Question

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

Answer 1

Hint:In this question, we are given an equation in terms of x and we need to solve this equation to get the value of x which satisfies the equation using completing the square method. We will first remove the coefficient of ${{x}^{2}}$ to get an equation of the form ${{x}^{2}}+bx+c=0$. Then we add a term y to both sides of the equation such that we can form a whole square term using y. The y for an equation of the form ${{x}^{2}}+bx+c=0$ is given by ${{\left( \dfrac{b}{2} \right)}^{2}}$. After rearranging and taking square roots we will obtain two linear equations in terms of x. These will give us the value of x.

Complete step-by-step answer:
Here we are given the equation as $3{{x}^{2}}-2x-2=0$. We need to solve it by completing the square. For this let us add a certain element y on both sides of the equation. We know that, for an equation of the form ${{x}^{2}}+bx+c=0$ we add and subtract ${{\left( \dfrac{b}{2} \right)}^{2}}$ element to complete the square.
So let us first change $3{{x}^{2}}-2x-2=0$ in the form as ${{x}^{2}}+bx+c$. Let us divide both sides by 3 we get $\dfrac{3{{x}^{2}}-2x-2}{3}=0$.
Separating the terms we get $\dfrac{3{{x}^{2}}}{3}-\dfrac{2x}{3}-\dfrac{2}{3}=0\Rightarrow {{x}^{2}}-\dfrac{2x}{3}-\dfrac{2}{3}=0$.
Comparing with ${{x}^{2}}+bx+c=0$ we have $b=\dfrac{-2}{3}\text{ and }c=\dfrac{-2}{3}$.
Now let us add the term ${{\left( \dfrac{b}{2} \right)}^{2}}$ here which will be ${{\left( \dfrac{\dfrac{-2}{3}}{2} \right)}^{2}}$.
Simplifying the term we get \[{{\left( \dfrac{-2}{6} \right)}^{2}}\Rightarrow {{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{1}{9}\].
So let us add \[\dfrac{1}{9}\] to both sides of the equation we get \[{{x}^{2}}-\dfrac{2x}{3}-\dfrac{2}{3}+\dfrac{1}{9}=\dfrac{1}{9}\].
Rearranging the terms we get \[{{x}^{2}}-\dfrac{2x}{3}+\dfrac{1}{9}=\dfrac{1}{9}+\dfrac{2}{3}\].
As we can see the terms \[{{x}^{2}}-\dfrac{2x}{3}+\dfrac{1}{9}\] looks like in the form ${{a}^{2}}-2ab+{{b}^{2}}$ where $b=\dfrac{1}{3},a=x$. So we can write it as equal to ${{\left( a-b \right)}^{2}}$. Hence we get the equation as \[{{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{1}{9}+\dfrac{2}{3}\].
On the right side let us take an LCM of 9 and simplify the terms we get \[{{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{1+6}{9}=\dfrac{7}{9}\].
Hence the equation becomes \[{{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{7}{9}\].
Now let us take square root on both sides of the equation so we get \[{{\left( {{\left( x-\dfrac{1}{3} \right)}^{2}} \right)}^{\dfrac{1}{2}}}=\pm {{\left( \dfrac{7}{9} \right)}^{\dfrac{1}{2}}}\].
Simplifying the powers we have $x-\dfrac{1}{3}=\pm \sqrt{\dfrac{7}{9}}$.
As we know that $\sqrt{9}=3$ so we get $x-\dfrac{1}{3}=\pm \dfrac{\sqrt{7}}{3}$.
Hence we have obtained two linear equations for x. These are $x-\dfrac{1}{3}=\dfrac{\sqrt{7}}{3}\text{ and }x-\dfrac{1}{3}=-\dfrac{\sqrt{7}}{3}$.
Let us add \[\dfrac{1}{3}\] to both sides of the equation we get $x-\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{\sqrt{7}}{3}+\dfrac{1}{3}\text{ and }x-\dfrac{1}{3}+\dfrac{1}{3}=-\dfrac{\sqrt{7}}{3}+\dfrac{1}{3}$.
Simplifying the equation we get $x=\dfrac{\sqrt{7}+1}{3}\text{ and }x=\dfrac{1-\sqrt{7}}{3}$.
These are the required values of x which satisfy the equation.

Note: Students should make sure that coefficient of ${{x}^{2}}$ is 1 before selecting the term y which is to be added. Students should take care of the signs while solving the equation and do not forget about negative signs after taking square root. They can also check their answers by substituting values of x in the original equation to see if the left side is equal to the right side.