Find the roots of the following quadratic equations,

Solution:

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

$\Rightarrow 4 x^{2}-14 x+6=0$

$\Rightarrow(2 x)^{2}-7(2 x)+6=0$

$\Rightarrow y^{2}-7 y+6=0$ where $y=2 x$

$\Rightarrow\left\{\mathbf{y}^{2}-\frac{7}{2} \mathbf{y}-\frac{7}{2} \mathbf{y}+\left(-\frac{7}{2}\right)^{2}\right\}-\left(-\frac{7}{2}\right)^{2}+6=0$

$\Rightarrow\left\{y-\frac{7}{2}\right\}^{2}-\frac{49}{4}+6=0 \Rightarrow\left\{y-\frac{7}{2}\right\}^{2}-\frac{25}{4}=0$

$\Rightarrow\left(2 x-\frac{7}{2}\right)^{2}=\frac{25}{4} \Rightarrow 2 x-\frac{7}{2}=\pm \sqrt{\frac{25}{4}}$

$\Rightarrow 2 x-\frac{7}{2}=\pm \frac{5}{2}$

$\Rightarrow 2 x-\frac{7}{2}=\frac{5}{2}$ or $\quad 2 x-\frac{7}{2}=-\frac{5}{2}$

$\Rightarrow 2 x=\frac{\mathbf{5}}{\mathbf{2}}+\frac{\mathbf{7}}{\mathbf{2}}$ or $\quad 2 x=-\frac{\mathbf{5}}{\mathbf{2}}+\frac{\mathbf{7}}{\mathbf{2}}$

$\Rightarrow 2 x=6 \quad$ or $\quad 2 x=1$

Hence, the roots of the given quadratic

equation are $\frac{\mathbf{1}}{\mathbf{2}}$ and 3 .

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

$\Rightarrow 2 x^{2}+x=4$

On dividing both sides of the equation by 2 ,

we obtain $\Rightarrow \quad x^{2}+\frac{1}{2} x=2$

On adding $\left(\frac{\mathbf{1}}{\mathbf{4}}\right)^{\mathbf{2}}$ to both sides of the equation, we obtain

$\Rightarrow(x)^{2}+2 \times x \times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}=2+\left(\frac{1}{4}\right)^{2}$

$\Rightarrow\left(x+\frac{1}{4}\right)^{2}=\frac{33}{16}$

$\Rightarrow x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x=\frac{\pm \sqrt{33}}{4}-\frac{1}{4} \Rightarrow x=\frac{\sqrt{33}-1}{4}$ or $\frac{-\sqrt{33}-1}{4}$

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

$\Rightarrow(2 x)^{2}+2 \times 2 x \times \sqrt{3}+(\sqrt{3})^{2}=0$

$\Rightarrow(\mathbf{2} \mathbf{x}+\sqrt{\mathbf{3}})^{\mathbf{2}}=\mathbf{0}$

$\Rightarrow(\mathbf{2 x}+\sqrt{\mathbf{3}})=0$ and $(\mathbf{2 x}+\sqrt{\mathbf{3}})=0$

$\Rightarrow x=\frac{-\sqrt{3}}{2}$ and $x=\frac{-\sqrt{3}}{2}$

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

$\Rightarrow 2 x^{2}+x=-4$

On dividing both sides of the equation by 2, we obtain

$\Rightarrow x^{2}+\frac{1}{2} x=-2$

$\Rightarrow x^{2}+2 \times x \times \frac{1}{4}=-2$

On adding $\left(\frac{\mathbf{1}}{\mathbf{4}}\right)^{2}$ to both sides of the equation, we obtain

$\Rightarrow(\mathrm{x})^{2}+2 \times \mathrm{x} \times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}=\left(\frac{1}{4}\right)^{2}-2$

$\Rightarrow\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}-2$

$\Rightarrow\left(x+\frac{1}{4}\right)^{2}=-\frac{31}{16}$

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