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I would like to solve following equation:

$$15 (x+2)^{-4} = 11(x+2)^{-2} +4$$ I would first remove the negative power by adding $(x+2)^4$

Then I get $$15 = 11(x+2)^2 + 4(x+2)^4\\ 11(x+2)^2 + 4(x+2)^4 -15 = 0$$ should I do now a quadratic equation where $d=(x+2)^2$
$$11d + 4d^2 -15 =0$$ But this would be very much work The other way is to expand them and factorize them. but this is even more work. Isnt there are simple way?

AlexR
  • 24,905
  • try $(x+2)^{-2} = d$ instead so that you get to solve a factorable quadratic : $15d^2 -11d-4 = 0$ – AgentS Oct 21 '14 at 19:14

1 Answers1

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$$15(x+2)^{-4} = 11(x+2)^{-2} + 4$$ Multiply by $(x+2)^{4}$ $$15 = 11(x+2)^2 + 4(x+2)^4$$ Let $y = (x+2)^2$ $$15 = 11y + 4y^2$$ Subtract $15$ $$4y^2 + 11y - 15=0$$ Factor $$(4y + 15)(y - 1)=0$$ Consider both cases: $$(4y+15)=0$$ $$ y = -\frac{15}{4}$$ and $$(y-1)=0$$ $$y=1$$ So, the two solutions for $y$ are $-\frac{15}{4}$ and $1$. Substitute the values for $y$ into the expression containing $x$. $$1 = (x+2)^2$$ $$\pm 1 = x+2$$ $$x = -2 \pm 1$$ $$x = -1, -3$$ and $$-\frac{15}{4} = (x+2)^2$$ $$\pm \frac{\sqrt{15}i}{2} = (x+2)$$ $$x = -2 \pm \frac{\sqrt{15}i}{2}$$ So all of the solution for $x$ are:
$-1, -3, -2 + \dfrac{\sqrt{15}i}{2},$ and $-2 - \dfrac{\sqrt{15}i}{2}$

Cole Hansen
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  • To my knowledge, this is the "easiest" solution. I don't know of any other way that could be interpreted as "easier". – Cole Hansen Oct 21 '14 at 19:31