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) $3x^2 - 4\sqrt3x + 4 = 0$
(iii) $2x^2-6x + 3 = 0$

To do:

We have to find the nature of the roots of the given quadratic equations and find them.

Solution:

(i) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2, b=-3$ and $c=5$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

Therefore,

$D=(-3)^2-4(2)(5)=9-40=-31$.

As $D<0$, the given quadratic equation has no real roots.

(ii) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=3, b=-4\sqrt3$ and $c=4$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

Therefore,

$D=(-4\sqrt3)^2-4(3)(4)=16(3)-12(4)$

$=48-48$

$=0$

As $D=0$, the given quadratic equation has real and equal roots.

 $x=\frac{-b \pm \sqrt{D}}{2 a}$

$=\frac{-(-4 \sqrt{3}) \pm 0}{2 \times 3}$

$=\frac{4 \sqrt{3}}{6}$

Hence, the roots of the given quadratic equation are $\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}$

(iii) Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2, b=-6$ and $c=3$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

Therefore,

$D=(-6)^2-4(2)(3)=36-24=12>0$.

As $D>0$, the given quadratic equation has real and distinct roots.

 $x=\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}$

$=\frac{-(-6) \pm \sqrt{12}}{2 \times 2}$

$=\frac{6 \pm \sqrt{4 \times 3}}{2 \times 2}$

$=\frac{6 \pm 2 \sqrt{3}}{4}$

$=\frac{2(3 \pm \sqrt{3})}{4}$

$=\frac{3 \pm \sqrt{3}}{2}$

Hence, the roots of the given quadratic equation are $\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}$.