# Proof that given equation(quartic) doesn't have real roots

$$(x^2-9)(x-2)(x+4)+(x^2-36)(x-4)(x+8)+153=0$$ I need to prove that the above equation doesn't have a real solution. I tried breaking it up into an $(\alpha)(\beta)\cdots=0$ expression, but no luck. Wolfram alpha tells me that the equation doesn't have real roots, but I'm sure there's simpler way to solve this than working trough the quartic this gives.

• I don't know if that could work but maybe you could get some information when considering intervals like $(-\infty, -8), (-8, -6), (-6, -4) etc.$ (where signs of things in parenthesis change). – I want to make games Oct 27 '15 at 20:37
• This problem can be reduced to showing that $\sum\limits_{k=1}^{2}(x+3k)(x-3k)(x-2k)(x+4k) > 0$, which I haven't figured out yet. – Clarinetist Oct 27 '15 at 21:02

$$P(x) = ({x^2} - 9)(x - 2)(x + 4) + ({x^2} - 36)(x - 4)(x + 8) + 153\tag{1}$$

Now consider theses

\eqalign{ & f(x) = ({x^2} - 9)(x - 2)(x + 4) \cr & f({x \over 2}) = \left( {{{\left( {{x \over 2}} \right)}^2} - 9)} \right)\left( {\left( {{x \over 2}} \right) - 2} \right)\left( {\left( {{x \over 2}} \right) + 4} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 4}\left( {{x^2} - 36} \right){1 \over 2}\left( {x - 2} \right){1 \over 2}\left( {x - 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {16}}\left( {{x^2} - 36} \right)\left( {x - 2} \right)\left( {x - 8} \right) \cr}\tag{2}

combine $(1)$ and $(2)$ to get

$$P(x) = f(x) + 16f({x \over 2}) + 153\tag{3}$$

Next, we try to find the range of $f(x)$. For this purpose, consider this

\eqalign{ & f(x) = ({x^2} - 9)(x - 2)(x + 4) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {x^4} + 2{x^3} - 17{x^2} - 18x + 72 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{x^2} + x - 9} \right)^2} - 9 \cr}\tag{4}

Now, by $(4)$ you can conclude that

$$\left\{ \matrix{ f(x) \ge - 9 \hfill \cr f({x \over 2}) > - 9 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\left\{ \matrix{ f(x) > - 9 \hfill \cr f({x \over 2}) \ge - 9 \hfill \cr} \right.\tag{5}$$

Notice the equality signs! Can you figure out why this happens? Then using $(5)$ you can conclude that

$$\left\{ \matrix{ f(x) \ge - 9 \hfill \cr 16f({x \over 2}) > - 144 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\left\{ \matrix{ f(x) > - 9 \hfill \cr 16f({x \over 2}) \ge - 144 \hfill \cr} \right.\tag{6}$$

and then summing up either of the relations $(6)$ will lead to

$$f(x) + 16f({x \over 2}) > - 153\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,f(x) + 16f({x \over 2}) + 153 > 0\,\,\,\,\,\, \to \,\,\,\,\,\,\,P(x) > 0\tag{7}$$

I think we are done now! :)

Expand $x^2-9=(x-3)(x+3)$ and $x^2-36=(x-6)(x+6)$. Let $t=x+\frac 1 2$, $u=x+1$. Then $$LHS=(t^2-(2.5)^2)(t^2-(3.5)^2)+(u^2-25)(u^2-49)+153\geq -(\frac {3.5^2-2.5^2} 2)^2-(\frac {49-25} 2)^2+153=0$$ with equality iff $t^2=\frac {2.5^2+3.5^2} 2$ and $u^2=\frac {25+49} 2$ which is incompatible. Hence $LHS>0$