Question

The roots $\alpha$ and $\beta$ of the quadratic equation $x^2-5x+3(k-1)=0$ are such that $\alpha -\beta =11$ Find the value of $k$.

Answer 1

Given $x^2-5x+3(k-1)=0$ and its roots are $\alpha ,\beta$ such that

$\alpha -\beta =11$ ...........(i)
and also sum of roots $\alpha +\beta =-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$

using conditions we have $\alpha +\beta =(-)\frac{-5}{1}$
$\Rightarrow \alpha +\beta =5$ .........(ii)
Similar way,
product of roots are$\alpha \times \beta =\frac{\text{constant}}{\text{coefficient of}{x}^2}$

$\alpha \beta =3(k-1)$ ........(iii)

Since, $\alpha -\beta =\sqrt{{\alpha }^2+{\beta }^2-2\alpha \beta }=\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$

$\Rightarrow (11{)}^2=(\alpha +\beta {)}^2-4\alpha \beta$

$\Rightarrow 121-25=-12(k-1)$
$\Rightarrow 96=-12(k-1)$
$\Rightarrow \frac{96}{-12}=k-1$
$\Rightarrow -8=k-1$
$\Rightarrow k=-8+1$

$\therefore k=-7$

Hence, the value of $k$ is $-7$.