Question

sonya found that ( x = -\frac{4}{3} ) is one solution of ( (6x + 4)^2 = )
the other solution is ( x = )
-16
-8
8
16

Explanation:

Step1: Find the constant term

Given \( x = -\frac{4}{3} \) is a solution of \( (6x + 4)^2 = k \) (let the constant be \( k \)). Substitute \( x = -\frac{4}{3} \) into \( 6x + 4 \):
\( 6\times(-\frac{4}{3}) + 4 = -8 + 4 = -4 \)
Then \( k = (-4)^2 = 16 \), so the equation is \( (6x + 4)^2 = 16 \).

Step2: Solve the equation

Take square roots on both sides: \( 6x + 4 = \pm\sqrt{16}=\pm4 \)
We already know one solution when \( 6x + 4 = -4 \) (since \( x = -\frac{4}{3} \) gives \( 6x + 4 = -4 \)). Now solve for the other solution when \( 6x + 4 = 4 \):
Subtract 4 from both sides: \( 6x = 4 - 4 = 0 \)
Divide by 6: \( x = \frac{0}{6} = 0 \)

Answer:

The constant term is \( 16 \), and the other solution is \( x = 0 \)