Question

How to complete square $3{{x}^{2}}-6x-4=0$?

Answer 1

Hint: Now we are given with a quadratic equation of the form $a{{x}^{2}}+bx+c$ . We will first divide the whole equation by a and then add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ in the equation. Now we will simplify the equation with the help of formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . Now we have a complete square in the equation. We will now rearrange the equation and hence take all the constant terms to right. Now taking square root of the equation we will get two linear equation in x. We solve the equation to find the value of x and hence we have the root of the given quadratic.

Complete step-by-step answer:
Now the given equation $3{{x}^{2}}-6x-4=0$ is a quadratic equation in x of the form $a{{x}^{2}}+bx+c$ where a = 3, b = -6 and c = -4.
Now to solve the quadratic using complete square we first need the coefficient of ${{x}^{2}}$ to be 1.
Hence we will divide the equation by 3 hence we get,
$\Rightarrow {{x}^{2}}-2x-\dfrac{4}{3}=0$
Now we will add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ which in this case is ${{\left( \dfrac{-6}{6} \right)}^{2}}=1$
Hence we have,
$\Rightarrow {{x}^{2}}-2x+1-1-\dfrac{4}{3}=0$
Now we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ Hence we get,
$\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{4}{3}+1$
Taking LCM in the equation we get,
$\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{4+3}{3}$
$\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{7}{3}$
Now we will take the square root on both sides. Hence we get,
$\Rightarrow x-1=\dfrac{\pm \sqrt{7}}{\sqrt{3}}$
Hence the value of x is $1-\sqrt{\dfrac{7}{3}}$ and $1+\sqrt{\dfrac{7}{3}}$ .

Note: Now note that while taking the square root of any quantity we get two solutions. One positive solution and one negative solution. This is because the square of a negative number is equal to the square of the same positive number. For example we have ${{\left( -2 \right)}^{2}}={{2}^{2}}=4$ .