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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) \(2x^2 + x + 4 = 0\)

Answer :

(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.