Question

Solve by completing square method: \[{{x}^{2}}+3x+6=0\]

Answer 1

Hint: We will be applying the completing square method to find \[x\]. We complete the square by adding the same numbers to both sides of the quadratic equation and then try to get it in a square form.

Complete step-by-step answer:
Before proceeding with the question, we should know about completing the square method.
To solve quadratic equation of form \[a{{x}^{2}}+bx+c=0\] by completing the square:
(i) Transform the quadratic equation so that the constant term \[c\] is alone on the right side.
(ii) If \[a\], the leading coefficient is not equal to 1, divide both sides by \[a\].
(iii) Add the square of half the coefficient of \[x\] term, \[{{\left[ \dfrac{b}{2a} \right]}^{2}}\]to both sides of the equation.
(iv) Factor the left side as the square of a binomial.
(v) And now solve for \[x\].
So now following the above mentioned method we get,
 \[{{x}^{2}}+3x+6=0.....(1)\]
Subtracting 6 from both sides of equation (1) we get,
 \[\,\Rightarrow {{x}^{2}}+3x=-6.......(2)\]
Here coefficient: \[a=1\], \[b=3\] and \[c=6\]. Now adding \[{{\left[ \dfrac{3}{2} \right]}^{2}}\] to both sides of equation (2) we get,
 \[\,\Rightarrow {{x}^{2}}+3x+{{\left[ \dfrac{3}{2} \right]}^{2}}=-6+{{\left[ \dfrac{3}{2} \right]}^{2}}......(3)\]
Solving both sides of equation (3) we get;
 \[\begin{align}
  & \,\Rightarrow {{x}^{2}}+3x+\dfrac{9}{4}=-6+\dfrac{9}{4} \\
 & \,\Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{-15}{4}...........(4) \\
\end{align}\]
Taking the square root of both the sides of equation (4) we get,
 \[\begin{align}
  & \,\Rightarrow \left( x+\dfrac{3}{2} \right)=\pm \sqrt{\dfrac{-15}{4}} \\
 & \,\Rightarrow \left( x+\dfrac{3}{2} \right)=\pm \dfrac{\sqrt{-15}}{2}.......(5) \\
\end{align}\]
As \[\sqrt{-15}=\sqrt{15}\,i\] because \[i=\sqrt{-1}\]. So applying this in equation (5) and solving we get,
 \[\begin{align}
  & \,\Rightarrow \left( x+\dfrac{3}{2} \right)=\pm \dfrac{\sqrt{15}\,i}{2} \\
 & \,\Rightarrow x=-\dfrac{3}{2}\pm \dfrac{\sqrt{15}\,i}{2} \\
 & \,\Rightarrow x=\dfrac{-3+\sqrt{15}\,i}{2},\,\dfrac{-3-\sqrt{15}\,i}{2} \\
\end{align}\]
So \[x=\dfrac{-3+\sqrt{15}\,i}{2},\,\dfrac{-3-\sqrt{15}\,i}{2}\] is the answer.

Note: In this type of question where it is mentioned that we use the square method to solve, we have to memorize the steps involved in the method. We can make mistakes if we do not know that \[i=\sqrt{-1}\]. Also we may in a hurry write only one value of \[x\] where we get two answers of \[x\].