Question

How do you solve $3{{x}^{2}}-12x+5=0$ by 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 $3{{x}^{2}}-12x+5=0$
On transferring the term $5$ from the left-hand side to the right-hand side, we get:
$\Rightarrow 3{{x}^{2}}-12x=-5$
Now to reduce the term $3{{x}^{2}}$, we will divide both the sides of the expression by $3$.
On dividing, we get:
$\Rightarrow {{x}^{2}}-\dfrac{12}{3}x=-\dfrac{5}{3}$
On simplifying the left-hand side of the expression, we get:
$\Rightarrow {{x}^{2}}-4x=-\dfrac{5}{3}$
Now on the left-hand side, we have two terms which are ${{x}^{2}}$and $-4x$, 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 and subtract $4$, to get $2ab=4$.
Therefore, the expression can be written as:
$\Rightarrow {{x}^{2}}-4x+4-4=-\dfrac{5}{3}$
On completing the square, we get:
$\Rightarrow {{\left( x-2 \right)}^{2}}-4=-\dfrac{5}{3}$
On transferring $4$from the left-hand side to the right-hand side, we get:
$\Rightarrow {{\left( x-2 \right)}^{2}}=4-\dfrac{5}{3}$
On taking the lowest common multiple and simplifying, we get:
$\Rightarrow {{\left( x-2 \right)}^{2}}=\dfrac{7}{3}$
On taking the square root on both the sides, we get:
$\Rightarrow x-2=\pm \sqrt{\dfrac{7}{3}}$
On transferring $2$from the left-hand side to the right-hand side, we get:
$\Rightarrow x=2\pm \sqrt{\dfrac{7}{3}}$

Therefore, the two solution to the equation are ${{x}_{1}}=2+\sqrt{\dfrac{7}{3}}$ and $\Rightarrow {{x}_{2}}=2-\sqrt{\dfrac{7}{3}}$.

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.