Question

How do you solve \[3{{x}^{2}}+6x+6=0\] by completing the square?

Answer 1

Hint: Divide both the sides with 3 to simplify the equation. Now, assume the obtained expression as y and compare it with the general form given as: - \[y=a{{x}^{2}}+bx+c\]. Find the respective values of a, b and c. Now, find the discriminant of the given quadratic equation by using the formula: - \[D={{b}^{2}}-4ac\], where D = discriminant. Now, write the expression as: - \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\] and substitute it equal to 0 to find the two values of x. If the value of D is negative then the roots will be of the form of complex numbers.

Complete step by step answer:
Here, we have been provided with the quadratic equation: - \[3{{x}^{2}}+6x+6=0\] and we are asked to solve it. That means we have to find the values of x. We have been asked to use completing the square method.
As we can see that 3 is common in all the terms, so dividing both the sides with 3 we get,
$\Rightarrow {{x}^{2}}+2x+2=0$
Now, we know that any quadratic equation of the form \[y=a{{x}^{2}}+bx+c\] can be simplified as \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\], using completing the square method. Here, ‘D’ denotes the discriminant. So, on assuming ${{x}^{2}}+2x+2=y$ and comparing it with the general quadratic equation, we get,
\[\Rightarrow \] a = 1, b = 2, c = 2
Applying the formula for discriminant of a quadratic equation given as, \[D={{b}^{2}}-4ac\], we get,
\[\begin{align}
  & \Rightarrow D={{\left( 2 \right)}^{2}}-4\left( 1 \right)\left( 2 \right) \\
 & \Rightarrow D=4-8 \\
 & \Rightarrow D=-4 \\
\end{align}\]
As we can see that the value of D is negative it means that we will not get any real roots but we need to find the complex roots. Therefore, substituting the values in the simplified form of y, we get,
\[\begin{align}
  & \Rightarrow y=1\left[ {{\left( x+\left( \dfrac{2}{2\times 1} \right) \right)}^{2}}-\dfrac{\left( -4 \right)}{4\times {{\left( 1 \right)}^{2}}} \right] \\
 & \Rightarrow y=\left[ {{\left( x+1 \right)}^{2}}+1 \right] \\
\end{align}\]
Substituting y = 0, we get,
\[\Rightarrow \left[ {{\left( x+1 \right)}^{2}}+1 \right]=0\]
\[\begin{align}
  & \Rightarrow {{\left( x+1 \right)}^{2}}+1=0 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}=-1 \\
\end{align}\]
Taking square root both the sides, we get,
\[\begin{align}
  & \Rightarrow \left( x+1 \right)=\pm \sqrt{-1} \\
 & \Rightarrow x+1=\pm i \\
 & \Rightarrow x=-1\pm i \\
\end{align}\]
Considering the two signs one – by – one, we get,
\[\Rightarrow x=\left( -1+i \right)\] or \[x=\left( -1-i \right)\]

Hence, the solutions of the given quadratic equation are: \[x=\left( -1+i \right)\] and \[x=\left( -1-i \right)\].

Note: One may note that here we are not getting any real roots of the given equation because the value of the discriminant is less than 0. Remember that we get the real roots of a quadratic only when the discriminant value is greater than or equal to 0. One may also use the quadratic formula to solve the question and check if we are getting the same values of x. Note that the general expression of this completing the square method is also known as the vertex form. Generally, this form is used in coordinate geometry of parabola to find the vertex of the parabola which is given as \[\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)\]. The quadratic formula: - \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] is derived from completing the square method.