Question

How do you solve $3{{x}^{2}}-6x-1=0$ using completing the square?

Answer 1

Hint: In this question we will first reduce the term $3{{x}^{2}}$ into the simpler form and then do the necessary addition and subtraction so that the square can be completing which means we will reverse the formula of expansion ${{(a-b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and then simplify further to solve for the value of $x$.

Complete step-by-step answer:
We have the expression as $3{{x}^{2}}-6x-1=0$
On transferring the term $5$ from the left-hand side to the right-hand side, we get:
$\Rightarrow 3{{x}^{2}}-6x=1$
Now to reduce the term $3{{x}^{2}}$, we will divide both the sides of the expression by $3$.
On dividing, we get:
$\Rightarrow \dfrac{3{{x}^{2}}}{3}-\dfrac{6}{3}x=\dfrac{1}{3}$
On simplifying the left-hand side of the expression, we get:
$\Rightarrow {{x}^{2}}-2x=\dfrac{1}{3}$
Now on the left-hand side, we have two terms which are ${{x}^{2}}$ and $-2x$, now these two terms can be simplified if the square is completed which means if a constant value is added such that it is can be simplified as the whole square form as ${{(a-b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
Now we know the coefficient of ${{x}^{2}}$ is $1$ therefore, we will add value $1$ on both sides, to get $2ab=4$ .
Therefore, the expression can be written as:
$\Rightarrow {{x}^{2}}-2x+1=\dfrac{1}{3}+1$
On simplifying the right-hand side, we get:
$\Rightarrow {{x}^{2}}-2x+1=\dfrac{4}{3}$
On completing the square, we get:

$\Rightarrow {{(x-1)}^{2}}=\dfrac{4}{3}$
On taking the square root on both the sides, we get:
\[\Rightarrow x-1=\pm \sqrt{\dfrac{4}{3}}\]
On transferring $1$ from the left-hand side to the right-hand side, we get:
\[\Rightarrow x=1\pm \sqrt{\dfrac{4}{3}}\]
Therefore, the two solution to the equation are ${{x}_{1}}=1+\sqrt{\dfrac{4}{3}}$ and ${{x}_{2}}=1-\sqrt{\dfrac{4}{3}}$.

Note: It is to be noted that these types of equations are called polynomial equations and a polynomial equation with a degree two is called a quadratic equation.
A quadratic equation can be solved by splitting the middle term or by using the quadratic formula which is $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.