Question

Quadratic equation. Find the value of m so that the roots of the equation (4 - m) x^2 + (2m + 4)x + (8m + 1) = 0 may be equal.

Answer 1

The quadratic has one root with multiplicity 2 if the discriminant is 0, which is

$(2m+4)^2-4(4-m)(8m+1)$

(That is, for a quadratic $ax^2+bx+c$, the discriminant is $b^2-4ac$.)

Set the discriminant equal to 0 and solve for $m$:

$(2m+4)^2-4(4-m)(8m+1)=0$

$4m^2+16m+16+32m^2-124m-16=0$

$36m^2-108=0$

$36m(m-3)=0$

$\implies m=0\text{ or }m=3$