Question

# Find the roots of the quadratic equation $2{{x}^{2}}-x-4=0$ by the method of completing the square.

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Hint: In this question, we have to follow certain steps such that we get the left-side of the quadratic equation as a perfect square. The first step in completing the square method is to make the coefficient of ${{x}^{2}}$ unity.
We will solve the quadratic equation by using the square method. There are three steps to be followed in order to apply this method. The steps that we will follow while solving quadratic through completing the square method are enlisted below.

Step 1: We will divide each term by coefficient of ${{x}^{2}}$.
Step 2: Now we will transpose constant terms to the right-hand side.
Step 3: Now we will add a constant on both sides to make the left-hand side a perfect square.
After obtaining the perfect square, we will simplify it to obtain the roots of the quadratic equation.
We will solve the given quadratic equation by completing the square method using the above steps.

The given quadratic equation is $2{{x}^{2}}-x-4=0$. This is of the form $a{{x}^{2}}+bx+c=0$. Comparing both the equations, we will get that a = 2, b = -1, c = -4.

Step 1: We will divide each term by the coefficient of ${{x}^{2}}$ i.e. a = 2 in this case.
\begin{align} & \Rightarrow {{x}^{2}}-\dfrac{x}{2}-\dfrac{4}{2}=0 \\ & \Rightarrow {{x}^{2}}-\dfrac{x}{2}-2=0 \\ \end{align}

Step 2: Now we will transpose the constant term, i.e -2 to the right-hand side.
$\Rightarrow {{x}^{2}}-\dfrac{x}{2}=2$
Now, we can write the term $\dfrac{x}{2}$ as below,
\begin{align} & \Rightarrow {{x}^{2}}-2\times \dfrac{x}{4}=2 \\ & \Rightarrow {{x}^{2}}-2\times x\times \dfrac{1}{4}=2 \\ \end{align}

Step 3: Now we will add ${{\left( \dfrac{1}{4} \right)}^{2}}$ on both sides to make the left-hand side a perfect square.
${{x}^{2}}-2\times x\times \dfrac{1}{4}+{{\left( \dfrac{1}{4} \right)}^{2}}=2+{{\left( \dfrac{1}{4} \right)}^{2}}$
Since we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we will write the above as,
\begin{align} & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=2+\dfrac{1}{16} \\ & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=\dfrac{\left( 2\times 16 \right)+1}{16} \\ & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=\dfrac{33}{16} \\ \end{align}
Taking square root on both sides, we will get,
\begin{align} & \Rightarrow \left( x-\dfrac{1}{4} \right)=\pm \sqrt{\dfrac{33}{16}} \\ & \Rightarrow x=\pm \dfrac{\sqrt{33}}{4}+\dfrac{1}{4} \\ & \Rightarrow x=\dfrac{1}{4}\left( 1\pm \sqrt{33} \right) \\ \end{align}

Therefore, we will get two values for x considering positive and negative value as $\dfrac{1}{4}\left( 1+\sqrt{33} \right)$ and $\dfrac{1}{4}\left( 1-\sqrt{33} \right)$ respectively.

Hence, the roots of the quadratic equation are $\dfrac{1}{4}\left( 1+\sqrt{33} \right)$ and $\dfrac{1}{4}\left( 1-\sqrt{33} \right)$.

Note: We can get the roots of the quadratic equation by using the formula or the factorisation method, but here we have been asked to find the roots by completing the square method. Once we have obtained the answer, we can cross-check the roots of the quadratic equation with roots obtained using $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ formula. If roots are the same then the answer is correct otherwise it’s wrong.