Ex.4.3 Q1 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

Find the roots of the following quadratic equations, if they exist, by the method of completing the square.

(i) \(2x^2-7x+3=0\)

(ii) \(2x^2+x-4=0\)

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

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

 Video Solution
Quadratic Equations
Ex 4.3 | Question 1

Text Solution

What is the unknown?

Roots of the quadratic equation.

Reasoning:

Steps required to solve a quadratic equation by applying the ‘completing the square’ method are given below:

Let the given quadratic equation be: \(ax{}^\text{2}+bx+c=0\)

(i) Divide all the terms by \(a\)

\[\begin{align}\frac{{a{x^2}}}{a} + \frac{{bx}}{a} + \frac{c}{a} &= 0\\{x^2} + \frac{{bx}}{a} + c &= 0\end{align}\]

(ii) Move the constant term \(\begin{align}\frac{c}{a}\end{align}\) to the right side of the equation:

\[{x^2} + \frac{b}{a}x = - \frac{c}{a}\]

(iii) Complete the square on the left side of the equation by adding \(\begin{align} \frac{{{b^2}}}{{4{a^2}}}.\end{align}\) Balance this by adding the same value to the right side of the equation.

Steps:

(i) \(2x^2-7x+3=0\)

Divide \(2x{}^\text{2}-7x+3=0\) by \(2:\)

\[\begin{align}{x^2} - \frac{7}{2}x + \frac{3}{2}& = 0\\{x^2} - \frac{7}{2}x &= - \frac{3}{2}\\\end{align}\]

Since \(\begin{align}\frac{7}{2} \div 2 = \frac{7}{4} \end{align}\), \(\begin{align}{\left( {\frac{7}{4}} \right)^2} \end{align}\) should be added to both sides of the equation:

\[\begin{align}{x^2} - \frac{7}{2}x + {\left( {\frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + {\left( {\frac{7}{4}} \right)^2}\\{\left( {x - \frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + \frac{{49}}{{16}}\\ &= \frac{{ - 24 + 49}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} & = \frac{{25}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} &= {\left( { \pm \frac{5}{4}} \right)^2}\\x - \frac{7}{4} = \frac{5}{4} &\qquad x - \frac{7}{4} = \frac{{ - 5}}{4}\\x = \frac{5}{4} + \frac{7}{4} &\qquad x = \frac{{ - 5}}{4} + \frac{7}{4}\\
x = \frac{{12}}{4} &\qquad  x = \frac{2}{4}\\x = 3&\qquad x = \frac{1}{2}\end{align}\]

Roots are \(\begin{align}3, \;\frac{1}{2}.\end{align}\)

(ii) \(2x^2+x-4=0\)

\[\begin{align}{x^2} + \frac{x}{2} - 2 &= 0\\{x^2} + \frac{x}{2} &= 2\\{x^2} + \frac{x}{2} &= 2\end{align}\]

Since, \(\begin{align} \frac{{\rm{1}}}{{\rm{2}}} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4},\;
{\left( {\frac{1}{4}} \right)^2} \end{align}\) should be added on both sides

\[\begin{align}{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= 2 + {\left( {\frac{1}{4}} \right)^2}\\
{\left( {{x^2} + \frac{1}{4}} \right)^2} &= 2 + \frac{1}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{32 + 1}}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{33}}{{16}}\\\left( {x + \frac{1}{4}} \right) &= \pm \frac{{\sqrt {33} }}{4}\\\left( {x + \frac{1}{4}} \right) = \frac{{\sqrt {33} }}{4} &\qquad \left( {x + \frac{1}{4}} \right) = - \frac{{\sqrt {33} }}{4}\\
x = \frac{{\sqrt {33} }}{4} - \frac{1}{4} &\qquad x = - \frac{{\sqrt {33} }}{4} - \frac{1}{4}\\x = \frac{{\sqrt {33} - 1}}{4} &\qquad x = \frac{{ - \sqrt {33} - 1}}{4}\end{align}\]

Roots are \(\begin{align} \frac{{\sqrt {33} - 1}}{4},\frac{{ - \sqrt {33} - 1}}{4}\end{align}\)

iii) \(4x+ 4\sqrt {\left( 3 \right)} x +3=0\)

\[\begin{align}x^2+ \sqrt 3 x +\frac{{\rm{3}}}{{\rm{4}}}&= 0\\{x^2} + \sqrt 3 x& = - \frac{3}{4}\\{x^2} + \sqrt 3 x + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}& = - \frac{3}{4} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}\end{align}\]

\(\begin{align}{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \end{align}\) is added on both sides

\[\begin{align}{{\left( {x + \frac{{\sqrt 3 }}{2}} \right)}^2} &= - \frac{3}{4} + \frac{3}{4}\\{\left( {x + \frac{{\sqrt 3 }}{2}} \right)^2} &= 0\\x = - \frac{{\sqrt 3 }}{2} &\quad x= - \frac{{\sqrt 3 }}{2}\end{align}\]

Roots are \(\begin{align} - \frac{{\sqrt 3 }}{2},\, - \frac{{\sqrt 3 }}{2}\end{align}\)

iv) \(2x^{2}+x+4=0\)

\[\begin{align}{x^2} + \frac{x}{2} + 2 &= 0\\{x^2} + \frac{x}{2}& = - 2\\{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= - 2 + {\left( {\frac{1}{4}} \right)^2}\\{\left( {x + \frac{1}{4}} \right)^2} &= - 2 + \frac{1}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 32 + 1}}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 31}}{{16}} < 0\end{align}\]

Square of any real number can’t be negative. 

\(\therefore\;\)Real roots don’t exist.