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Question: How do you solve the quadratic equation \(2{{x}^{2}}+4x=-3\) ?...

How do you solve the quadratic equation 2x2+4x=−32{{x}^{2}}+4x=-3 ?

Explanation

Solution

We solve the problem by “completing the square” method. Finally, after some rearrangements, we get an equation of the form a(x+d)2+e=0a{{\left( x+d \right)}^{2}}+e=0. Taking square roots on both sides, we get two complex roots of xx .

Complete step-by-step solution:
The given equation is
2x2+4x=−32{{x}^{2}}+4x=-3
We will solve the quadratic equation by completing the square. Squaring a quadratic expression simply means representing the quadratic equation as the sum of squares of a linear expression plus a constant term. Let us consider the general quadratic expression ax2+bx+ca{{x}^{2}}+bx+c. Squaring it would mean representing it in the form a(x+d)2+ea{{\left( x+d \right)}^{2}}+e .
Let us first add 33 to both sides of the given equation. The equation thus becomes,
⇒2x2+4x+3=0\Rightarrow 2{{x}^{2}}+4x+3=0
As we can see that the coefficient of x2{{x}^{2}} is 22 . So, we take 22 common from the first two terms of the equation. The equation thus becomes
⇒2(x2+2x)+3=0\Rightarrow 2\left( {{x}^{2}}+2x \right)+3=0
If we observe closely now, we can see that add we add 11 within the bracket, the entire expression in the bracket becomes x2+2x+1{{x}^{2}}+2x+1 which is nothing but (x+1)2{{\left( x+1 \right)}^{2}}. If we add 11 inside the bracket, then we also have to subtract 22 outside the bracket.
⇒2(x2+2x+1)+3−2=0 ⇒2(x+1)2+1=0 \begin{aligned} & \Rightarrow 2\left( {{x}^{2}}+2x+1 \right)+3-2=0 \\\ & \Rightarrow 2{{\left( x+1 \right)}^{2}}+1=0 \\\ \end{aligned}
We subtract 11 from both sides of the equation.
⇒2(x+1)2=−1\Rightarrow 2{{\left( x+1 \right)}^{2}}=-1
We divide both sides of the equation by 22 .
⇒(x+1)2=−12\Rightarrow {{\left( x+1 \right)}^{2}}=-\dfrac{1}{2}
Now, we can clearly see that taking square roots on both sides will yield complex terms as according to the above equation, the square of a number is negative. So, we can write after taking square roots,
⇒(x+1)=±12i\Rightarrow \left( x+1 \right)=\pm \dfrac{1}{\sqrt{2}}i
Subtracting 11 from both sides, we get
⇒x=−1±12i\Rightarrow x=-1\pm \dfrac{1}{\sqrt{2}}i
Therefore, we can conclude that the solution for 2x2+4x=−32{{x}^{2}}+4x=-3 is x=−1±12ix=-1\pm \dfrac{1}{\sqrt{2}}i.

Note: Special attention must be paid while arranging the terms of the equation and adding and subtracting some terms for the completion of the square. Most of the students commit mistakes here. We can also use Sridhar Acharya’s formula x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} to find the roots of the equation.