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. $a{x}^2+a(b-1)x+{(a-1)}^2=0$
• B. $b{x}^2+a(b-1)x+{(b-1)}^2=0$
• C. $x^2+ax+bv=0$
• D. $ab{x}^2+bx+a=0$

Answer 1

Answer: (B)

$b{x}^2+a(b-1)x+{(b-1)}^2=0$

Given: $x^2+ax+b=0$

Then, sum of roots $=\alpha +\beta =-a......\left(i\right)$

Product of roots $=\alpha .\beta =b........\left(ii\right)$

Now,
$(\alpha -\frac{1}{\beta })+(\beta -\frac{1}{\alpha })=(\alpha +\beta )-\left(\frac{\alpha +\beta }{\alpha \beta }\right)$

$=-a-\frac{(-a)}{b}$ [from eqs. $\left(i\right)$ and $\left(ii\right)$]

$=-a+\frac{a}{b}=\frac{a}{b}(1-b)...\left(iii\right)$
and $(\alpha +\beta )-\left(\frac{\alpha +\beta }{\alpha \beta }\right)=\alpha \beta -1-1+\frac{1}{\alpha \beta }$

$=b+\frac{1}{b}-\mathrm{2.....}\left(iv\right)$ from eq $\left(ii\right)$

$=\frac{1}{b}(b^2-2b+1)=\frac{1}{b}{(b-1)}^2$

$\therefore$ Required quadratic equation whose roots are $(\alpha -\frac{1}{\beta })$ and $(\beta -\frac{1}{\alpha })$ is

$x^2-\{(\alpha -\frac{1}{\beta })+(\beta -\frac{1}{\alpha })\}x+\{(\alpha -\frac{1}{\beta })+(\beta -\frac{1}{\alpha })\}=0$

On putting the values from eqs. $\left(i\right)$ and $\left(ii\right)$ we get

$x^2-\frac{a}{b}(1-b)x+\frac{1}{b}{(b-1)}^2=0$

$\Rightarrow b{x}^2+a(b-1)x+{(b-1)}^2=0$