Question

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

Answer 1

Hint: For solving this question we will use the formulas ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ . And we will try to write each quadratic equation in the form of the complete square. After that, we will easily solve for the roots of the given quadratic equations.

Complete step-by-step answer:
Given:
We have to find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) $2{{x}^{2}}-7x+3=0$ (ii) $2{{x}^{2}}+x-4=0$
(iii) $4{{x}^{2}}+4\sqrt{3}x+3=0$ (iv) $2{{x}^{2}}+x+4=0$
Now, before we proceed we should know that, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.We will be using these formulas to find the roots of the given quadratic equations. Basically ,we will try to do some simple arithmetic operations, so that we can apply the whole square formula easily and then, we will easily write the roots of the given quadratic equations. For more clarity take a look at the example $a{{x}^{2}}+bx+c=0$ .
First, we will divide the entire equation by the coefficient of ${{x}^{2}}$ . So, we get:
${{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0$
Now, we will write ${{x}^{2}}={{\left( x \right)}^{2}}$ , $\dfrac{b}{a}x=2\times \dfrac{b}{2a}\times x$ and $\dfrac{c}{a}=\dfrac{c}{a}+{{\left( \dfrac{b}{2a} \right)}^{2}}-{{\left( \dfrac{b}{2a} \right)}^{2}}$ . Then,
$\begin{align}
  & {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}+2\times \dfrac{b}{2a}\times x+{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}-{{\left( \dfrac{b}{2a} \right)}^{2}}=0 \\
\end{align}$
Now, as we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so, we can write ${{\left( x \right)}^{2}}+2\times \dfrac{b}{2a}\times x+{{\left( \dfrac{b}{2a} \right)}^{2}}={{\left( x+\dfrac{b}{2a} \right)}^{2}}$ in the above equation. Then,
$\begin{align}
  & {{\left( x \right)}^{2}}+2\times \dfrac{b}{2a}\times x+{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}-{{\left( \dfrac{b}{2a} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}}{4{{a}^{2}}}-\dfrac{c}{a} \\
 & \Rightarrow {{\left( \dfrac{2ax+b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}} \\
 & \Rightarrow \dfrac{{{\left( 2ax+b \right)}^{2}}}{4{{a}^{2}}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}} \\
 & \Rightarrow {{\left( 2ax+b \right)}^{2}}=\left( {{b}^{2}}-4ac \right) \\
\end{align}$
Now, we will take the square root on the both sides in the above equation. Then,
$\begin{align}
  & {{\left( 2ax+b \right)}^{2}}=\left( {{b}^{2}}-4ac \right) \\
 & \Rightarrow 2ax+b=\pm \sqrt{{{b}^{2}}-4ac} \\
 & \Rightarrow 2ax=-b\pm \sqrt{{{b}^{2}}-4ac} \\
 & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
\end{align}$
Now, from the above result we can say that roots of the equation $a{{x}^{2}}+bx+c=0$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now, we will be using this method of completing square to find the roots of the given quadratic equations
(i) $2{{x}^{2}}-7x+3=0$ :
First, we will divide the entire equation by the coefficient of ${{x}^{2}}$ , i.e. 2. So, we get:
${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$
Now, we will write ${{x}^{2}}={{\left( x \right)}^{2}}$ , $\dfrac{7}{2}x=2\times \dfrac{7}{4}\times x$ and $\dfrac{3}{2}=\dfrac{3}{2}+{{\left( \dfrac{7}{4} \right)}^{2}}-{{\left( \dfrac{7}{4} \right)}^{2}}$ . Then,
$\begin{align}
