sqrt{69}.$$  This is nearly identical to your solution.
A: We have a monic biquadratic equation with real coefficients $ \  x^4 - 7x^2 - 5 \ = \ 0 \ $ for which $ \ (-7)^2  -  4·(-5) \ = \ 69 \ > \ 0 \ \ , $ so there are either two or four real roots.  Since $ \ 69 \ > \ (-7)^2 \ \ , $ but the middle coefficient is negative, there will only be two real roots.  (These conclusions follow from the expression for $ \ x^2 \ . ) \ $
The polynomial can then be factored as $$ \ (x - r)·(x + r)·(x - \alpha - i\beta)·(x - \alpha + i\beta) \ \ = \ \ (x^2 - r^2)·(x^2 - 2\alpha x + \alpha^2 + \beta^2) \ \ . $$  As you already noted, the sum of all of the zeroes and the sum $ \ S \ $ of the real zeroes is zero, since the polynomial function is even. But it must also be the case that the real part $ \ \alpha \ $ of the complex zeroes must equal zero, as the product of the factors would have terms with odd powers of $ \ x  \ $ otherwise.  Hence, the factorization of our polynomial is just $ \ (x^2 - r^2)·(x^2  + \beta^2) \ \ . $
This leaves us to solve, in order to "extract" the real roots, the system of equations
$$ \beta^2 \ - \ r^2 \ = \ -7 \ \ , \ \ \beta^2·r^2 \ = \ 5 \ \ \Rightarrow \ \ (\beta^2 \ + \ r^2)^2 \ = \ (-7)^2 \ + \ 4·5 \ \ = \ \ 69 $$
$$ \ \ \Rightarrow \ \ (\beta^2 \ + \ r^2) \ - \ (\beta^2 \ - \ r^2) \ \ = \ \ 2·r^2 \ \ = \ \ \sqrt{69} \ + \ 7 \ \ . $$
[We are, in a sense, applying the quadratic formula without explicitly writing it.]  This quantity is then $ \ -2P \ , $ so the value sought in the question is
$$ 2P^2 \ - \ 3PS \ + \ 2S \ \ = \ \ \frac{( \ -2P \ )^2}{2} \ - \ 0 \ + \ 0 \ \ = \ \ \frac{( \ \sqrt{69} \ + \ 7 \ )^2}{2} $$ $$ = \ \ \frac{69 \ + \ 2·7·\sqrt{69} \ + \ 49}{2} \ \ = \ \ 59 \ + \ 7 \sqrt{69} \ \ , \ \ $$
which is choice $ \ \mathbf{(4)} \ \ . $
Although not requested, we also have the four roots of the equation
$$ \pm \ r \ \ = \ \ \pm \ \sqrt{\frac{7 \ + \ \sqrt{69}}{2}} $$
and
$$ 2·\beta^2 \ \ = \ \ \sqrt{69} \ - \ 7 \ \ \Rightarrow \ \ \pm \ i·\beta \ \ = \ \ \pm \ i·\sqrt{\frac{-7 \ + \ \sqrt{69}}{2}} \ \ . $$
