Question

If $\alpha ,\beta$ are the roots of the equation $5x^2+3x-2=0,$ then the equation whose roots are $7\alpha ,7\beta$ is $a{x}^2+bx+c=0.$ Then the value of $\frac{-c-3b}{a}$ is

• A. 7

Answer 1

The correct option is A 7
Given : $5x^2+3x-2=0$ with $\alpha ,\beta$ as roots.

$\text{Sum of roots}=\alpha +\beta =\frac{-3}{5}\cdots \left(i\right)$

$\text{Product of roots}=\alpha .\beta =\frac{-2}{5}\cdots \left(ii\right)$

Now we have to find the equation whose roots are $7\alpha ,7\beta$
$\therefore \text{Sum of roots}=7\alpha +7\beta =\frac{-b}{a}$
$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=7(\alpha +\beta )$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=7(\frac{-3}{5})\left[\text{From}\right(i\left)\right]$

$\Rightarrow \text{Sum of roots}=\frac{-b}{a}=(\frac{-21}{5})$

Now, $\text{Product of roots}=7\alpha .7\beta =\frac{c}{a}$
$\Rightarrow \text{Productof roots}=\frac{c}{a}=49(\alpha .\beta )$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=49(\frac{-2}{5})\left[\text{From}\right(ii\left)\right]$

$\Rightarrow \text{Productof roots}=\frac{c}{a}=(\frac{-98}{5})$

$\therefore \frac{-c-3b}{a}=-(\frac{c}{a})+3(\frac{-b}{a})$

$\Rightarrow \frac{-c-3b}{a}=-(\frac{-98}{5})+3(\frac{-21}{5})$

$\Rightarrow \frac{-c-3b}{a}=\frac{35}{5}=7$