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Question Number 187835 by pascal889 last updated on 22/Feb/23

Commented by Frix last updated on 22/Feb/23

$$\sqrt{x-a}+\sqrt{x-b}=\sqrt{c}$$$$x=\frac{{(a-b)}^2}{4c}+\frac{a+b}{2}+\frac{c}{4}$$

Answered by HeferH last updated on 22/Feb/23

$$x=2$$

Answered by Rasheed.Sindhi last updated on 22/Feb/23

$$\sqrt{x-1}+\sqrt{x-2}=1$$$$(\sqrt{x-1}+\sqrt{x-2})(\sqrt{x-1}-\sqrt{x-2})$$$$=1(\sqrt{x-1}-\sqrt{x-2})$$$$\sqrt{x-1}-\sqrt{x-2}={\left(\sqrt{x-1}\right)}^2-{\left(\sqrt{x-2}\right)}^2$$$$=x-1-x+2=1$$$$\{\begin{array}{l}\sqrt{x-1}+\sqrt{x-2}=1\\ \sqrt{x-1}-\sqrt{x-2}=1\end{array}$$$$\Rightarrow 2\sqrt{x-1}=2\Rightarrow x-1=1\Rightarrow x=2$$

Answered by Rasheed.Sindhi last updated on 22/Feb/23

$$\sqrt{x-1}=a,\sqrt{x+2}=b$$$$a+b=1.....A$$$$a^2-b^2={\left(\sqrt{x-1}\right)}^2-{\left(\sqrt{x-2}\right)}^2$$$$=x-1-x+2=1$$$$\frac{a^2-b^2}{a+b}=\frac{1}{1}$$$$a-b=1.....B$$$$A+B:2a=2\Rightarrow a=1$$$$\sqrt{x-1}=1\Rightarrow x-1=1\Rightarrow x=2$$

Answered by manxsol last updated on 22/Feb/23

$$\sqrt{x-a}+\sqrt{x-b}=\sqrt{c}I$$$$\text{Extra left or missing right}$$$$\sqrt{x-a}-\sqrt{x-b}=\frac{b-a}{\sqrt{c}}II$$$$+I+II$$$$2\sqrt{x-a}=\sqrt{c}+\frac{b-a}{\sqrt{c}}$$$$2\sqrt{c}\sqrt{x-a}=(c+b-a)$$$$4\mathit{c}(\mathit{x}-\mathit{a})={(\mathit{c}+\mathit{b}-\mathit{a})}^2$$$$condition$$$$x⩾a$$$$x⩾b$$$$c+b⩾a$$$$exercise$$$$a=1$$$$b=2$$$$c=1$$$$4\mathit{c}(\mathit{x}-\mathit{a})={(\mathit{c}+\mathit{b}-\mathit{a})}^2$$$$4\left(1\right)(x-1)={(2+1-1)}^2$$$$4(x-1)=4$$$$x-1=1$$$$x=2$$$$$$

Answered by SEKRET last updated on 23/Feb/23

$$\sqrt{\mathbf{x}-1}=\mathbf{a}\sqrt{\mathbf{x}-2}=\mathbf{b}$$$$\{\begin{array}{l}\mathbf{a}+\mathbf{b}=1\\ \mathbf{a}-\mathbf{b}=1\end{array}\mathbf{a}=1\mathbf{b}=0\mathbf{x}=2$$$$$$

Commented by manxsol last updated on 23/Feb/23

$$excellent$$

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