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

Answer:  (b)  $-{\displaystyle \frac{b}{\sqrt{ac}}}$ .

Solution:

Step 1: Simplify the given expression by using laws of exponent

$\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\frac{\sqrt{\alpha }}{\sqrt{\beta }}+\frac{\sqrt{\beta }}{\sqrt{\alpha }}$          [∵ $\sqrt{\frac{\mathit{a}}{\mathit{b}}}\mathbf{=}\frac{\sqrt{\mathit{a}}}{\sqrt{\mathit{b}}}$]

$\Rightarrow \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\frac{(\sqrt{\alpha }{)}^2+(\sqrt{\beta }{)}^2}{\sqrt{\alpha \beta }}$

$\Rightarrow \sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\frac{\alpha +\beta }{\sqrt{\alpha \beta }}$    ....(1)

Step 2: Use the concept of sum and product of zeroes of a polynomial

If $\alpha$ and $\beta$ are the roots of the polynomial $f(x)=a{x}^2+bx+c$, then

$\alpha +\beta =-\frac{b}{a}$ and $\alpha \times \beta =\frac{c}{a}$

Substituting the value of $\alpha$ and $\beta$  in R.H.S of (1), we get

$\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=\frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}$      ...(2)

Step 3: Simplify the R.H.S of (2) to get the result

${\displaystyle \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}}={\displaystyle \frac{-\frac{b}{a}}{\frac{\sqrt{c}}{\sqrt{a}}}}$

$={\displaystyle \frac{-b\times \sqrt{a}}{a\times \sqrt{c}}}$

$={\displaystyle \frac{-b}{a^{1-\frac{1}{2}}\times \sqrt{c}}}$

$={\displaystyle \frac{-b}{a^{\frac{1}{2}}\times \sqrt{c}}}$

$={\displaystyle \frac{-b}{\sqrt{a}\times \sqrt{c}}}$

$={\displaystyle \frac{-b}{\sqrt{ac}}}$

$\therefore {\displaystyle \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}}}=-{\displaystyle \frac{b}{\sqrt{ac}}}$

So, $\sqrt{\frac{\alpha }{\beta }}+\sqrt{\frac{\beta }{\alpha }}=-{\displaystyle \frac{b}{\sqrt{ac}}}$

Hence, the correct answer is option (b)  $-{\displaystyle \frac{b}{\sqrt{ac}}}$ .

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