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How can I solve this equation?

$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$

Simply bringing it to a common denominator does not lead me to success

What I tried

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Blue
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sleepydogbark
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3 Answers3

2

We have that for $(4+x)(4+2x)(4+3x)\neq0 $

$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 \iff 18 x^3 + 165 x^2 + 378 x + 255=0$$

$$\iff 6 x^3 + 35 x^2 + 126 x + 85=0$$

and by rational root theorem we can find that $x=-\frac 5 3$ is a root and then we obtain

$$6 x^3 + 35 x^2 + 126 x + 85=(3 x + 5) (2 x^2 + 15 x + 17)$$

user
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2

If you insist on not dealing with a cubic equation, then maybe this can work. Notice: $$\dfrac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)} = \dfrac{(1+x)(-1-2x)(1+3x)}{(4+x)(-4-2x)(4+3x)} = \dfrac{a(b-c)(c-4)}{b(a-c)(c-1)} = 4$$ where $c = 3x+5$, $a = x+1$ and $b = x+4.$ If you expand this quadratic in $c:$ $$c^2(a-4b)+c(3ab+4b-4a) = 0$$ and this one is easy to solve.

But I must say that learning how to factor cubic if they have nice solutions by the Rational Root Theorem etc is much more useful than trying to find clever substitutions. The other answer already covers that.

dezdichado
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  • @lonestudent not really. In fact, even substituting $c = 3x+1000$ works because of the presence of $x,2x,3x$ since $x+2x = 3x$ and thus it eliminates one order. If you think no one can observe things like this beforehand, then you obviously haven't done enough math competitions in your high school. – dezdichado Sep 19 '21 at 18:58
  • I don't think you have a clue as to what I am saying. The three linear terms being present means you can reduce the degree by one with literally any substitution because they are affinely dependent. As to why $3x+5$, it is literally staring in your face: $$(1+x)+(4+2x) = 3x+5 = (1+2x)+(4+x).$$ I cannot explain it any simpler. – dezdichado Sep 19 '21 at 19:36
  • I deleted my comments because I didn't get what I wanted from you. – lone student Sep 20 '21 at 03:24
-1

\begin{equation} \frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)} =4 \\ \text{We multiply both sides by teh common denominator}\\ (1+x)(1+2x)(1+3x\quad =\quad 4\big((4+x)(4+2x)(4+3x)\big) \\ \text{we expand both sides} \\ x^3 + 11 x^2 + 6 x + 1\quad =\quad 24 x^3 + 176 x^2 + 384 x + 256\\ \text{we collect terms} \\ 18 x^3 + 165 x^2 + 378 x + 255=0\\ \text{we factor} \\ 3 (3 x + 5) (2 x^2 + 15 x + 17)=0\\ \\ \text{we solve} \\ x = -5/3 \qquad\quad\space\space \space \approx -1.6667\\ x = \frac{-15 + \sqrt{89}}{4} \quad\approx -1.3915\\ x = \frac{-15 - \sqrt{89}}{4} \quad\approx -6.1085 \end{equation}

poetasis
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