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Question Number 206477 by hardmath last updated on 15/Apr/24

$$\mathrm{If}\mathrm{a}>\mathrm{b}>0\mathrm{and}{4\mathrm{a}}^2+{\mathrm{b}}^2=4\mathrm{ab}$$$$\mathrm{Find}:\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}=?$$

Answered by MATHEMATICSAM last updated on 16/Apr/24

$$4a^2+b^2=4ab$$$$\Rightarrow 4a^2-4ab+b^2=0$$$$\Rightarrow {(2a-b)}^2=0$$$$\Rightarrow 2a-b=0$$$$\Rightarrow 2a=b$$$$\Rightarrow \frac{a}{b}=\frac{1}{2}$$$$\Rightarrow \frac{a+b}{a-b}=\frac{1+2}{1-2}=-3[\mathrm{Comp}.\mathrm{and}\mathrm{Div}.]$$$$\Rightarrow \frac{a-b}{a+b}=-\frac{1}{3}\left(\mathrm{Ans}\right)$$

Answered by A5T last updated on 15/Apr/24

$$4a^2-4ab+b^2=0$$$$\Rightarrow a=\frac{4b\underset{-}{+}\sqrt{16b^2-16b^2}}{8}=\frac{b}{2}$$$$\Rightarrow \frac{a-b}{a+b}=\frac{-\frac{b}{2}}{\frac{3b}{2}}=\frac{-1}{3}$$

Answered by Rasheed.Sindhi last updated on 15/Apr/24

$${4\mathrm{a}}^2-4\mathrm{ab}+{\mathrm{b}}^2=0$$$${(2\mathrm{a}-\mathrm{b})}^2=0$$$$\mathrm{b}=2\mathrm{a}$$$$\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{a}-2\mathrm{a}}{\mathrm{a}+2\mathrm{a}}=\frac{-\mathrm{a}}{3\mathrm{a}}=-\frac{1}{3}$$

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