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Question Number 65664 by mathmax by abdo last updated on 01/Aug/19

$$solve\frac{\sqrt{1-x}-\sqrt{2x+1}}{\sqrt{1-x}+\sqrt{2x+1}}=\frac{x+1}{3}$$

Answered by MJS last updated on 01/Aug/19

$$\frac{\sqrt{1-x}-\sqrt{2x+1}}{\sqrt{1-x}+\sqrt{2x+1}}=\frac{{(\sqrt{1-x}-\sqrt{2x+1})}^2}{(\sqrt{1-x}+\sqrt{2x+1})(\sqrt{1-x}-\sqrt{2x+1})}=$$$$=\frac{x+2-2\sqrt{1-x}\sqrt{2x+1}}{-3x}=-\frac{x+2}{3x}+\frac{2\sqrt{1-x}\sqrt{2x+1}}{3x}$$$$-\frac{x+2}{3x}+\frac{2\sqrt{1-x}\sqrt{2x+1}}{3x}=\frac{x+1}{3}$$$$2\sqrt{1-x}\sqrt{2x+1}=x^2+2x+2$$$$\mathrm{squaring}$$$$x(x^3+4x^2+16x+4)=0$$$$x_1=0\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{false}$$$$x^3+4x^2+16x+4=0$$$$x=t-\frac{4}{3}$$$$t=\sqrt[3]{\frac{170}{27}+\frac{2}{9}\sqrt{1713}}-\sqrt[3]{-\frac{170}{27}+\frac{2}{9}\sqrt{1713}}$$$$x=-\frac{4}{3}+\sqrt[3]{\frac{170}{27}+\frac{2}{9}\sqrt{1713}}-\sqrt[3]{-\frac{170}{27}+\frac{2}{9}\sqrt{1713}}$$$$x\approx -.266582$$

Commented by mathmax by abdo last updated on 01/Aug/19

$$thankyousir.$$

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