If α, β are the zeroes of the polynomial \(f(x)=ax^2+bx+c\), then \(\sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}=?\)

Answer:  (b)  \(-\dfrac{b}{\sqrt{ac} }\) . Solution: Step 1: Simplify the given expression by using laws of exponent \(\sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}= \frac{\sqrt{α}}{\sqrt{β}}+\frac{\sqrt{β}}{\sqrt{α}}\)          [∵ \(\boldsymbol{\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}}\)] \(\Rightarrow \sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}= \frac{(\sqrt{α})^2+(\sqrt{β})^2}{\sqrt{αβ}}\) \(\Rightarrow \sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}= \frac{α+β}{\sqrt{αβ}}\)    ....(1) Step 2: Use the concept of sum and product of zeroes of a polynomial If \(α\) and \(β\) are the roots of the polynomial \(f(x)=ax^2+bx+c\), then \( \alpha + \beta = -\frac{b}{a}\) and \( \alpha \times \beta = \frac{c}{a}\) Substituting the value of \(α\) and \(β\)  in R.H.S of (1), we get \( \sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}= \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}\)      ...(2) Step 3: Simplify the R.H.S of (2) to get the result \(\dfrac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}=\dfrac{-\frac{b}{a}}{\frac{\sqrt{c}}{\sqrt{a}}}\) \(=\dfrac{-b\times \sqrt{a}}{a \times \sqrt{c}}\) \(=\dfrac{-b}{a^{1-\frac{1}{2}} \times \sqrt{c}}\) \(=\dfrac{-b}{a^{\frac{1}{2}} \times \sqrt{c}}\) \(=\dfrac{-b}{\sqrt{a} \times \sqrt{c}}\) \(=\dfrac{-b}{\sqrt{ac} }\) \(\therefore\dfrac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}=-\dfrac{b}{\sqrt{ac} }\) So, \(\sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}=-\dfrac{b}{\sqrt{ac} }\) Hence, the correct answer is option (b)  \(-\dfrac{b}{\sqrt{ac} }\) .  

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