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I need some tips with solving this algebraic equation $$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$

I've tried subtracting: $$\frac{8x}{x+2}$$ and also setting $$\sqrt x = t$$

Where "t" is a substitution to make things simpler.

This is how it ends up:

$$x + 2 = 6\sqrt x - \frac{8x}{x+2}$$

or

$$t^2 + 2 = 6t - \frac{8t^2}{t^2+2}$$

Unfortunately, I couldn't find a way from here without getting polynomials of the fourth degree or equations with $$x\sqrt x$$

I'd just like to clarify that I'm not looking for the solution here. I'd just like it if I could have some pointers or tips on where to go from here, or even if I did something wrong.

Thanks in advance!!!

Blue
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Dinosaurs are cool
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    Ultimately, you'll get a fourth-degree polynomial equation in $t$. However, it factors into two quadratics. – Blue Jun 20 '20 at 23:28
  • By the way ... The work you've done is correct. (Subtracting-out $8x$ on the left so that the fraction reduces is even insightful.) However, it's not particularly helpful to have the fractional expression hanging around. Since you'll just have to clear denominators anyway, you might as well have started by cross-multipliing in the beginning to get $$x^2+12x+4 = 6 \sqrt{x} (x+2)$$ Then make the $t$ substitution and go from there. – Blue Jun 20 '20 at 23:31

3 Answers3

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Middle term splitting will always remain my favorite method for factorising polynomials!

\begin{align*} &\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x\\ \iff &\dfrac{(x+2)^2+8x}{x+2}=6\sqrt x\\ \iff &(x+2)^2-6\sqrt x(x+2)+8x=0\\ \iff &(x+2)^2-4\sqrt x(x+2)-2\sqrt x(x+2)+8x=0\\ \iff &(x+2-4\sqrt x)(x+2-2\sqrt x)=0 \end{align*} Now apply the quadratic formula.

Martund
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Substitute $\sqrt x=t$ to get fourth degree polynomial equation which further can be factorized into two quadratics polynomials $$\frac{t^4 + 12t^2 + 4}{t^2+2} = 6t$$ $$t^4-6t^3+12t^2-12t+4=0$$ $$(t^2-4t+2)(t^2-2t+2)=0$$ $$t^2-4t+2=0, \ \ \ \ t^2-2t+2=0$$ Solving above quadratic equations, we get $$t=2\pm \sqrt2, \ \ \ t=1\pm i$$

Harish Chandra Rajpoot
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  • How did you factorise? – UmbQbify Jun 20 '20 at 23:46
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    @AshWhole, See https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x and https://math.stackexchange.com/questions/403025/equation-with-high-exponents – lab bhattacharjee Jun 21 '20 at 01:24
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Start with $$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$ Then, $$x^2+12x+4=6\sqrt{x}(x+2)$$ Square both sides, $$ x^4 + 24 x^3 + 152 x^2 + 96 x + 16 = 36x^3+144x^2+144x$$ Therefore we want to find the roots of the equation $$x^4-12x^3+8x^2 - 48x +16 =0$$ Although you could use the quartic formula it's a real pain. Wolfram Alpha finds $$x=6 \pm 4\sqrt{2} \ ; x= \pm 2i.$$

K.defaoite
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