Question

How do you solve the equation ${{x}^{2}}+2x+6=0$ by completing the square?

Answer 1

Hint: To solve the equation ${{x}^{2}}+2x+6=0$ by completing the square method. For that, we are required to first rearrange this equation in such a manner so that we can make some part of the equation as a perfect square of the following form: ${{\left( a+b \right)}^{2}}\And {{\left( a-b \right)}^{2}}$. And then simplify the given equation.

Complete step by step answer:
The equation given in the above problem is as follows:
${{x}^{2}}+2x+6=0$
As you can see that the degree of the above equation is 2 so the above equation is a quadratic equation. And now, we are going to make some part of the above equation as a perfect square. Rewriting the above equation we get,
${{x}^{2}}+2.1.\dfrac{1}{2}x+6=0$
If you carefully look at the above equation then you will find that the first two terms in the L.H.S of the above equation has the potential to convert into perfect square and which can be achieved by adding and subtracting $\dfrac{1}{4}$ in the L.H.S of the above equation and we get,
$\Rightarrow {{x}^{2}}+2.1.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=0$
Now, writing $\dfrac{1}{4}$ written with the plus sign in the above equation as ${{\left( \dfrac{1}{2} \right)}^{2}}$ then we get,
$\Rightarrow {{x}^{2}}+2.1.\dfrac{1}{2}x+{{\left( \dfrac{1}{2} \right)}^{2}}-\dfrac{1}{4}+6=0$
Solving the last two terms in the L.H.S of the above equation will give us:
$\begin{align}
  & \Rightarrow {{x}^{2}}+2.1.\dfrac{1}{2}x+{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{6\left( 4 \right)-1}{4}=0 \\
 & \Rightarrow {{x}^{2}}+2.1.\dfrac{1}{2}x+{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{24-1}{4}=0 \\
 & \Rightarrow {{x}^{2}}+2.1.\dfrac{1}{2}x+{{\left( \dfrac{1}{2} \right)}^{2}}+\dfrac{23}{4}=0 \\
\end{align}$
Now, as you can see that first three terms in the L.H.S of the above equation is of the form:
${{x}^{2}}+2.1.\dfrac{1}{2}x+{{\left( \dfrac{1}{2} \right)}^{2}}={{\left( x+\dfrac{1}{2} \right)}^{2}}$
Using the above relation in the equation written just above this identity then we get,
$\Rightarrow {{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{23}{4}=0$
Now, we can write $\dfrac{23}{4}$ as $-\left( -\dfrac{23}{4} \right)$ then the above equation will look as follows:
${{\left( x+\dfrac{1}{2} \right)}^{2}}-\left( -\dfrac{23}{4} \right)=0$
Rewriting the above equation we get,
$\Rightarrow {{\left( x+\dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{23}}{2}i \right)}^{2}}=0$ ………….. Eq. (1)
In the above equation, $''i''$ is the iota in a complex number and we know that ${{i}^{2}}=-1$ . Now, we can use the following identity expansion in the above equation:
$\Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Substituting $a=\left( x+\dfrac{1}{2} \right)$ and $b=\dfrac{\sqrt{23}}{2}i$ in the above equation then we get,
$\Rightarrow {{\left( x+\dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{23}}{2}i \right)}^{2}}=\left( x+\dfrac{1}{2}-\dfrac{\sqrt{23}}{2}i \right)\left( x+\dfrac{1}{2}+\dfrac{\sqrt{23}}{2}i \right)$
Using the above relation in eq. (1) we get,
$\Rightarrow \left( x+\dfrac{1}{2}-\dfrac{\sqrt{23}}{2}i \right)\left( x+\dfrac{1}{2}+\dfrac{\sqrt{23}}{2}i \right)=0$
Equating each bracket to 0 we get,
$\begin{align}
  & \Rightarrow \left( x+\dfrac{1}{2}-\dfrac{\sqrt{23}}{2}i \right)=0 \\
 & \Rightarrow x=\dfrac{\sqrt{23}}{2}i-\dfrac{1}{2} \\
 & \Rightarrow \left( x+\dfrac{1}{2}+\dfrac{\sqrt{23}}{2}i \right)=0 \\
 & \Rightarrow x=-\left( \dfrac{1}{2}+\dfrac{\sqrt{23}}{2}i \right) \\
\end{align}$
Hence, we got the solutions of the above equation as follows:
$\begin{align}
  & \Rightarrow x=\dfrac{\sqrt{23}}{2}i-\dfrac{1}{2}; \\
 & x=-\left( \dfrac{1}{2}+\dfrac{\sqrt{23}}{2}i \right) \\
\end{align}$

Note: As you can see that the solutions we are getting are imaginary because $''i''$ (or iota) complex number is coming and we can verify whether the solutions we are getting are imaginary or not by analyzing the discriminant (or D) of the given quadratic equation.
We know that the discriminant of the following quadratic equation is as follows:
$a{{x}^{2}}+bx+c=0$
$D={{b}^{2}}-4ac$
Now, calculating the value of D from given equation ${{x}^{2}}+2x+6=0$ we get,
$\begin{align}
  & \Rightarrow D={{2}^{2}}-4\left( 1 \right)\left( 6 \right) \\
 & \Rightarrow D=4-24 \\
 & \Rightarrow D=-20 \\
\end{align}$
As you can see that the value of discriminant we are getting is negative so the solutions of this quadratic equation are imaginary.