Question

Two real numbers $\alpha \&\beta$ are such that $\alpha +\beta =3,\alpha -\beta =4$, then $\alpha \&\beta$ are the roots of the quadratic equation:

• A. $4x^2-12x-7=0$
• B. $4x^2-12x+7=0$
• C. $4x^2-12x+25=0$
• D. none of these

Answer 1

The correct option is A $4x^2-12x-7=0$

Given that,

$\alpha +\beta =3$

$\alpha -\beta =4$

By solving, we get

$\alpha =\frac{7}{2}$ and $\beta =\frac{-1}{2}$

The quadratic equation whose roots are $\alpha =\frac{7}{2}$ and $\beta =\frac{-1}{2}$ is

$x^2-(\alpha +\beta )x+\alpha \beta =0$

$⟹x^2-(\frac{7}{2}+\frac{-1}{2})x+\left(\frac{7}{2}\right)\left(\frac{-1}{2}\right)=0$

$⟹x^2-3x-\frac{7}{4}=0$

$\therefore 4x^2-12x-7=0$

Option A is correct.