The equation $x^4-8x^3+24x^2-32x-14=0$ has two real roots $x_1, x_2$ and two non-real roots $x_3, x_4$. Find the value of $x_1x_2+x_3x_4$.
I took $x_1, x_2$ as conjugate pairs and $x_3$ and $x_4$ conjugate pairs but what I got was not enough to solve the question I then tried by trail and error and got 8 as the answer but I am highly doubtful about it. Please if anyone who can come up with a proper method.
On
I will provide a solution that does not require explicitly finding the roots, but let's fix a misconception first.
If $f$ is a polynomial with real coefficients, then only the non-real roots of $f$ come in conjugate pairs. The theorem is as follows:
Let $f\in\mathbb{R}[x]$. For any $\alpha\in\mathbb{C}$, we have $f(\alpha)=0$ if and only if $f(\overline\alpha)=0$.
Apply this to a real root $\alpha$, you get nothing. This of course makes sense because you can have $f=(x-a)(x-b)g$ where $a,b\in\mathbb{R}$ and $g$ is an irreducible quadratic real polynomial. Then $f$ still has real coefficients, but $a$ and $b$ are arbitrary.
Back to your problem. Let $f(x)=x^4-8x^3+24x^2-32x-14$.
For any polynomial $g$ and a polynomial expression $h(x_1,\dots,x_n)$ symmetric in the roots $x_1,\dots,x_n$ of $g$, we know that we can express $h$ as a polynomial in the coefficients of $g$. This is known as the symmetric function theorem. For example, if $x_1,x_2$ are roots of $x^2-ax+b$, then we have $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=a^2-2b$. For more details, see wiki and an exposition article here.
Now, in your particular problem, the expression $x_1x_2+x_3x_4$ is not symmetric (for example, swapping $x_1$ and $x_3$ results in a different polynomial expression). But also notice that this is not "completely asymmetric". For example, swapping $x_1$ and $x_2$ does not change the expression. This motivates us to consider how many different expressions there can be under all $24$ permutations. Surprisingly, only $3$, namely: $$ x_1x_2+x_3x_4,\ x_1x_3+x_2x_4,\ x_1x_4+x_2x_3 $$ As noted by @lhf, if you sum the $3$ expressions, you get $24$ because the sum is now symmetric! Can we do better? Sure, consider the following polynomial [1]: $$ R_3(x)=(x-(x_1x_2+x_3x_4))(x-(x_1x_3+x_2x_4))(x-(x_1x_4+x_2x_3)) $$ Since $R_3$ includes all $3$ expressions, no matter how you permute $x_1,\dots,x_4$, the resulting polynomial will be unchanged. Thus, $R_3$ is a polynomial whose coefficients are symmetric polynomials of $x_1,\dots,x_4$! This means that we can express the coefficients of $R_3$ as polynomials in the coefficients of $f$, thanks to the symmetric function theorem.
Let $$ \begin{align*} R_3(x) &= x^3+Ax^2+Bx+C. \end{align*} $$ Then $$ A=-(x_1x_2+x_3x_4+x_1x_3+x_2x_4+x_1x_4+x_2x_3)=-24. $$ For $B$ and $C$, it's long when written out, but the idea is really simple (take advantage of symmetry): $$ \begin{align*} B &= x_{1}^{2} x_{2} x_{3} + x_{1} x_{2}^{2} x_{3} + x_{1} x_{2} x_{3}^{2} + x_{1}^{2} x_{2} x_{4} + x_{1} x_{2}^{2} x_{4} + x_{1}^{2} x_{3} x_{4} \\ &\quad +x_{2}^{2} x_{3} x_{4} + x_{1} x_{3}^{2} x_{4} + x_{2} x_{3}^{2} x_{4} + x_{1} x_{2} x_{4}^{2} + x_{1} x_{3} x_{4}^{2} + x_{2} x_{3} x_{4}^{2} \\ &={\left(x_{1} + x_{2} + x_{3}\right)} x_{1} x_{2} x_{3} +{\left(x_{1} + x_{2} + x_{4}\right)} x_{1} x_{2} x_{4}\\ &\quad +{\left(x_{1} + x_{3} + x_{4}\right)} x_{1} x_{3} x_{4} +{\left(x_{2} + x_{3} + x_{4}\right)} x_{2} x_{3} x_{4} \\ &= (x_1+x_2+x_3+x_4)(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)-4x_1x_2x_3x_4 \\ &= 8\cdot 32 - 4\cdot (-14) \\ &= 312 \end{align*} $$ and $$ \begin{align*} C &= -x_{1}^{2} x_{2}^{2} x_{3}^{2} - x_{1}^{2} x_{2}^{2} x_{4}^{2} - x_{1}^{2} x_{3}^{2} x_{4}^{2} - x_{2}^{2} x_{3}^{2} x_{4}^{2} \\ &\quad- x_{1} x_{2} x_{3} x_{4}^{3} - x_{1}^{3} x_{2} x_{3} x_{4} - x_{1} x_{2}^{3} x_{3} x_{4} - x_{1} x_{2} x_{3}^{3} x_{4}. \end{align*} $$ Note that $$ \begin{align*} &\quad x_{1}^{2} x_{2}^{2} x_{3}^{2} + x_{1}^{2} x_{2}^{2} x_{4}^{2} + x_{1}^{2} x_{3}^{2} x_{4}^{2} + x_{2}^{2} x_{3}^{2} x_{4}^{2} \\ &= (x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)^2\\ &\quad -(2 \, x_{1}^{2} x_{2}^{2} x_{3} x_{4} + 2 \, x_{1}^{2} x_{2} x_{3}^{2} x_{4} + 2 \, x_{1} x_{2}^{2} x_{3}^{2} x_{4} + 2 \, x_{1}^{2} x_{2} x_{3} x_{4}^{2} + 2 \, x_{1} x_{2}^{2} x_{3} x_{4}^{2} + 2 \, x_{1} x_{2} x_{3}^{2} x_{4}^{2}) \\ &= (x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)^2 \\ &\quad - 2\, x_{1} x_{2} x_{3} x_{4}{\left(x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} + x_{1} x_{4} + x_{2} x_{4} + x_{3} x_{4}\right)} \\ &= 32^2-2\cdot (-14)\cdot 24 \\ &= 1696 \end{align*} $$ and $$ \begin{align*} &\quad x_{1}^{3} x_{2} x_{3} x_{4} + x_{1} x_{2}^{3} x_{3} x_{4} + x_{1} x_{2} x_{3}^{3} x_{4} + x_{1} x_{2} x_{3} x_{4}^{3} \\ &= x_{1} x_{2} x_{3} x_{4}{\left(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}\right)} \\ &= x_1x_2x_3x_4 \\ &\quad \cdot ((x_1+x_2+x_3+x_4)^2- 2\, {\left(x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} + x_{1} x_{4} + x_{2} x_{4} + x_{3} x_{4}\right)}) \\ &= -14\cdot (8^2-2\cdot 24) \\ &= -224 \end{align*} $$
Substituting this back into $C$, we have $$ \begin{align*} C = -1696 - (-224) = -1472. \end{align*} $$
The dust settles, and we are left with: $$ R_3(x) = x^3-24x^2+312x-1472. $$ We want to see if we can factorize $R_3$. We can see if there is a rational root of $R_3$. By rational root theorem, any rational root must be a divisor of $1472=2^6\cdot 23$, and hence an integer. If $\alpha$ is a such a root, then we have $$ \begin{align*} \alpha^3&\equiv 0\ (\text{mod}\ 2) \\ \alpha^3-2&\equiv 0\ (\text{mod}\ 3) \\ \alpha^3-\alpha^2+2\alpha-2&\equiv 0\ (\text{mod}\ 5) \end{align*} $$ After a few testings, we have: $$ \begin{align*} \alpha&\equiv 0\ (\text{mod}\ 2) \\ \alpha&\equiv 2\ (\text{mod}\ 3) \\ \alpha&\equiv 3\ (\text{mod}\ 5) \end{align*} $$ Using Chinese remainder theorem, it's not hard to find that the smallest positive solution is $\alpha=8$, which happens to be a root of $R_3$!
Then we have $$ R_3(x)={\left(x - 8\right)}{\left(x^{2} - 16 \, x + 184\right)}. $$ The discriminant of the quadratic factor is $16^2-4\cdot 184=-480<0$, so $x=8$ is the only real root of $R_3$.
Finally, recall that the roots of $R_3$ are: $$ x_1x_2+x_3x_4,\ x_1x_3+x_2x_4,\ x_1x_4+x_2x_3 $$ Since we know that $x_1x_2+x_3x_4$ is real, it has to be the unique real root of $R_3$, so we have $x_1x_2+x_3x_4=8$.
[1]: This is known as the cubic resolvent of $f$. If you know a little Galois theory, you will immediately see that the coefficients of $R_3$ can be expressed as polynomials of coefficients of $f$.
