Question

How do you solve $ 2{{x}^{2}}-5x-3=0 $ by completing the square?

Answer 1

Hint: In the problem, they have mentioned using the method of completing the square. We know that the method of completing the square is used to solve or find the roots of a quadratic equation. So first we will check whether the given equation is a quadratic equation or not. If the given equation is a quadratic equation and which is in form of $ a{{x}^{2}}+bx+c=0 $ , then we will go to the further steps. In the first step, we will divide the given equation with the coefficient of $ {{x}^{2}} $ to get the coefficient of $ {{x}^{2}} $ as $ 1 $ . In the step two we will add and subtract the square of the half of the coefficient of $ x $ in the above equation and rearrange the obtained equation, then we will apply the formula either $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $ or $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ based the obtained equation in step two. In step three we will simplify and move the constants to another side of the equation. In step four we will apply square roots on both sides of the equation and simplify it to get the value of $ x $ .

Complete step by step answer:
Given equation, $ 2{{x}^{2}}-5x-3=0 $
We can clearly see that the above equation is a quadratic equation, so we can use the method of completing the square to solve the given equation.
Comparing the given equation with $ a{{x}^{2}}+bx+c=0 $ , then we will get
 $ a=2 $ , $ b=-5 $ , $ c=-3 $ .
Dividing the given equation with $ 2 $ , then we will get
 $ \begin{align}
  & \dfrac{2{{x}^{2}}-5x-3}{2}=\dfrac{0}{2} \\
 & \Rightarrow \dfrac{2}{2}{{x}^{2}}-\dfrac{5}{2}x-\dfrac{3}{2}=0 \\
 & \Rightarrow {{x}^{2}}-\dfrac{5}{2}x-\dfrac{3}{2}=0 \\
\end{align} $
In the above equation the coefficient of $ x $ is $ \dfrac{5}{2} $ . Adding the square of half of the coefficient of $ x $ in the above equation, then we will get
 $ \begin{align}
  & {{x}^{2}}-\dfrac{5}{2}x-\dfrac{3}{2}+{{\left( \dfrac{1}{2}.\dfrac{5}{2} \right)}^{2}}-{{\left( \dfrac{1}{2}.\dfrac{5}{2} \right)}^{2}}=0 \\
 & \Rightarrow {{x}^{2}}-\dfrac{5}{2}x-\dfrac{3}{2}+{{\left( \dfrac{5}{4} \right)}^{2}}-{{\left( \dfrac{5}{4} \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}{4} \right)\left( x \right)+{{\left( \dfrac{5}{4} \right)}^{2}}-\dfrac{3}{2}-\dfrac{25}{16}=0 $
We know that $ {{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}} $ , then we will get
 $ \Rightarrow {{\left( x-\dfrac{5}{4} \right)}^{2}}-\dfrac{3}{2}-\dfrac{25}{16}=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}{4} \right)}^{2}}=\dfrac{3}{2}+\dfrac{25}{16} \\
 & \Rightarrow {{\left( x-\dfrac{5}{4} \right)}^{2}}=\dfrac{8\times 3+25}{16} \\
 & \Rightarrow {{\left( x-\dfrac{5}{4} \right)}^{2}}=\dfrac{49}{16} \\
\end{align} $
Applying square root on both side of the above equation, then we will get
 $ \begin{align}
  & \Rightarrow \sqrt{{{\left( x-\dfrac{5}{4} \right)}^{2}}}=\sqrt{\dfrac{49}{16}} \\
 & \Rightarrow x-\dfrac{5}{4}=\pm \dfrac{7}{4} \\
\end{align} $
From the above equation, we will get
 $ x-\dfrac{5}{4}=\dfrac{7}{4} $ or $ x-\dfrac{5}{4}=-\dfrac{7}{4} $
Simplifying the above equations, then
 $ \begin{align}
  & x=\dfrac{5}{4}+\dfrac{7}{4} \\
 & \Rightarrow x=\dfrac{5+7}{4} \\
 & \Rightarrow x=\dfrac{12}{4} \\
 & \Rightarrow x=3 \\
\end{align} $ or $ \begin{align}
  & x=\dfrac{5}{4}-\dfrac{7}{4} \\
 & \Rightarrow x=\dfrac{5-7}{4} \\
 & \Rightarrow x=\dfrac{-2}{4} \\
 & x=-\dfrac{1}{2} \\
\end{align} $
 $ \therefore $ The solution for the given equation is $ x=3 $ or $ x=-\dfrac{1}{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.