Question

Let $\alpha ,\beta$ are the roots of the quadratic equation $2x^2-mx-8=0$. If $m_1$ and $m_2$ are the two values of $m$ for which $|\alpha -\beta |=m-1$ then $(m_1+m_2)$ is equal to

• A. $3$
• B. $\frac{8}{3}$
• C. $-\frac{3}{8}$
• D. $-8$

Answer 1

The correct option is B $\frac{8}{3}$

${2x}^2-mx-8=0$

Let $\alpha ,\beta$ be the roots of the above equation.

So, $\alpha +\beta =m/2,\alpha \beta =-4$

${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\beta \alpha m^2/4={\alpha }^2+{\beta }^2-8{\alpha }^2+{\beta }^2=m^2/4+8$

${(\alpha -\beta )}^2={\alpha }^2+{\beta }^2-2\beta \alpha {(\alpha -\beta )}^2=m^2/4+8+8=m^2/4+16\alpha -\beta =\sqrt{\frac{m^2+64}{4}}=\frac{\sqrt{m^2+64}}{2}$

Given |α−β |=m−1.

So, $\frac{\sqrt{m^2+64}}{2}=m-1\sqrt{m^2+64}=2m-2$

Squaring on both sides,

$m^2+64={4m}^2+4-8m$

${3m}^2-8m-60=0$

${3m}^2-18m+10m-60=03m(m-6)+10(m-6)=0(3m+10)(m-6)=0m=6,-10/3$

So, m is 6, -10/3.

The sum of both values of m are 8/3.

Hence option B.