Question

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

Answer 1

Hint: Firstly, to solve this equation by completing the square, we need to send the constant to the RHS. Now take the constant which is in front of ${{x}^{2}}$ common on the LHS side and send it to the RHS. Now if we consider, $a{{x}^{2}}+bx+c=0$ as the general equation of a quadratic equation, add ${{\left( \dfrac{b}{2} \right)}^{2}}$ on both sides of the equation. Now a square will be formed on the LHS. Group them together and then solve to get the value of $x$.

Complete step-by-step solution:
The given equation is, $3{{x}^{2}}-2x-3=0$
We need to solve this equation by completing the square.
The first step is to send the constant to the RHS.
Upon rearranging the terms, we get,
$\Rightarrow 3{{x}^{2}}-2x=3$
Now take the constant in front of ${{x}^{2}}$ common on the LHS side.
$\Rightarrow 3\left( {{x}^{2}}-\dfrac{2x}{3} \right)=3$
Now simplify by cancelling out the common terms on both sides.
$\Rightarrow \left( {{x}^{2}}-\dfrac{2x}{3} \right)=1$
Now if we consider, $a{{x}^{2}}+bx+c=0$ as the general equation of a quadratic equation, we need to add ${{\left( \dfrac{b}{2} \right)}^{2}}$ on both sides of the equation to form a square on the LHS.
Here we consider the most recent quadratic we get on solving.
Which is ${{x}^{2}}-\dfrac{2x}{3}-1=0$
Here $b=\dfrac{-2}{3}$
Hence, ${{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{\dfrac{-2}{3}}{2} \right)}^{2}}=\dfrac{1}{9}$
Therefore add $\dfrac{1}{9}$ on both sides of the equation to form a square.
$\Rightarrow \left( {{x}^{2}}-\dfrac{2x}{3}+\dfrac{1}{9} \right)=1+\dfrac{1}{9}$
Now we can see that there is a square on the LHS of the equation.
Upon rewriting it into a square we get,
$\Rightarrow {{\left( x-\dfrac{1}{3} \right)}^{2}}=\dfrac{10}{9}$
Now apply square root on both sides of the equation.
$\Rightarrow \sqrt{{{\left( x-\dfrac{1}{3} \right)}^{2}}}=\sqrt{\dfrac{10}{9}}$
Now evaluate further.
$\Rightarrow x-\dfrac{1}{3}=\pm \dfrac{\sqrt{10}}{3}$
Now solve for $x$
$\Rightarrow x= \dfrac{\pm \sqrt{10}}{3}+\dfrac{1}{3}$
$\Rightarrow x= \dfrac{\pm \sqrt{10}+1}{3}$
Hence the solution for the equation $3{{x}^{2}}-2x-3$ is $x=\dfrac{\sqrt{10}+1}{3};\dfrac{-\sqrt{10}+1}{3}$

Note: Never forget to take two conditions whenever there is a $\pm$ symbol. The expression is written twice once with $+\;$ and another time with $-\;$ . You can always cross-check your answer by placing the value of $x\;$ back in the equation. If you get LHS = RHS then your answer is correct.