Question

If the discriminant of $3x^2+2x+a=0$ is double the discriminant of $x^2-4x+2=0$ then find ${}^{\prime }{a}^{\prime }$.

Answer 1

Given,

The discriminant of $3x^2+2x+a=0$ is double the discriminant of $x^2-4x+2=0$

We know that the discriminant of a quadratic equation $a{x}^2+bx+c=0$ is given by: $b^2-4ac$

Hence,

$(2{)}^2-4\times 3\times a=2\times ((-4{)}^2-4\times 1\times 2)$

$4-12a=2\times (16-8)$

$4-12a=16$

$12a=4-16$

$12a=-12$

$a=-1$