Quadratic equation - solving for x

Question

Question: Solve the equation $$(x+2)(x+3)(x+8)(x+12) = 4x^2$$

I tried to solve the equation by expanding the LHS and then equating it to the RHS, but that just doesn't seem to be feasible. I am probably missing a key point here, which would make the question a lot easier. Help please!

Answer 1

I've just got an easier way. Setting $A=x^2+24$ gives you $$\begin{align}(x+2)(x+3)(x+8)(x+12)=4x^2&\iff (x+2)(x+12)(x+3)(x+8)=4x^2\\&\iff (x^2+14x+24)(x^2+11x+24)=4x^2\\&\iff (A+14x)(A+11x)=4x^2\\&\iff A^2+25xA+150x^2=0\\&\iff (A+10x)(A+15x)=0.\end{align}$$

Answer 2

What I thought at first was to find some integers $x$ (at least two integers) to satisfy the equality. Luckily, I got $$(-4+2)(-4+3)(-4+8)(-4+12)=4\cdot (-4)^2,$$ $$(-6+2)(-6+3)(-6+8)(-6+12)=4\cdot (-6)^2.$$ Hence, you can divide $(x+2)(x+3)(x+8)(x+12)-4x^2$ by $(x+4)(x+6)$ to find the other solutions.

Answer 3

If you don't see an obvious way to solve such problem (or you are not as lucky as mathlove :-)), there is always a brute-force approach that can (hopefully) work. You can always find all rational roots with this "mechanical" method.

Expand your left polynomial and subtract $4x^2$, to get the equation $$x^4+25x^3+198x^2+600x+576=0$$ Then, if $p \over q$ is a rational solution, you can show easily that $p$ divides $576$, and $q$ divides $1$ (the coefficient of highest degree).

Thus, a rational solution is an integer, and you have to test "only" the divisors of $576=2^6\cdot3^2$, which are:

$$\{1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 64, 72, 96, 144, 192, 288, 576\}$$

Plus the same with negative sign.

While (very) cumbersome, you will nevertheless find the roots $-4$ and $-6$. After factoring them out (with long division), there remains only a quadratic equation to solve, which is easy.

Answer 4

I have nothing against the suggestions made by others. But a little thought can save a lot of work.

The function $4x^2$ is non-negative and well known.

The quartic function will be large and positive for large positive or large negative $x$. Note that it is positive for $x\gt -2; -3\gt x \gt -8; -12 \gt x$ and negative for $-2\gt x\gt -3; -8 \gt x\gt -12$.

Hence we have a classic $W$-shaped quartic and a quick sketch shows that there will be points where the two functions are equal at $x\lt -12; x\gt -2$ where the quartic is initially below the quadratic but ultimately rises above it. It is easily seen that the quartic is positive at $x=0$, so that there is a root between $-2$ and $0$ and no roots for positive $x$. A similar test for $x\lt -12$ will identify the approximate root there, bearing in mind that the quartic is increasing rapidly as $x$ decreases.

The nature of the other roots depends on what happens in the middle of the $W$ in the interval $-3\gt x \gt -8$ and whether the quartic rises above the quadratic (giving two roots), touches it (giving one double root) or remains below it (leading to two non-real complex roots, and an irreducible quadratic factor over $\mathbb R$).

In the absence of any easily found roots, numerical methods or the general solution of the quartic will be necessary. As it happens on testing the middle section of the $W$ using the convenient product form of the quartic at integer values we find that there are two roots in this interval, and locate them as $-4, -6$.

Note that the information from a basic sketch and computing easy points will exclude most of the possible rational roots in any event.

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Now to pursue a suggestion I made in a comment. Using the information we have about existing roots, we let $y=x+5$ and the equation becomes: $$(y-3)(y-2)(y+3)(y+7)=4(y-5)^2$$ or $$(y^2-9)(y^2+5y-14)=4y^2-40y+100$$

This reduces to $$y^4+5y^3-27y^2-5y+26=y^4-27y^2+26+5y^3-5y=(y^2-1)(y^2-26)+5y(y^2-1)=0$$ which can then be easily solved.