Question

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

Answer 1

We are given an equation as $2{{x}^{2}}-3x-3=0$ and to find the solution we will first know about the type of the equation. Once we have the type of the equation then we will see more ways to find the solution. Completing the square method is one in which we make a complete square and then solve further. So, we use this to solve our problem.

Complete step-by-step solution:
We are given an equation as $2{{x}^{2}}-3x-3=0.$ We can see that this equation has the highest power as 2, so it is clearly a two-degree polynomial or a quadratic polynomial. We have to find the solution to it, which means we have to find the value of x which will satisfy the given equation. As the power is 2, so the maximum it can have is 2 solutions. We have to use completing the square method to find the solution. The steps to find the solution using completing the square method is
Step 1: Divide all the terms by the coefficient of ${{x}^{2}}$ as in our equation $2{{x}^{2}}-3x-3=0$ the coefficient of ${{x}^{2}}$ is 2. So, we divide all the terms by 2, we get,
${{x}^{2}}-\dfrac{3}{2}x-\dfrac{3}{2}=0$
Step 2: Move the number (constant terms) to the right side. As our constant term is $\dfrac{-3}{2},$ so we add $\dfrac{3}{2}$ on both the sides to remove it. So, we get,
${{x}^{2}}-\dfrac{3}{2}x-\dfrac{3}{2}+\dfrac{3}{2}=\dfrac{3}{2}$
So, we get,
$\Rightarrow {{x}^{2}}-\dfrac{3}{2}x=\dfrac{3}{2}$
Step 3: Complete the square on the left side of the equation and balance this by adding the same value on the right side. So in our case, ${{x}^{2}}-\dfrac{3}{2}x=\dfrac{3}{2},$ we add ${{\left( \dfrac{3}{4} \right)}^{2}}$ to complete the square. Also, we add it to the right side to keep the equation balanced. So, we get,
$\Rightarrow {{x}^{2}}-\dfrac{3}{2}x+\dfrac{9}{16}=\dfrac{3}{2}+\dfrac{9}{16}$
So simplifying this, we get,
$\Rightarrow {{\left( x-\dfrac{3}{4} \right)}^{2}}=\dfrac{24+9}{16}=\dfrac{33}{16}$
And we have
$\Rightarrow {{\left( x-\dfrac{3}{4} \right)}^{2}}=\dfrac{33}{16}$
Taking square root on both the sides, we get,
$\Rightarrow x-\dfrac{3}{4}=\pm \sqrt{\dfrac{33}{16}}$
As $\sqrt{16}=4,$ so we get,
$\Rightarrow x-\dfrac{3}{4}=\pm \dfrac{\sqrt{33}}{4}$
Now, adding $\dfrac{3}{4}$ on both the sides, we get,
$\Rightarrow x-\dfrac{3}{4}+\dfrac{3}{4}=\pm \dfrac{\sqrt{33}}{4}+\dfrac{3}{4}$
On solving we get,
$\Rightarrow x=\dfrac{\pm \sqrt{33}+3}{4}$
Therefore, the solutions are $x=\dfrac{+\sqrt{33}+3}{4}$ and $x=\dfrac{-\sqrt{33}+3}{4}.$

Note: We can also find the solution to this by plotting $2{{x}^{2}}-3x-3=0$ on the graph. The point where the graph cut the x-axis will be defining our solution as here the right side is zero. So here the solution is also referred to as zero of the given polynomial, so that is why when we find the solution, it will lie on the x-axis only as they are zero of the equation and zero always lie on the x-axis.