  & {{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}-2\times \dfrac{7}{4}\times x+{{\left( \dfrac{7}{4} \right)}^{2}}+\dfrac{3}{2}-{{\left( \dfrac{7}{4} \right)}^{2}}=0 \\
\end{align}$
Now, as we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ so, we can write ${{\left( x \right)}^{2}}-2\times \dfrac{7}{4}\times x+{{\left( \dfrac{7}{4} \right)}^{2}}={{\left( x-\dfrac{7}{4} \right)}^{2}}$ in the above equation. Then,
\[\begin{align}
  & {{\left( x \right)}^{2}}-2\times \dfrac{7}{4}\times x+{{\left( \dfrac{7}{4} \right)}^{2}}+\dfrac{3}{2}-{{\left( \dfrac{7}{4} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x-\dfrac{7}{4} \right)}^{2}}+\dfrac{3}{2}-\dfrac{49}{16}=0 \\
 & \Rightarrow {{\left( x-\dfrac{7}{4} \right)}^{2}}-\dfrac{25}{16}=0 \\
 & \Rightarrow {{\left( x-\dfrac{7}{4} \right)}^{2}}=\dfrac{25}{16} \\
 & \Rightarrow x-\dfrac{7}{4}=\pm \sqrt{\dfrac{25}{16}} \\
 & \Rightarrow x=\dfrac{7}{4}\pm \dfrac{5}{4} \\
 & \Rightarrow x=\dfrac{7}{4}\pm \dfrac{5}{4} \\
 & \Rightarrow x=\dfrac{7+5}{4},\dfrac{7-5}{4} \\
 & \Rightarrow x=\dfrac{12}{4},\dfrac{2}{4} \\
 & \Rightarrow x=3,\dfrac{1}{2} \\
\end{align}\]
Now, from the above result, we conclude that $x=3,\dfrac{1}{2}$ will be the roots of the quadratic equation $2{{x}^{2}}-7x+3=0$ .
(ii) $2{{x}^{2}}+x-4=0$ :
First, we will divide the entire equation by the coefficient of ${{x}^{2}}$ , i.e. 2. So, we get:
${{x}^{2}}+\dfrac{x}{2}-2=0$
Now, we will write ${{x}^{2}}={{\left( x \right)}^{2}}$ , $\dfrac{x}{2}=2\times \dfrac{1}{4}\times x$ and $-2=-2+{{\left( \dfrac{1}{4} \right)}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}$ . Then,
$\begin{align}
  & {{x}^{2}}+\dfrac{x}{2}-2=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}+2\times \dfrac{1}{4}\times x+{{\left( \dfrac{1}{4} \right)}^{2}}-2-{{\left( \dfrac{1}{4} \right)}^{2}}=0 \\
\end{align}$
Now, as we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so, we can write ${{\left( x \right)}^{2}}+2\times \dfrac{1}{4}\times x+{{\left( \dfrac{1}{4} \right)}^{2}}={{\left( x+\dfrac{1}{4} \right)}^{2}}$ in the above equation. Then,
\[\begin{align}
  & {{\left( x \right)}^{2}}+2\times \dfrac{1}{4}\times x+{{\left( \dfrac{1}{4} \right)}^{2}}-2-{{\left( \dfrac{1}{4} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}-2-\dfrac{1}{16}=0 \\
 & \Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}-\dfrac{33}{16}=0 \\
 & \Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{33}{16} \\
 & \Rightarrow \left( x+\dfrac{1}{4} \right)=\pm \sqrt{\dfrac{33}{16}} \\
 & \Rightarrow x=-\dfrac{1}{4}\pm \sqrt{\dfrac{33}{16}} \\
 & \Rightarrow x=-\dfrac{1}{4}\pm \dfrac{\sqrt{33}}{4} \\
 & \Rightarrow x=\dfrac{-1\pm \sqrt{33}}{4} \\
\end{align}\]
Now, from the above result, we conclude that $x=\dfrac{\sqrt{33}-1}{4},-\dfrac{\left( \sqrt{33}+1 \right)}{4}$ will be the roots of the quadratic equation $2{{x}^{2}}+x-4=0$ .
(iii) $4{{x}^{2}}+4\sqrt{3}x+3=0$ :
First, we will divide the entire equation by the coefficient of ${{x}^{2}}$ , i.e. 4. So, we get:
${{x}^{2}}+\sqrt{3}x+\dfrac{3}{4}=0$
Now, we will write ${{x}^{2}}={{\left( x \right)}^{2}}$ , $\sqrt{3}x=2\times \dfrac{\sqrt{3}}{2}\times x$ and $\dfrac{3}{4}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$ . Then,
$\begin{align}
  & {{x}^{2}}+\sqrt{3}x+\dfrac{3}{4}=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}+2\times \dfrac{\sqrt{3}}{2}\times x+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=0 \\
\end{align}$
