Question

1. what are the solutions to the equation: ( y = 3x^2 - 2x - 8 )?

a. ( x = -4 ) is the only solution
b. ( x = -2 ) and ( x = \frac{4}{3} )
c. ( x = 4 ) and ( x = -6 )
d. ( x = 2 ) and ( x = -\frac{4}{3} )

Explanation:

Step1: Set equation to zero

To find solutions, set \( y = 0 \): \( 3x^2 - 2x - 8 = 0 \).

Step2: Factor the quadratic

Factor \( 3x^2 - 2x - 8 \). We need two numbers \( a,b \) such that \( a \times b = 3 \times (-8)= -24 \) and \( a + b = -2 \). The numbers are \( -6 \) and \( 4 \). Rewrite middle term: \( 3x^2 - 6x + 4x - 8 = 0 \). Group: \( 3x(x - 2) + 4(x - 2) = 0 \). Factor: \( (3x + 4)(x - 2) = 0 \).

Step3: Solve for x

Set each factor to zero:

• \( 3x + 4 = 0 \Rightarrow 3x = -4 \Rightarrow x = -\frac{4}{3} \)
• \( x - 2 = 0 \Rightarrow x = 2 \)

Answer:

d. \( x = 2 \) and \( x = -\frac{4}{3} \)