If α, β are the zeroes of the polynomial \(f(x)=ax^2+bx+c\), then \(\sqrt{\frac{α}{β}}+\sqrt{\frac{β}{α}}=?\)
(a) \(b\)
(b) \(\frac{-b}{\sqrt{ac}}\)
(c) \(-\frac{\sqrt{b}}{ac}\)
(d) \(\frac{1}{ac}\)
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) \(-\dfrac{b}{\sqrt{ac} }\) .
