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I can not solve the equation:$$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0.$$ Can someone help me. Thanky veru much.

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First I solve the equation:$x^3-2x^2-5x+6=0.$ Becouse the sum of the coefficinet of the given equation is $0,$ $(1+(-2)+(-5)+6=0)$, one zero is 1. Now we have a divide $$(x^3-2x^2-5x+6):(x-1)=x^2-x-6$$ i.e., $$x^3-2x^2-5x+6=(x-1)(x^2-x-6)=(x-1)(x+2)(x-3),$$ becouse $$x^2-x-6=0\Rightarrow x_{1,2}=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-6)}}{2}=\frac{1\pm \sqrt{1+24}}{2}=\frac{1\pm5}{2}; x_1=-2;x_2=3.$$ Now we solve the equation: $x^2+5x+6=0\Rightarrow x_1=-3,x_2=-2\Rightarrow x^2+5x+6=(x+2)(x+3).$ Now we have: $$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0$$ $$\sqrt[4]{x^3-2x^2-5x+6}=-\sqrt{x^2+5x+6}/^4$$ $$x^3-2x^2-5x+6=(x^2+5x+6)^2$$ $$(x-1)(x+2)(x-3)=[(x+2)(x+3)]^2$$ $$(x-1)(x+2)(x-3)=(x+2)^2(x+3)^2$$ $$(x-1)(x+2)(x-3)-(x+2)^2(x+3)^2=0$$ $$(x+2)[(x-1)(x-3)-(x+2)(x+3)^2]=0$$ $$(x+2)(x^3+7x^2+25x+14)=0$$ $$x+2=0\Rightarrow x=-2$$ $$x^3+7x^2+25x+14=0$$ (you try to solve this equation, subs. $x=\frac{3y-7}{3},$ but will not earn real solution)

I hope you help

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Hint:

$\sqrt A$ (and $\sqrt[4]A$), when defined, are non-negative. Hence the equation is equivalent to the system $$\begin{cases}x^3-2x^2-5x+6=0,\\x^2+5x+6=0. \end{cases}$$

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  • $\begingroup$ the second line needs an $=0$ as well I think $\endgroup$ – Aydin Gerek Nov 8 '16 at 16:47
  • $\begingroup$ @Aydin Gerek: Yes. Thanks for pointing the typo! $\endgroup$ – Bernard Nov 8 '16 at 16:50
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$$\sqrt[4]{x^3-2x^2-5x+6}=\sqrt[4]{(x+2)(x-1)(x-3)}\\\sqrt{x^2+5x+6}=\sqrt{(x+2)(x+3)}$$ Both quantities must be zero then we get the system $$\begin{cases}(x+2)(x-1)(x-3)=0\\(x+2)(x+3)=0\end{cases}$$ It is apparent that $(x+2)=0\iff x=-2$ is the only solution.

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