Question

How do you use completing the square method to solve 4x2+5x=−1?

Answer 1

The given equation is actually,

$\mathsf{4x^2 + 5x = -1 }$

Consider it.

Here the coefficient of x² is 4. We have to get it as 1. For this, divide both sides by 4.

$\displaystyle \begin{aligned}&\mathsf{\frac{4x^2 + 5x}{4}=\frac{-1}{4}}\\ \\ \Longrightarrow\ \ &\mathsf{x^2+\frac{5}{4}x=-\frac{1}{4}}\end{aligned}$

Now consider the coefficient of x, i.e., 5/4. The next step is to add the square of half of this 5/4 to both sides of the equation.

→  Half of 5/4 is 5/8.

→  Square of 5/8 is 25/64.

So, add 25/64 to both sides.

$\displaystyle \mathsf{x^2+\frac{5}{4}x+\frac{25}{64}=-\frac{1}{4}+\frac{25}{64}}$

Now, when we factorize the LHS and equalize the denominators of the fractions in the RHS, we get,

$\displaystyle \begin{aligned}&\mathsf{\left(x+\frac{5}{8}\right)^2=-\frac{16}{64}+\frac{25}{64}}\\ \\ \Longrightarrow\ \ &\mathsf{\left(x+\frac{5}{8}\right)^2=\frac{25-16}{64}}\\ \\ \Longrightarrow\ \ &\mathsf{\left(x+\frac{5}{8}\right)^2=\frac{9}{64}}\\ \\ \Longrightarrow\ \ &\mathsf{x+\frac{5}{8}=\sqrt{\frac{9}{64}}}\\ \\ \Longrightarrow\ \ &\mathsf{x+\frac{5}{8}=\pm \frac{3}{8}}\\ \\ \Longrightarrow\ \ &\mathsf{x=\pm \frac{3}{8}-\frac{5}{8}}\end{aligned}$

$\displaystyle \begin{aligned}\Longrightarrow\ \ &\mathsf{x=\frac{3-5}{8}}\ &OR\ \ \ \ \ \ &\mathsf{x=\frac{-3-5}{8}}\\ \\ \Longrightarrow\ \ &\mathsf{x=\frac{-2}{8}}\ &OR\ \ \ \ \ \ &\mathsf{x=\frac{-8}{8}}\\ \\ \Longrightarrow\ \ &\mathsf{x=\mathbf{-\frac{1}{4}}}\ &OR\ \ \ \ \ \ &\mathsf{x=\mathbf{-1}}\end{aligned}$

Hence we found out the solutions by 'completing the square'!