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

How do you complete the square to solve 2x2+16x+42=02{{x}^{2}}+16x+42=0 ?

Explanation

Solution

The equation that we need to complete the square of is 2x2+16x+42=02{{x}^{2}}+16x+42=0 . First, we divide both sides by 22 and get x2+8x+21=0{{x}^{2}}+8x+21=0 . After that, we move the 2121 to the RHS, add 42=16{{4}^{2}}=16 on both sides of the equation, and then compare it with (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} to get (x+4)2=−5{{\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,
2x2+16x+42=02{{x}^{2}}+16x+42=0
First, we divide the coefficient of every term by the coefficient of x2{{x}^{2}} , which is 22 . So, we get,
⇒x2+8x+21=0\Rightarrow {{x}^{2}}+8x+21=0
Subtracting 2121 from both sides of the equation, we get,
⇒x2+8x=−21....(ii)\Rightarrow {{x}^{2}}+8x=-21....\left( ii \right)
Now, we will consider the coefficient of x in the above expression, which is 88 . We will take the half of this coefficient, which is,
82=4\dfrac{8}{2}=4
We will now square this term and add the result to both sides of equation (i) and get,
⇒x2+8x+(4)2=−21+(4)2 ⇒x2+2(4x)+16=−5....(ii) \begin{aligned} & \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{aligned}
We can see that equation (ii) is of the form (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} , where a is x and b is 44 . Thus, we can write equation (ii) as,
⇒(x+4)2=−5\Rightarrow {{\left( x+4 \right)}^{2}}=-5
Taking square roots on both sides, we get,
⇒x+4=±5i ⇒x=−4±5i \begin{aligned} & \Rightarrow x+4=\pm \sqrt{5}i \\\ & \Rightarrow x=-4\pm \sqrt{5}i \\\ \end{aligned}
Thus, we can conclude that the solution is −4±5i-4\pm \sqrt{5}i .

Note: The problem can also be solved using the Sridhar Acharya method which is x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} where “a” is 22 , “b” is 1616 and “c” is 4242 . 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.