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Question Number 4889 by 123456 last updated on 19/Mar/16

$$f\left(x\right)=x^2-2x+3$$$$g\left(x\right)=2x^2+7x-1$$$$h\left(x\right)=x^2+10x-7$$$$f\left(g\left(h\left(x\right)\right)\right)=0,x=?$$

Answered by Yozzii last updated on 19/Mar/16

$$f\left(g\left(h\left(x\right)\right)\right)=0.Letu=g\left(h\left(x\right)\right).$$$$\therefore f\left(u\right)=0\Rightarrow u^2-2u+3=0$$$$\therefore u=\frac{2\pm \sqrt{2^2-4\times 1\times 3}}{2}=\frac{2\pm 2\sqrt{-2}}{2}$$$$u=1\pm i\sqrt{2},u_1=1+i\sqrt{2},u_2=1-i\sqrt{2}.$$$$\therefore g\left(h\left(x\right)\right)=u_r(r=1,2)$$$$Letn=h\left(x\right).\therefore g\left(n\right)=u_r.$$$$2n^2+7n-1=u_r$$$$2n^2+7n-1-u_r=0$$$$n=\frac{-7\pm \sqrt{49-4\times 2(-1-u_r)}}{4}$$$$n=\frac{-7\pm \sqrt{57+8u_r}}{4}$$$$\therefore n_1=\frac{-7+\sqrt{57+8u_r}}{4},n_2=\frac{-7-\sqrt{57+8u_r}}{4}$$$$r=1,2.\left(4rootsforn\right)$$$$\therefore h\left(x\right)=n_j(j=1,2)$$$$x^2+10x-7=n_j$$$$x=\frac{-10\pm \sqrt{{10}^2-4\left(1\right)(-7-n_j)}}{2}$$$$x=\frac{-10\pm \sqrt{4\times 25+4(7+n_j)}}{2}$$$$x_j=-5\pm \sqrt{32+n_j}$$$$x_{1,j}=-5+\sqrt{32+n_j},x_{2,j}=-5-\sqrt{32+n_j}$$$$x_{1,r}=-5+\sqrt{32+\frac{-7\pm \sqrt{57+8u_r}}{4}}$$$$x_1=-5+\sqrt{32+\frac{-7+\sqrt{65+i8\sqrt{2})}}{4}}$$$$x_2=-5+\sqrt{32+\frac{-7+\sqrt{65-8i\sqrt{2}}}{4}}$$$$x_3=-5+\sqrt{32+\frac{-7-\sqrt{65-8i\sqrt{2}}}{4}}$$$$x_4=-5+\sqrt{32+\frac{-7-\sqrt{65+8i\sqrt{2}}}{4}}$$$$-------------------$$$$x_{2,r}=-5-\sqrt{32+\frac{-7\pm \sqrt{57+8u_r}}{4}}$$$$x_5=-5-\sqrt{32+\frac{-7+\sqrt{65+8i\sqrt{2}}}{4}}$$$$x_6=-5-\sqrt{32+\frac{-7+\sqrt{65-8i\sqrt{2}}}{4}}$$$$x_7=-5-\sqrt{32+\frac{-7-\sqrt{65-8i\sqrt{2}}}{4}}$$$$x_8=-5-\sqrt{32+\frac{-7-\sqrt{65+8i\sqrt{2}}}{4}}.$$$$$$$$$$

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