Question

How do you solve the quadratic equation by completing the square: \[2{{x}^{2}}-3x+1=0\]?

Answer 1

Hint: Assume the expression \[2{{x}^{2}}-3x+1=y\] and compare it with the general form given as: - \[y=a{{x}^{2}}+bx+c\]. Find the respective values of a, b and c. Now, find the discriminant of the given quadratic equation by using the formula: - \[D={{b}^{2}}-4ac\], where D = discriminant. Now, write the expression as: - \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\] and substitute it equal to 0 to find the two values of x.

Complete step by step answer:
Here, we have been provided with the quadratic equation: - \[2{{x}^{2}}-3x+1=0\] and we are asked to solve it. That means we have to find the values of x. We have been asked to use completing the square method.
Now, we know that any quadratic equation of the form \[y=a{{x}^{2}}+bx+c\] can be simplified as \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\], using completing the square method. Here, ‘D’ denotes the discriminant. So, on assuming \[2{{x}^{2}}-3x+1=y\] and comparing it with the general quadratic equation, we get,
\[\Rightarrow \] a = 2, b = -3, c = 1
Applying the formula for discriminant of a quadratic equation given as, \[D={{b}^{2}}-4ac\], we get,
\[\begin{align}
  & \Rightarrow D={{\left( -3 \right)}^{2}}-4\left( 2 \right)\left( 1 \right) \\
 & \Rightarrow D=9-8 \\
 & \Rightarrow D=1 \\
\end{align}\]
Therefore, substituting the values in the simplified form of y, we get,
\[\begin{align}
  & \Rightarrow y=2\left[ {{\left( x+\left( \dfrac{-3}{2\times 2} \right) \right)}^{2}}-\dfrac{1}{4\times {{\left( 2 \right)}^{2}}} \right] \\
 & \Rightarrow y=2\left[ {{\left( x-\dfrac{3}{4} \right)}^{2}}-\dfrac{1}{16} \right] \\
\end{align}\]
Substituting y = 0, we get,
\[\Rightarrow 2\left[ {{\left( x-\dfrac{3}{4} \right)}^{2}}-\dfrac{1}{16} \right]=0\]
\[\begin{align}
  & \Rightarrow \left[ {{\left( x-\dfrac{3}{4} \right)}^{2}}-\dfrac{1}{16} \right]=0 \\
 & \Rightarrow {{\left( x-\dfrac{3}{4} \right)}^{2}}=\dfrac{1}{16} \\
\end{align}\]
Taking square root both the sides, we get,
\[\begin{align}
  & \Rightarrow \left( x-\dfrac{3}{4} \right)=\pm \sqrt{\dfrac{1}{16}} \\
 & \Rightarrow x-\dfrac{3}{4}=\pm \dfrac{1}{4} \\
 & \Rightarrow x=\dfrac{3}{4}\pm \dfrac{1}{4} \\
\end{align}\]
Considering the two signs one – by – one, we get,
\[\Rightarrow x=\left( \dfrac{3}{4}+\dfrac{1}{4} \right)\] or \[x=\left( \dfrac{3}{4}-\dfrac{1}{4} \right)\]
\[\Rightarrow x=1\] or \[x=\dfrac{1}{2}\]

Hence, the solutions of the given quadratic equation are x = 1 and \[x=\dfrac{1}{2}\].

Note: One may also use the middle term split method to solve the question and check if we are getting the same values of x. Note that the general expression of this completing the square method is also known as the vertex form. Generally, this form is used in coordinate geometry of parabola to find the vertex of the parabola which is given as \[\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)\]. Note that the quadratic formula: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] is derived from completing the square method.