Question

For what positive values of '$m$' roots of given equation is equal, distinct, imaginary $m{k}^2-3k+1=0$

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

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

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

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

(i) If the roots are equal then $b^2-4ac=0$, therefore,

$b^2-4ac=0\Rightarrow (-3{)}^2-(4\times m\times 1)=0\Rightarrow 9-4m=0\Rightarrow -4m=-9\Rightarrow 4m=9\Rightarrow m=\frac{9}{4}$

(ii) If the roots are distinct then $b^2-4ac>0$, therefore,

$b^2-4ac>0\Rightarrow (-3{)}^2-(4\times m\times 1)>0\Rightarrow 9-4m>0\Rightarrow -4m>-9\Rightarrow 4m<9\Rightarrow m<\frac{9}{4}$

(iii) If the roots are imaginary then $b^2-4ac<0$, therefore,

$b^2-4ac<0\Rightarrow (-3{)}^2-(4\times m\times 1)<0\Rightarrow 9-4m<0\Rightarrow -4m<-9\Rightarrow 4m>9\Rightarrow m>\frac{9}{4}$

Hence $m=\frac{9}{4}$ if the roots are equal, $m<\frac{9}{4}$ if the roots are distinct and $m>\frac{9}{4}$ if the roots are imaginary.