Question

If $m,n$ are the roots of the equation $4x^2-3x-1=0,$ then the equation whose roots are $\frac{6}{m},\frac{6}{n}$ is

• A. $x^2+18x+144=0$
• B. $x^2+18x-144=0$
• C. $x^2-18x+144=0$
• D. $x^2-18x-144=0$

Answer 1

The correct option is B $x^2+18x-144=0$

Given: $4x^2-3x-1=0$ roots are $m,n$

$\text{Sum of roots}=m+n=\frac{3}{4}\cdots \left(i\right)$

$\text{Product of roots}=m.n=\frac{-1}{4}\cdots \left(ii\right)$

Now we have to find the equation whose roots are $\frac{6}{m},\frac{6}{n}$

$\therefore \text{Sum of roots}=\frac{6}{m}+\frac{6}{n}=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{6(m+n)}{m.n}=\frac{-b}{a}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=\frac{6\times \frac{3}{4}}{\frac{-1}{4}}$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=-18\cdots \left(iii\right)$

Now, $\text{Product of roots}=\frac{6}{m}.\frac{6}{n}=\frac{c}{a}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=\frac{36}{m.n}$

$\Rightarrow \text{Product of roots}=\frac{c}{a}=\frac{36}{(\frac{-1}{4})}$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=-144\cdots \left(iv\right)$

Now we have the sum & products of roots for the required equation.
And we can write the equation as: $x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$

$\Rightarrow x^2-(-18)x+(-144)=0$

$\Rightarrow x^2+18x-144=0$