Question

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

Answer 1

We will use the concept of quadratic equation to solve the equation $2{{x}^{2}}-2x-2=0$. As the question asks us to use the method of complete square so we will use it. We will first convert the given equation into complete square by first dividing the whole equation by 2 and then we will write x terms as $x=2\times \dfrac{1}{2}\times x$ and add and subtract with ${{\left( \dfrac{1}{2} \right)}^{2}}$, then we will get the equation of the form ${{x}^{2}}-2\times \dfrac{1}{2}\times x+{{\left( \dfrac{1}{2} \right)}^{2}}-1-{{\left( \dfrac{1}{2} \right)}^{2}}=0$. Then we can write the equation as ${{\left( x-\dfrac{1}{2} \right)}^{2}}-\dfrac{5}{4}=0$ and then we will solve for x.

Complete answer:
We will use the concept of complete square from the quadratic equation to solve the above question. In the complete square method we first convert all the x terms in the equation in perfect squares and then we will solve it to get the value of x.
Since, we know from the question that we have to solve the quadratic equation $2{{x}^{2}}-2x-2=0$ by using the complete square method.
At first, we will divide the given equation by x, then we will get:
$\Rightarrow \dfrac{2{{x}^{2}}-2x-2}{2}=0$
$\Rightarrow \dfrac{2{{x}^{2}}}{2}-\dfrac{2x}{2}-\dfrac{2}{2}=0$
$\Rightarrow {{x}^{2}}-x-1=0$
Now, we will try to convert the above equation into complete square form i.e. ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Since, we can write x as $2\times \dfrac{1}{2}\times x$.
Here, in $2\times \dfrac{1}{2}\times x$ we can say that a = x and b = $\dfrac{1}{2}$.
Since, in equation ${{x}^{2}}-x-1=0$ we do not have ${{b}^{2}}$ term i.e. ${{\left( \dfrac{1}{2} \right)}^{2}}$, so we will add and subtract ${{\left( \dfrac{1}{2} \right)}^{2}}$ in the equation ${{x}^{2}}-x-1=0$.
$\Rightarrow {{x}^{2}}-2\times \dfrac{1}{2}\times x-1+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}=0$
$\Rightarrow {{x}^{2}}-2\times \dfrac{1}{2}\times x+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}-1=0$
$\Rightarrow {{x}^{2}}-2\times \dfrac{1}{2}\times x+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}-1=0$
Since, we can write ${{x}^{2}}-2\times \dfrac{1}{2}\times x+{{\left( \dfrac{1}{2} \right)}^{2}}$ as a complete square form i.e. ${{\left( x-\dfrac{1}{2} \right)}^{2}}$ so we can write the above equation as:
$\Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}-1=0$
$\Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\dfrac{5}{4}=0$
Now, we will take $\dfrac{5}{4}$ to the RHS and then we will get:
$\Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}=\dfrac{5}{4}$
Now, we will remove the square and take it as square root to the RHS, then we will get:
$\Rightarrow x-\dfrac{1}{2}=\pm \sqrt{\dfrac{5}{4}}$
So, after taking $\dfrac{1}{2}$ to the RHS we will get:
$\Rightarrow x=\pm \sqrt{\dfrac{5}{4}}+\dfrac{1}{2}$
$\Rightarrow x=\pm \dfrac{\sqrt{5}}{2}+\dfrac{1}{2}$
$\therefore x=\dfrac{1\pm \sqrt{5}}{2}$
Hence, solution of equation $2{{x}^{2}}-2x-2=0$ is $\dfrac{1+\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}$.
This is our required solution.

Note: Students are required to note that we can verify the solution of given quadratic equation by using the discriminant formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ which is the solution of general quadratic equation $a{{x}^{2}}+bx+c=0$.