On
It might be remarked that if one missed the "binomial-fourth-power" coefficients in $ \ f(x) \ = \ x^4 - 8x^3 + 24x^2 - 32x - 14 \ \ $ (which leads to Vinanth S Bharadwaj's very direct argument), "depressing" the polynomial takes us to a similar place: $$ f(x+2) \ \ = \ \ F(y) \ \ = \ \ y^4 \ - \ 30 \ \ , $$ for which the zeroes are arranged on the real and imaginary axes at a radius $ \ \rho \ \ . \ $ While it is easy to obtain $ \ \rho \ = \ 30^{1/4} \ \ , \ $ we don't actually care what its value is. Returning to $ \ f(x) \ \ , \ $ we find that the products of the real and imaginary zeroes are $ \ x_1·x_2 \ = \ (2 + \rho)·(2 - \rho) \ = \ 4 - \rho^2 \ \ $ and $ \ x_3·x_4 \ = \ (2 + i\rho)·(2 - i\rho) $ $ = \ 4 + \rho^2 \ \ . \ $ The sum we seek is then $ \ x_1x_2 \ + \ x_3x_4 \ = \ 8 \ \ . $
Going to somewhat more trouble, we could use the complex-conjugacy of $ \ x_3 \ , \ x_4 = \overline{x_3} \ = \ \gamma \pm i·\delta \ \ $ to write two quadratic factors for $ \ f(x) \ = \ (x^2 + ax + b)·(x^2 - 2·\gamma·x + c) \ \ , \ $ with $ \ c \ = \ z_{3,4}·\overline{z_{3,4} } \ \ . $ Without presenting the ensuing calculations for a system of four equations, we obtain $$ f(x) \ = \ (x^2 - 4x + 4 - \sqrt{30})·(x^2 - 4x + 4 + \sqrt{30}) \ \ . $$ Once again, we don't need to determine the zeroes, but can simply observe that $ \ x_1·x_2 \ $ and $ \ x_3·x_4 \ $ are equal to the constant terms of these factors (we won't even need to say which pair* corresponds to which constant). The sum of said constant terms is $ \ 8 \ \ . $
$ \ ^{*} $ though we see that the second factor is irreducible over $ \ \mathbb{R} \ . $
On
For the sake of completeness, the following is a computer-aided way to brute force it.
Let $\,x_3 = z\,$ be one of the (non-real) complex roots, and let $\,t=|z|^2 \ge 0\,$. Because the polynomial has real coefficients, the other complex root is $\,x_4=\bar z =\dfrac{t}{z}\,$. Therefore the equalities hold true:
$$ \begin{cases} \begin{align} \;\; z^4 - 8 z^3 + 24 z^2 - 32 z - 14 &= 0 \\ \;\; t^4 - 8 t^3 z + 24 t^2 z^2 - 32 t z^3 - 14 z^4 &= 0 \end{align} \end{cases} \tag{*} $$
Eliminating $\,z\,$ between the two equations using polynomial resultants gives (courtesy WA):
$$ (t^2 - 8 t - 14)^2 (t^4 - 16 t^3 + 36 t^2 - 1696 t + 196) (t^4 - 16 t^3 + 156 t^2 + 224 t + 196)^2 = 0 $$
Since equations $(*)$ have two roots in common, $z$ and $\bar z$, $\,t\,$ must be a double root of the resultant. The last factor has no real roots, so $\,t\,$ must be a root of the quadratic factor i.e. $\,t=4 \pm \sqrt{30}\,$, and $\,t \ge 0 \implies t = 4 + \sqrt{30}\,$.
Then from Vieta's relations $\,x_1x_2 = -\dfrac{14}{x_3x_4}=-\dfrac{14}{t}=-\dfrac{14}{4+\sqrt{30}}=4-\sqrt{30}\,$, and therefore $\,x_1x_2+x_3x_4$ $\require{cancel}= (4-\cancel{\sqrt{30}})+(4+\cancel{\sqrt{30}})=8\,$.
The equation is very similar to $(x-k)^4$. You can derive this equation and check the similarity to your equation and $k$ comes out to be $2$.
You can then use completing square method to solve the equation.
$x^4-8x^3+24x^2-32x+16-16-14=0$
$(x-2)^4-30=0$
$x = 2\pm\sqrt[\leftroot{-2}\uproot{2}4]{30}$ and $x = 2\pm i\sqrt[\leftroot{-2}\uproot{2}4]{30}$
You can solve this and the answer comes out to be 8