Now, as we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so, we can write ${{\left( x \right)}^{2}}+2\times \dfrac{\sqrt{3}}{2}\times x+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}={{\left( x+\dfrac{\sqrt{3}}{2} \right)}^{2}}$ in the above equation. Then,
$\begin{align}
  & {{\left( x \right)}^{2}}+2\times \dfrac{\sqrt{3}}{2}\times x+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x+\dfrac{\sqrt{3}}{2} \right)}^{2}}=0 \\
\end{align}$
Now, as we know that square of zero is zero. Then,
\[\begin{align}
  & {{\left( x+\dfrac{\sqrt{3}}{2} \right)}^{2}}=0 \\
 & \Rightarrow x+\dfrac{\sqrt{3}}{2}=0 \\
 & \Rightarrow x=-\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Now, from the above result, we conclude that the quadratic equation $4{{x}^{2}}+4\sqrt{3}x+3=0$ will have two equal roots $x=-\dfrac{\sqrt{3}}{2}$ .
(iv) $2{{x}^{2}}+x+4=0$
First, we will divide the entire equation by the coefficient of ${{x}^{2}}$ , i.e. 2. So, we get:
${{x}^{2}}+\dfrac{x}{2}+2$
Now, we will write ${{x}^{2}}={{\left( x \right)}^{2}}$ , $\dfrac{x}{2}=2\times \dfrac{1}{4}\times x$ and $2=2+{{\left( \dfrac{1}{4} \right)}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}$ . Then,
$\begin{align}
  & {{x}^{2}}+\dfrac{x}{2}+2=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}+2\times \dfrac{1}{4}\times x+{{\left( \dfrac{1}{4} \right)}^{2}}+2-{{\left( \dfrac{1}{4} \right)}^{2}}=0 \\
\end{align}$
Now, as we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so, we can write ${{\left( x \right)}^{2}}+2\times \dfrac{1}{4}\times x+{{\left( \dfrac{1}{4} \right)}^{2}}={{\left( x+\dfrac{1}{4} \right)}^{2}}$ in the above equation. Then,
$\begin{align}
  & {{\left( x \right)}^{2}}+2\times \dfrac{1}{4}\times x+{{\left( \dfrac{1}{4} \right)}^{2}}+2-{{\left( \dfrac{1}{4} \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}+2-\dfrac{1}{16}=0 \\
 & \Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}+\dfrac{31}{16}=0 \\
 & \Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}=-\dfrac{31}{16} \\
\end{align}$
Now, as we know that the square of a real number is never negative so, there will be no real roots of the given quadratic equation. Then,
$\begin{align}
  & {{\left( x+\dfrac{1}{4} \right)}^{2}}=-\dfrac{31}{16} \\
 & \Rightarrow x+\dfrac{1}{4}=\pm \sqrt{-\dfrac{31}{16}} \\
\end{align}$
Now, we will write $\sqrt{-\dfrac{31}{16}}=i\sqrt{\dfrac{31}{16}}$ in the above equation. Then,
\[\begin{align}
  & x+\dfrac{1}{4}=\pm \sqrt{-\dfrac{31}{16}} \\
 & \Rightarrow x=-\dfrac{1}{4}\pm i\sqrt{\dfrac{31}{16}} \\
 & \Rightarrow x=-\dfrac{1}{4}\pm i\dfrac{\sqrt{31}}{4} \\
 & \Rightarrow x=\dfrac{-1\pm i\sqrt{31}}{4} \\
\end{align}\]
Now, from the above result, we conclude that $x=\dfrac{-1+i\sqrt{33}}{4},-\dfrac{\left( 1+i\sqrt{33} \right)}{4}$ will be the roots of the quadratic equation $2{{x}^{2}}+x+4=0$ .

Note: Here, the student should first understand what is asked in the problem and proceed in the right direction to get the correct answer quickly. Moreover, we should apply each formula with proper values and find the roots of the given quadratic equations, by the method of completing the square only. Also, it must be remembered that before starting the method of completing squares, make sure that the coefficient of ${{x}^{2}}$ is 1. If it is not 1, then divide the entire equation by the coefficient of ${{x}^{2}}$ , then proceed with the solution.