Question

If $\alpha, \beta$ are the roots of the quadratic equation $x^2+ax+b=0, (b\ne 0)$; then the quadratic equation whose roots are $\alpha -\frac{1}{\beta}, \beta - \frac{1}{\alpha}$ is

• A. $ax^2+a(b-1)x+(a-1)^2=0$
• B. $bx^2+a(b-1)x+(b-1)^2=0$
• C. $x^2+ax+b = 0$
• D. $abx^2+bx+a = 0$

Answer 1

The Correct Option is B

The correct option is(B): bx 2+a(b−1)x+(b−1)2=0.

Given equation is, \(x^{2}+a x+b=0,(b \neq 0)\)
its roots are \(\alpha\) and \(\beta\). 
Then, sum of roots \(=\alpha+\beta=-a\) ....(i) 
Product of roots \(=\alpha \cdot \beta=b\) .....(ii) 
Now,
\(\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \beta}\right)\)
\(=-a-\frac{(-a)}{b}\) [from Eqs.(i) and (ii)]
\(=-a+\frac{a}{b}=\frac{a}{b}(1-b)\)
and \(\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}\)
\(=b+\frac{1}{b}-2[\text { from E (ii) }]\) ....(iv) 
\(=\frac{1}{b}\left(b^{2}-2 b+1\right)=\frac{1}{b}(b-1)^{2}\)
\(\therefore\) Required of quadratic equation whose roots are \(\left(\alpha-\frac{1}{\beta}\right)\) and \(\left(\beta-\frac{1}{\alpha}\right)\) is
\(x^{2}-\left\{\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right\} x\)
\(+\left\{\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)\right\}=0\)
On putting the values from Eqs. (i) and (ii), we get
\(x^{2}-\frac{a}{b}(1-b) x+\frac{1}{b}(b-1)^{2}=0\)
\(\Rightarrow \quad b x^{2}+a(b-1) x+(b-1)^{2}=0, b \neq 0\)