Question

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2x^2-3x+5=0$
(ii) $3x^2-4\sqrt{3}x+4=0$
(iii) $2x^2-6x+3=0$

Answer 1

i)

$x^2-3x+5=0a=2,b=-3,c=5$

Discriminant, $D=b^2-4ac=9-40=-31$

Since $D<0,$ the roots are imaginary for the given equation.

ii)

$3x^2-4\sqrt{3}x+4=0a=3,b=-4\sqrt{3},c=4$

Discriminant, $D$$=b^2-4ac=48-48=0$

Since $D=0,$ the roots are real and equal for the given equation.

${3x}^2-4\sqrt{3}x+4=0a=3,b=-4\sqrt{3},c=4x=\frac{[-b\pm \sqrt{b^2-4ac}]}{2a}x=\frac{4\sqrt{3}}{6}=\frac{2}{\sqrt{3}}$

iii)

$2x^2-6x+3=0a=2,b=-6,c=3$

Discriminant, $D=$ $b^2-4ac=36-24=12$

Since $D>0,$ roots are real but not equal.

$x=\frac{[-b\pm \sqrt{b^2-4ac}]}{2a}x=\frac{6\pm \sqrt{12}}{4}x=\frac{6\pm 2\sqrt{3}}{4}x=\frac{3\pm \sqrt{3}}{2}$