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Question Number 146315 by mnjuly1970 last updated on 12/Jul/21

Answered by Olaf_Thorendsen last updated on 13/Jul/21

$$\sqrt{\frac{x(x+5)}{x+2}}+\sqrt{x+3}=\frac{\sqrt{16x+32}}{x+2}$$$$\sqrt{\frac{x(x+5)}{x+2}}+\sqrt{x+3}=\frac{4}{\sqrt{x+2}}$$$$\sqrt{x(x+5)}+\sqrt{(x+3)(x+2)}=4$$$$\mathrm{Let}u=x+\frac{5}{2}$$$$\sqrt{(u-\frac{5}{2})(u+\frac{5}{2})}+\sqrt{(u+\frac{1}{2})(u-\frac{1}{2})}=4$$$$\sqrt{u^2-\frac{25}{4}}+\sqrt{u^2-\frac{1}{4}}=4$$$$\mathrm{Let}v=u^2-\frac{1}{4}$$$$\sqrt{v-6}=4-\sqrt{v}$$$$\Rightarrow v-6=16-8\sqrt{v}+v$$$$\sqrt{v}=\frac{11}{4}\Rightarrow v=\frac{121}{16}$$$$u^2=v+\frac{1}{4}=\frac{125}{16}\Rightarrow u=\pm \frac{5\sqrt{5}}{4}$$$$x=u-\frac{5}{2}=-\frac{5}{4}(2\pm \sqrt{5})$$$$x=-\frac{5}{4}(2+\sqrt{5})\approx -5,29$$$$(\mathrm{impossible}\mathrm{because}x+2>0)$$$$x=-\frac{5}{4}(2-\sqrt{5})\approx +0,295\mathrm{OK}$$$$$$$$\mathrm{Finally}x=-\frac{5}{4}(2-\sqrt{5})$$

Commented by mnjuly1970 last updated on 13/Jul/21

$$thankyoumrolaf$$

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