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Question Number 76340 by Master last updated on 26/Dec/19

Commented by behi83417@gmail.com last updated on 26/Dec/19

$$\mathrm{x}=-1,0$$

Answered by behi83417@gmail.com last updated on 26/Dec/19

$$-{3\mathrm{x}}^2+2\mathrm{x}+5=(1+\mathrm{x})(-3\mathrm{x}+5)$$$$\Rightarrow (1+\mathrm{x})(-3\mathrm{x}+5)=\sqrt{(1+\mathrm{x})(1-\mathrm{x})}({4\mathrm{x}}^2+8\mathrm{x}+5)$$$$1)[1+\mathbf{x}=0\Rightarrow \mathbf{x}=-1]$$$$2)(1+\mathrm{x}){(-3\mathrm{x}+5)}^2=(1-\mathrm{x}){({4\mathrm{x}}^2+8\mathrm{x}+5)}^2$$$$(1+\mathrm{x})({9\mathrm{x}}^2-30\mathrm{x}+25)=(1-\mathrm{x})({16\mathrm{x}}^4+{64\mathrm{x}}^2+25+{64\mathrm{x}}^3+{40\mathrm{x}}^2+80\mathrm{x})$$$$\Rightarrow {9\mathrm{x}}^2-30\mathrm{x}+25+{9\mathrm{x}}^3-{30\mathrm{x}}^2+25\mathrm{x}=$$$${16\mathrm{x}}^4+{64\mathrm{x}}^2+25+{64\mathrm{x}}^3+{40\mathrm{x}}^2+25\mathrm{x}$$$$-{16\mathrm{x}}^5-{64\mathrm{x}}^3-25\mathrm{x}-{64\mathrm{x}}^4-{40\mathrm{x}}^3-{80\mathrm{x}}^2$$$$\Rightarrow -{16\mathrm{x}}^5-{48\mathrm{x}}^4-{49\mathrm{x}}^3+{45\mathrm{x}}^2+5\mathrm{x}=0$$$$\Rightarrow -\mathrm{x}({16\mathrm{x}}^4+{48\mathrm{x}}^3+{49\mathrm{x}}^2-45\mathrm{x}-5)=0$$$$\Rightarrow [\mathbf{x}=0]$$$$16{\mathbf{x}}^4+48{\mathbf{x}}^3+49{\mathbf{x}}^2-45\mathbf{x}-5=0$$$$\Rightarrow \mathbf{x}=-0.1,0.623$$

Answered by mr W last updated on 26/Dec/19

$$-1⩽x⩽1$$$$(5-3x)(1+x)=\sqrt{(1-x)(1+x)}(4{(x+1)}^2+1)$$$$x=-1isasolution.$$$$forx\ne -1$$$$(5-3x)\sqrt{1+x}=\sqrt{(1-x)}(4{(x+1)}^2+1)$$$${(5-3x)}^2(1+x)=(1-x){(4{(x+1)}^2+1)}^2$$$$lett=x+1$$$${(8-3t)}^{2}t=(2-t){(4t^2+1)}^2$$$$(64-48t+9t^2)t=(2-t)(16t^4+8t^2+1)$$$$16t^5-32t^4+17t^3-64t^2+65t-2=0$$$$(t-1)(t^2+t+2)(16t^2-32t+1)=0$$$$t=1\Rightarrow x=0$$$$t=\frac{16\pm \sqrt{{16}^2-16}}{16}=1\pm \frac{\sqrt{15}}{4}\Rightarrow x=\pm \frac{\sqrt{15}}{4}$$$$t=\frac{-1\pm i\sqrt{7}}{2}\Rightarrow x=\frac{-3\pm i\sqrt{7}}{2}$$$$$$$$summaryofallsolutions:$$$$x=0$$$$x=-1$$$$x=\pm \frac{\sqrt{15}}{4}\approx \pm 0.9682$$$$x=\frac{-3\pm i\sqrt{7}}{2}$$

Commented by MJS last updated on 26/Dec/19

$$\mathrm{the}\mathrm{complex}\mathrm{solutions}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{fit}\mathrm{the}\mathrm{given}\mathrm{equation}$$

Commented by mr W last updated on 26/Dec/19

$$howtoseethis?$$$$dueto\sqrt{1-x^2}?$$

Commented by MJS last updated on 26/Dec/19

$$\mathrm{yes}.$$

Commented by mr W last updated on 26/Dec/19

$$thankssir!$$

Answered by MJS last updated on 26/Dec/19

$$\mathrm{squaring}\mathrm{and}\mathrm{transforming}$$$$\Rightarrow$$$$x(x+1)(x^2+3x+4)(x^2-\frac{15}{16})=0$$$$\Rightarrow$$$$x_1=-1$$$$x_2=-\frac{\sqrt{15}}{4}$$$$x_3=0$$$$x_4=\frac{\sqrt{15}}{4}$$$$x_{5,6}=-\frac{3}{2}\pm \frac{\sqrt{7}}{2}\mathrm{i}\left({\mathrm{don}}^{\prime }\mathrm{t}\mathrm{fit}\mathrm{the}\mathrm{given}\mathrm{equation}\right)$$$$$$$$\mathrm{other}\mathrm{path}$$$$\mathrm{let}x=\frac{2t}{t^2+1}$$$$\Rightarrow$$$$5+\frac{4t}{t^2+1}-\frac{12t^2}{{(t^2+1)}^2}=\frac{1-t^2}{t^2+1}(\frac{16t^2}{{(t^2+1)}^2}+\frac{16t}{t^2+1}+5)$$$$\Rightarrow$$$$t(t+1)(t^2+t+2)(t^2-\frac{3}{5})=0$$$$\Rightarrow$$$$t_1=-1\Rightarrow x_1=-1$$$$t_2=-\frac{\sqrt{15}}{5}\Rightarrow x_2=-\frac{\sqrt{15}}{4}$$$$t_3=0\Rightarrow x_3=0$$$$t_4=\frac{\sqrt{15}}{5}\Rightarrow x_4=\frac{\sqrt{15}}{4}$$$$t_{5,6}=-\frac{1}{2}\pm \frac{\sqrt{7}}{2}\mathrm{i}\Rightarrow x_{5,6}=-\frac{3}{2}\pm \frac{\sqrt{7}}{2}\mathrm{i}\left(\mathrm{false}\right)$$

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