 Q1. Find the roots of the following quadratic equations if they exist by the method of completing square.
(i) $$2x^2 - 7x + 3 = 0$$
(ii) $$2x^2 + x - 4 = 0$$
(iii) $$4x^2 + 4\sqrt{3}x + 3 = 0$$
(iv) $$2x^2 + x + 4 = 0$$

(i) $$2x^2 - 7x + 3 = 0$$
First step is to make coefficient of $$x^2 = 1$$
So on dividing equation by 2,
$$x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0$$
On dividing the middle term of the equation by 2x, we get the expression :
$$({{7}\over{2}}x)({{1}\over{2x}}) = {{7}\over{4}}$$
On adding and subtracting square of $${{7}\over{4}}$$ from the equation $$x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0$$
$$x^2 - {{7}\over{2}}x + {{3}\over{2}} + ({{7}\over{4}})^2 - ({{7}\over{4}})^2 = 0$$
$$x^2 + ({{7}\over{4}})^2 - {{7}\over{2}}x + {{3}\over{2}} - ({{7}\over{4}})^2 = 0$$
Using identity {$$(a-b)^2 = a^2 + b^2 -2ab$$}
$$(x - {{7}\over{4}})^2 + {{24 - 49}\over{16}} = 0$$
$$(x - {{7}\over{4}})^2 = {{49 - 24}\over{16}}$$
$$(x - {{7}\over{4}})^2 = {{25}\over{16}}$$
$$(x - {{7}\over{4}})^2 = ({{\pm5}\over{4}})^2$$
On taking square root on both sides:
$$x - {{7}\over{4}} = {{5}\over{4}}$$ and $${{-5}\over{4}}$$
$$x = {{5}\over{4}} + {{7}\over{4}} = {{12}\over{4}} = 3$$ and $${{-5}\over{4}} - {{7}\over{4}} = {{2}\over{4}} = {{1}\over{2}}$$
Hence, $$x = {{1} \over {2}} , 3$$

(ii) $$2x^2 + x - 4 = 0$$
First step is to make coefficient of $$x^2 = 1$$
So on dividing equation by 2
$$x^2 + {{x}\over{2}} - 2 = 0$$
On dividing the middle term of the equation by 2x, we get the expression :
$$({{x}\over{2}})({{1}\over{2x}}) = {{1}\over{4}}$$
On adding and subtracting square of $${{1}\over{4}}$$ from the equation $$x^2 + {{x}\over{2}} - 2 = 0$$
$$x^2 + {{x}\over{2}} - 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0$$
$$x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} - 2 - ({{1}\over{16}}) = 0$$
Using identity {$$(a+b)^2 = a^2 + b^2 + 2ab$$}
$$(x + {{1}\over{4}})^2 - {{33}\over{16}} = 0$$
$$(x + {{1}\over{4}})^2 = {{33}\over{16}}$$
On taking square root both sides:
$$x + {{1}\over{4}} = {{\sqrt{33}}\over{4}}$$ and $${{-\sqrt{33}}\over{4}}$$
$$x = {{\sqrt{33}}\over{4}} - {{1}\over{4}} = {{\sqrt{33} - 1}\over{4}}$$ and $${{-\sqrt{33}}\over{4}} - {{1}\over{4}} = {{2}\over{4}} = {{-\sqrt{33} - 1}\over{4}}$$
Hence, $$x = {{\sqrt{33} - 1}\over{4}} , {{-\sqrt{33} - 1}\over{4}}$$

(iii) $$4x^2 + 4\sqrt{3}x + 3 = 0$$
First step is to make coefficient of $$x^2 = 1$$
So on dividing equation by 4
$$x^2 + \sqrt{3}x + {{3} \over {4}} = 0$$
On dividing the middle term of the equation by 2x, we get the expression :
$$(\sqrt{3}x)({{1}\over{2x}})$$ = $${{\sqrt{3}}\over{2}}$$
On adding and subtracting square of $${{\sqrt{3}}\over{2}}$$ (half of the middle term) from the equation $$x^2 + \sqrt{3}x + {{3} \over {4}} = 0$$
$$x^2 + \sqrt{3}x + {{3} \over {4}} + ({{\sqrt{3}}\over{2}})^2 - ({{\sqrt{3}}\over{2}})^2 = 0$$
$$x^2 + ({{3}\over{4}})^2 - \sqrt{3}x + {{3} \over {4}} - ({{3}\over{4}}) = 0$$
Using identity {$$(a+b)^2 = a^2 + b^2 + 2ab$$}
$$(x + {{\sqrt{3}}\over{2}})^2 = 0$$
$$(x + {{\sqrt{3}}\over{2}}) (x + {{\sqrt{3}}\over{2}}) = 0$$
On taking square root both sides:
$$x + {{\sqrt{3}}\over{2}} = 0$$ and $$x + {{\sqrt{3}}\over{2}} = 0$$
$$x = -{{\sqrt{3}}\over{2}}$$ and $$x = -{{\sqrt{3}}\over{2}}$$

(iv) $$2x^2 + x + 4 = 0$$
First step is to make coefficient of $$x^2 = 1$$
So on dividing equation by 2
$$x^2 + {{x}\over{2}} + 2 = 0$$
On dividing the middle term of the equation by 2x, we get the expression :
$$({{x}\over{2}})({{1}\over{2x}}) = {{1}\over{4}}$$
On adding and subtracting square of $${{1}\over{4}}$$ from the equation $$x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0$$
$$x^2 + {{x}\over{2}}x + 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0$$
$$x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} + 2 - ({{1}\over{4}})^2 = 0$$
Using identity {$$(a + b)^2 = a^2 + b^2 -2ab$$}
$$(x + {{1}\over{4}})^2 + 2 - {{1}\over{16}} = 0$$
$$(x + {{1}\over{4}})^2 = {{1}\over{16}} - 2 = {{1 - 32}\over{16}}$$
$$(x + {{1}\over{4}})^2 = \frac{-31}{16}$$
On taking square root on both sides,
Right hand side does not exist since square root of negative number does not exist.
Therefore, there is no solution.