Question

How do you solve $3{{x}^{2}}+5x=2$ by completing the square?

Answer 1

Hint: In this problem we need to solve the given equation by using the completing square method. This method involves the converting the quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$ into $a{{\left( x+d \right)}^{2}}+e=0$ where $d=\dfrac{b}{2a}$, $e=c-\dfrac{{{b}^{2}}}{4a}$. So, we will first convert the given equation in the form of a standard quadratic equation which is $a{{x}^{2}}+bx+c=0$. After that we will perform some arithmetic operations to achieve the equation in required form which is $a{{\left( x+d \right)}^{2}}+e=0$. From this we can write the roots of the equation as $x=d\pm \sqrt{\dfrac{-e}{a}}$. Now we will simplify the above equation to get the required result.

Complete step by step answer:
Given equation, $3{{x}^{2}}+5x=2$
Shifting the constant which is in right hand side to left hand side in order to get $a{{x}^{2}}+bx+c=0$ from, the we will get
$\Rightarrow 3{{x}^{2}}+5x-2=0$
We can clearly see that the above equation is quadratic equation, so we can use the method of completing square to solve the given equation.
Comparing the given equation with $a{{x}^{2}}+bx+c=0$, then we will get
$a=3$, $b=5$, $c=-2$.
Dividing the given equation with $3$, then we will get
$\begin{align}
  & \dfrac{3{{x}^{2}}+5x-2}{3}=\dfrac{0}{3} \\
 & \Rightarrow \dfrac{3}{3}{{x}^{2}}+\dfrac{5}{3}x-\dfrac{2}{3}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{5}{3}x-\dfrac{2}{3}=0 \\
\end{align}$
In the above equation the coefficient of $x$ is $\dfrac{5}{3}$. Adding the square of half of the coefficient of $x$ in the above equation, then we will get
$\begin{align}
  & {{x}^{2}}+\dfrac{5}{3}x-\dfrac{2}{3}+{{\left( \dfrac{1}{2}.\dfrac{5}{3} \right)}^{2}}-{{\left( \dfrac{1}{2}.\dfrac{5}{3} \right)}^{2}}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{5}{3}x-\dfrac{2}{3}+{{\left( \dfrac{5}{6} \right)}^{2}}-{{\left( \dfrac{5}{6} \right)}^{2}}=0 \\
\end{align}$
Rearranging and rewriting the terms in the above equation, then we will get
$\Rightarrow {{x}^{2}}+2\left( \dfrac{5}{6} \right)\left( x \right)+{{\left( \dfrac{5}{6} \right)}^{2}}-\dfrac{2}{3}-\dfrac{25}{36}=0$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, then we will get
$\Rightarrow {{\left( x+\dfrac{5}{6} \right)}^{2}}-\dfrac{2}{3}-\dfrac{25}{36}=0$
Taking the constants in the above equation to the other side of the equation, then we will get
$\begin{align}
  & \Rightarrow {{\left( x+\dfrac{5}{6} \right)}^{2}}=\dfrac{2}{3}+\dfrac{25}{36} \\
 & \Rightarrow {{\left( x+\dfrac{5}{6} \right)}^{2}}=\dfrac{12\times 2+25}{36} \\
 & \Rightarrow {{\left( x+\dfrac{5}{6} \right)}^{2}}=\dfrac{49}{36} \\
\end{align}$
Applying square root on both side of the above equation, then we will get
$\begin{align}
  & \Rightarrow \sqrt{{{\left( x+\dfrac{5}{6} \right)}^{2}}}=\sqrt{\dfrac{49}{36}} \\
 & \Rightarrow x+\dfrac{5}{6}=\pm \dfrac{7}{6} \\
\end{align}$
From the above equation, we will get
$x+\dfrac{5}{6}=\dfrac{7}{6}$ or $x+\dfrac{5}{6}=-\dfrac{7}{6}$
Simplifying the above equations, then
$\begin{align}
  & x=-\dfrac{5}{6}+\dfrac{7}{6} \\
 & \Rightarrow x=\dfrac{-5+7}{6} \\
 & \Rightarrow x=\dfrac{2}{6} \\
 & \Rightarrow x=\dfrac{1}{3} \\
\end{align}$ or $\begin{align}
  & x=-\dfrac{5}{6}-\dfrac{7}{6} \\
 & \Rightarrow x=\dfrac{-5-7}{6} \\
 & \Rightarrow x=\dfrac{-12}{6} \\
 & x=-2 \\
\end{align}$

$\therefore $ The solution for the given equation is $x=\dfrac{1}{3}$ or $x=-2$.

Note: In the method of completing squares most of the students forget to consider the $\pm $ sign after applying the square root to the equation. It is not the correct way to solve the equation since if you don’t consider the sign then we only get one root, but for a quadratic equation we have two roots. So, it is necessary to consider the $\pm $ sign in method of completing squares.