Question

Discuss the nature of roots of the given equation: $3d^2-2d+1=0$

Answer 1

We know that while finding the root of a quadratic equation $a{x}^2+bx+c=0$ by quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

if $b^2-4ac>0$, then the roots are real and distinct and

if $b^2-4ac<0$, then the roots are imaginary.

Here, the given quadratic equation $3d^2-2d+1=0$ is in the form $a{x}^2+bx+c=0$ where $a=3,b=-2$ and $c=1$, therefore,

$b^2-4ac=(-2{)}^2-(4\times 3\times 1)=4-12=-8<0$

Since $b^2-4ac<0$

Hence the roots are imaginary or no real roots.