# 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
Ex 4.3 | Question 1

## Text Solution

What is the unknown?

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}\end{align}

\begin{align} \left( {x + \frac{1}{4}} \right) \!\!= \frac{{\sqrt {33} }}{4} ,& \quad \!\!\!\!\!\!\left( \!{x \!+\! \frac{1}{4}} \!\!\right) \!\!= - \frac{{\sqrt {33} }}{4}\\ x = \frac{{\sqrt {33} }}{4} - \!\! \frac{1}{4}, &\quad \!\!x \!\!= - \!\! \frac{{\sqrt {33} }}{4} - \! \frac{1}{4}\\x = \frac{{\sqrt {33} - 1}}{4}, &\quad \!\!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{3}{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.

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