Question

sonya found that $x = -\frac{4}{3}$ is one solution of $(6x + 4)^2 = \boxed{?}$. the other solution is $x = \boxed{?}$

Explanation:

Step1: Expand the equation

First, expand \((6x + 4)^2\). Using the formula \((a + b)^2=a^2 + 2ab + b^2\), we have \((6x + 4)^2=(6x)^2+2\times6x\times4 + 4^2 = 36x^2+48x + 16\). But maybe a better way is to recognize that if \(x =-\frac{4}{3}\) is a solution, we can first find the value of the right - hand side when \(x =-\frac{4}{3}\).

Substitute \(x =-\frac{4}{3}\) into \(6x + 4\): \(6\times(-\frac{4}{3})+4=-8 + 4=-4\). Then \((6x + 4)^2=(-4)^2 = 16\). So the equation is \((6x + 4)^2=16\).

Step2: Solve the equation \((6x + 4)^2=16\)

Take the square root of both sides: \(6x+4=\pm\sqrt{16}=\pm4\).

We already know that when \(6x + 4=-4\), we have the solution \(x =-\frac{4}{3}\) (by solving \(6x+4=-4\Rightarrow6x=-8\Rightarrow x =-\frac{4}{3}\)).

Now, solve \(6x + 4 = 4\):
Subtract 4 from both sides: \(6x+4 - 4=4 - 4\Rightarrow6x = 0\).
Divide both sides by 6: \(x=\frac{0}{6}=0\).

Answer:

The other solution is \(x = 0\).