Question

How do you complete the square to solve \[2{{x}^{2}}+16x+42=0\] ?

Answer 1

Hint: The equation that we need to complete the square of is \[2{{x}^{2}}+16x+42=0\] . First, we divide both sides by $2$ and get ${{x}^{2}}+8x+21=0$ . After that, we move the $21$ to the RHS, add ${{4}^{2}}=16$ on both sides of the equation, and then compare it with ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to get ${{\left( x+4 \right)}^{2}}=-5$ . Finally, square rooting and a little arithmetic gives us the required answer.

Complete step-by-step solution:
The given equation that we have is,
\[2{{x}^{2}}+16x+42=0\]
First, we divide the coefficient of every term by the coefficient of ${{x}^{2}}$ , which is $2$ . So, we get,
$\Rightarrow {{x}^{2}}+8x+21=0$
Subtracting $21$ from both sides of the equation, we get,
$\Rightarrow {{x}^{2}}+8x=-21....\left( ii \right)$
Now, we will consider the coefficient of x in the above expression, which is $8$ . We will take the half of this coefficient, which is,
$\dfrac{8}{2}=4$
We will now square this term and add the result to both sides of equation (i) and get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+8x+{{\left( 4 \right)}^{2}}=-21+{{\left( 4 \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+2\left( 4x \right)+16=-5....\left( ii \right) \\
\end{align}$
We can see that equation (ii) is of the form ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , where a is x and b is $4$ . Thus, we can write equation (ii) as,
$\Rightarrow {{\left( x+4 \right)}^{2}}=-5$
Taking square roots on both sides, we get,
$\begin{align}
  & \Rightarrow x+4=\pm \sqrt{5}i \\
 & \Rightarrow x=-4\pm \sqrt{5}i \\
\end{align}$
Thus, we can conclude that the solution is $-4\pm \sqrt{5}i$ .

Note: The problem can also be solved using the Sridhar Acharya method which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where “a” is $2$ , “b” is $16$ and “c” is $42$ . In our solution, we must be careful with the sign of the RHS term as only it decides whether the roots will be real or imaginary. Seeing imaginary roots, we must not get confused as they might be the required result.