Question

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

Answer 1

Hint: In this question, we have to find the value of x from a quadratic equation. Therefore, we will use completing the square method, to get the solution. We start solving this problem, by adding 3 on both sides of the equation and then we will make the coefficient of ${{x}^{2}}$ as 1. After the necessary calculations, we will add ${{\left( \dfrac{1}{2} \right)}^{2}}$ on both sides of the equation. After that, we will create the left-hand side of the equation as a perfect square. Then, we will take the square root on both sides and make the necessary calculations. In the end, we will get two equations, thus we solve them separately, which is the required solution for the problem.

Complete step by step solution:
According to the question, we have to find the value of x.
Thus, we will use the completing the square method to get the solution.
The equation given to us is $2{{x}^{2}}-2x-3=0$ ---------- (1)
We start solving this problem by adding 3 on both sides in the equation (1), we get
$\Rightarrow 2{{x}^{2}}-2x-3+3=0+3$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2{{x}^{2}}-2x=3$
Now, we will make the coefficient of ${{x}^{2}}$ as 1 that is we will divide 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{2}{2}{{x}^{2}}-\dfrac{2}{2}x=\dfrac{3}{2}$
On further simplification, we get
$\Rightarrow {{x}^{2}}-x=\dfrac{3}{2}$
Now, we will add ${{\left( \dfrac{b}{2} \right)}^{2}}$on both sides of the equation, here $b=1$ , thus we will add ${{\left( \dfrac{1}{2} \right)}^{2}}$on both sides in the above equation, we get
$\Rightarrow {{x}^{2}}-x+{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{3}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}$
On further solving the above equation, we get
$\Rightarrow {{x}^{2}}-x+{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{3}{2}+\dfrac{1}{4}$
Now, taking the LCM on the right-hand side of the equation, we get
$\Rightarrow {{x}^{2}}-x+{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{6+1}{4}$
$\Rightarrow {{x}^{2}}-x+{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{7}{4}$
So, we will make the perfect square on the left-hand side of the above equation, that is we will use the algebraic identity ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ , we get
$\Rightarrow {{\left( x-\left( \dfrac{1}{2} \right) \right)}^{2}}=\dfrac{7}{4}$
Now, we will take the square root on both sides in the above equation, we get
$\Rightarrow \sqrt{{{\left( x-\left( \dfrac{1}{2} \right) \right)}^{2}}}=\sqrt{\dfrac{7}{4}}$
On further solving the above equation, we get
$\Rightarrow \left( x-\left( \dfrac{1}{2} \right) \right)=\pm \dfrac{\sqrt{7}}{2}$
Now, we will split (+) and (-) sign to get two new equations separately, that is
$\Rightarrow \left( x-\left( \dfrac{1}{2} \right) \right)=+\dfrac{\sqrt{7}}{2}$ -------- (2) and
$\Rightarrow \left( x-\left( \dfrac{1}{2} \right) \right)=-\dfrac{\sqrt{7}}{2}$ ------- (3)
Now, we will solve equation (2), which is
$\Rightarrow \left( x-\left( \dfrac{1}{2} \right) \right)=+\dfrac{\sqrt{7}}{2}$
So, we will add $\dfrac{1}{2}$ on both sides in the above equation, we get
$\Rightarrow x-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{\sqrt{7}}{2}+\dfrac{1}{2}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=\dfrac{\sqrt{7}+1}{2}$
Thus, we will solve equation (3), which is
$\Rightarrow \left( x-\left( \dfrac{1}{2} \right) \right)=-\dfrac{\sqrt{7}}{2}$
So, we will add $\dfrac{1}{2}$ on both sides in the above equation, we get
$\Rightarrow x-\dfrac{1}{2}+\dfrac{1}{2}=-\dfrac{\sqrt{7}}{2}+\dfrac{1}{2}$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x=\dfrac{-\sqrt{7}+1}{2}$

Thus, for the equation $2{{x}^{2}}-2x-3=0$, the value of x are $\dfrac{\sqrt{7}+1}{2},\dfrac{-\sqrt{7}+1}{2}$

Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical errors. Do not forget to split (+) and (-) sign to get an accurate answer. Do step-by-step calculations to get an accurate answer.