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Question Number 112805 by Aina Samuel Temidayo last updated on 09/Sep/20

$$\mathrm{If}\mathrm{x}=\frac{\sqrt{\mathrm{a}+2\mathrm{b}}+\sqrt{\mathrm{a}-2\mathrm{b}}}{\sqrt{\mathrm{a}+2\mathrm{b}}-\sqrt{\mathrm{a}-2\mathrm{b}}},\mathrm{then}\mathrm{the}\mathrm{value}$$$$\mathrm{of}{\mathrm{bx}}^2-\mathrm{ax}+\mathrm{b}\mathrm{is}$$

Answered by nimnim last updated on 09/Sep/20

$$\frac{\mathrm{x}}{1}=\frac{\sqrt{\mathrm{a}+2\mathrm{b}}+\sqrt{\mathrm{a}-2\mathrm{b}}}{\sqrt{\mathrm{a}+2\mathrm{b}}-\sqrt{\mathrm{a}-2\mathrm{b}}}$$$$\Rightarrow \frac{\mathrm{x}+1}{\mathrm{x}-1}=\frac{2\sqrt{\mathrm{a}+2\mathrm{b}}}{2\sqrt{\mathrm{a}-2\mathrm{b}}}\left(\mathrm{by}\mathrm{componendo}\mathrm{\&}\mathrm{dividendo}\right)$$$$\Rightarrow \frac{{\mathrm{x}}^2+2\mathrm{x}+1}{{\mathrm{x}}^2-2\mathrm{x}+1}=\frac{\mathrm{a}+2\mathrm{b}}{\mathrm{a}-2\mathrm{b}}$$$$\Rightarrow \frac{2({\mathrm{x}}^2+1)}{4\mathrm{x}}=\frac{2\mathrm{a}}{4\mathrm{b}}\left(\mathrm{by}\mathrm{componendo}\mathrm{\&}\mathrm{dividendo}\right)$$$$\Rightarrow \frac{{\mathrm{x}}^2+1}{\mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}}$$$$\Rightarrow {\mathrm{bx}}^2+\mathrm{b}=\mathrm{ax}$$$$\Rightarrow {\mathrm{bx}}^2-\mathrm{ax}+\mathrm{b}=0★$$

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

$$\mathrm{Ok}.\mathrm{Thanks}.$$

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