Question

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

• A. $4x^2-12x-7=0$
• B. $4x^2-12x+7=0$
• C. $4x^2-12x-25=0$
• D. $Noneofthese$

Answer 1

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

$\alpha +\beta =3$

$\alpha -\beta =4$

Adding both

$2\alpha =7$

$\alpha =\frac{7}{2}$

Substituting $\alpha$

$\beta =-\frac{1}{2}$

$\alpha \beta =-\frac{7}{4}$

So the quadratic equation is

$x^2-3x-\frac{7}{4}=04x^2-12x-7=0$