Question

Find the nature of the roots of the following quadratic equation. If the real roots exist, find them;
$2x^2-6x+3=0$

Answer 1

$2x^2-6x+3=0$
Comparing this equation with $a{x}^2+bx+c=0$, we get
$a=2,b=-6,c=3$
$Discriminant=b^2-4ac$
$=(-6{)}^2-4(2\left)\right(3)$
$=36-24=12$
$As{b}^2-4ac>0$,
Distinct real roots exist for this equation.
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(2\left)\right(3)}}{2\left(2\right)}=\frac{6\pm \sqrt{12}}{4}=\frac{6\pm 2\sqrt{3}}{4}=\frac{3\pm \sqrt{3}}{2}Therefore,therootsare\frac{3+\sqrt{3}}{2}and\frac{3-\sqrt{3}}{2}$