Find the roots of the quadratic equations given in

Solution:

(i) $2 x^{2}-7 x+3=0$

$a=2, b=-7, c=3$

Discriminant $D=(-7)^{2}-4(2)(3)=25$

$\Rightarrow \sqrt{10}=\sqrt{25}=5$

Roots of the given quadratic equation are

$\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{\mathbf{2 a}}$

i.e., $\frac{\mathbf{7} \pm \mathbf{5}}{\mathbf{2} \times \mathbf{2}}$, i.e., the roots are 3 and $\frac{\mathbf{1}}{\mathbf{2}}$.

(ii) $2 x^{2}+x-4=0$

$a=2, b=1, c=-4$

$D=(1)^{2}-4(2)(-4)=33 \Rightarrow \sqrt{\mathbf{D}}=\sqrt{\mathbf{3 3}}$

Roots are given by $x=\frac{-\mathbf{b} \pm \sqrt{\mathbf{1}}}{2 \mathbf{a}}=\frac{-\mathbf{1} \pm \sqrt{\mathbf{3 3}}}{4}$

Hence, the two roots of the quadratic equation are $\frac{-\mathbf{1} \pm \sqrt{\mathbf{3 3}}}{\mathbf{4}}$

(iii) $4 x^{2}+4 \sqrt{3} x+3=0$

On comparing this equation with $a x^{2}+b x+c=0$, we obtain $a=4, b=4 \sqrt{\mathbf{3}}, c=3$ By using quadratic formula, we obtain

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\Rightarrow x=\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}$

$\Rightarrow x=\frac{-4 \sqrt{3} \pm 0}{8}$

$\therefore \quad x=\frac{-\sqrt{3}}{2}$ or $\frac{-\sqrt{3}}{2}$

(iv) $\quad 2 x^{2}+x+4=0$

On comparing this equation with

$a x^{2}+b x+c=0$

we obtain $\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=4$

By using quadratic formula, we obtain

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \Rightarrow x=\frac{-1 \pm \sqrt{1-32}}{4}$

$\Rightarrow x=\frac{-1 \pm \sqrt{-31}}{4}$

However, the square of a number cannot be negative. Therefore, there is no real root for the given equation.