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Question Number 167772 by mathlove last updated on 24/Mar/22

$$x+\sqrt{x}=5$$$$x+\frac{5}{\sqrt{x}}=?$$

Commented by mr W last updated on 24/Mar/22

$${\left(\sqrt{x}\right)}^2+\sqrt{x}-5=0$$$$\sqrt{x}=\frac{\sqrt{21}-1}{2}$$$$x+\frac{5}{\sqrt{x}}=5+\frac{5}{\sqrt{x}}-\sqrt{x}$$$$=5+\frac{10}{\sqrt{21}-1}-\frac{\sqrt{21}-1}{2}$$$$=5+\frac{\sqrt{21}+1}{2}-\frac{\sqrt{21}-1}{2}$$$$=6$$

Commented by MJS_new last updated on 24/Mar/22

$$x+\sqrt{x}=a\Rightarrow x+\frac{a}{\sqrt{x}}=a+1$$$$\mathrm{prove}:$$$$\left(1\right)$$$$a=x+\sqrt{x}=(\sqrt{x}+1)\sqrt{x}=\frac{(1+\sqrt{x})(1-\sqrt{x})\sqrt{x}}{1-\sqrt{x}}=$$$$=\frac{(1-x)\sqrt{x}}{1-\sqrt{x}}$$$$\left(2\right)$$$$a=\frac{(1-x)\sqrt{x}}{1-\sqrt{x}}$$$$a(1-\sqrt{x})=(1-x)\sqrt{x}$$$$a-a\sqrt{x}=(1-x)\sqrt{x}$$$$\frac{a}{\sqrt{x}}-a=1-x$$$$x-\frac{a}{\sqrt{x}}=a+1$$

Answered by mnjuly1970 last updated on 25/Mar/22

$$x+\frac{x+\sqrt{x}}{\sqrt{x}}=x+\sqrt{x}+1=6$$

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