Question

How do you solve \[2{{x}^{2}}-4x-3=0\] by completing the square?

Answer 1

Hint: In this problem, we have to solve the given quadratic equation by completing the square. We can first take the constant term to the right-hand side or we can add 3 on both sides. Then we have two terms on the left-hand side, which we should convert to a perfect square by adding the number 2. Then we will get a perfect square to solve for x.

Complete step by step answer:
We know that the given quadratic equation to be solved is,
\[2{{x}^{2}}-4x-3=0\] ……. (1)
Now we can add 3 on both sides in the above equation (1), we get
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}-4x-3+3=0+3 \\
 & \Rightarrow 2{{x}^{2}}-4x=3 \\
\end{align}\]
Now we can add 2 on both sides to get a perfect square equation, we get
\[\Rightarrow 2{{x}^{2}}-4x+2=3+2\]
Now we can take the common term 2 outside, in the left-hand side, we get
\[\Rightarrow 2\left( {{x}^{2}}-2x+1 \right)=5\]
We know that,
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
Now we can apply this formula in the left-hand side of the above equation, we get
\[\Rightarrow 2{{\left( x-1 \right)}^{2}}=5\]
We can now divide by 2 on both the sides and cancel similar terms, we get
 \[\begin{align}
  & \Rightarrow \dfrac{2{{\left( x-1 \right)}^{2}}}{2}=\dfrac{5}{2} \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{5}{2} \\
\end{align}\]
We can take square on both sides, we get
\[\Rightarrow \sqrt{{{\left( x-1 \right)}^{2}}}=\pm \sqrt{\dfrac{5}{2}}\]
We can cancel the square root on the left-hand side,
\[\Rightarrow x-1=\pm \sqrt{\dfrac{5}{2}}\]
Now we can add the number 1 on both sides, we get
\[\Rightarrow x=1\pm \sqrt{\dfrac{5}{2}}\]
Therefore, the value of x = \[1+\sqrt{\dfrac{5}{2}},1-\sqrt{\dfrac{5}{2}}\] .

Note:
Students make mistakes while finding the perfect square of the equation which should be concentrated. In this problem, we have given that we should solve by completing the square, in case it is not given, we can also use another method, a quadratic formula to solve and find the value of x.