Question

1. $0 = 3x^2 - 2x - 5$

$+2) = 0$
$+2 = 0$
$overline{x = -2}$

1. $3x^2 + 10x = 8$

$3x$

Explanation:

Problem 8: Solve \( 0 = 3x^2 - 2x - 5 \) (or \( 3x^2 - 2x - 5 = 0 \))

Step 1: Factor the quadratic equation

We need to find two numbers that multiply to \( 3\times(-5)= -15 \) and add up to \( -2 \). The numbers are \( -5 \) and \( 3 \).
Rewrite the middle term using these numbers:
\( 3x^2 + 3x - 5x - 5 = 0 \)
Factor by grouping:
\( 3x(x + 1) - 5(x + 1) = 0 \)
\( (3x - 5)(x + 1) = 0 \)

Step 2: Solve for \( x \) using the zero - product property

If \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
Case 1: \( 3x - 5 = 0 \)
Add 5 to both sides: \( 3x=5 \)
Divide both sides by 3: \( x=\frac{5}{3} \)
Case 2: \( x + 1 = 0 \)
Subtract 1 from both sides: \( x=-1 \)

Problem 11: Solve \( 3x^2 + 10x = 8 \) (rewrite as \( 3x^2 + 10x - 8 = 0 \))

Step 1: Factor the quadratic equation

We need to find two numbers that multiply to \( 3\times(-8)=-24 \) and add up to \( 10 \). The numbers are \( 12 \) and \( - 2 \).
Rewrite the middle term:
\( 3x^2+12x - 2x - 8 = 0 \)
Factor by grouping:
\( 3x(x + 4)-2(x + 4)=0 \)
\( (3x - 2)(x + 4)=0 \)

Step 2: Solve for \( x \) using the zero - product property

Case 1: \( 3x - 2 = 0 \)
Add 2 to both sides: \( 3x = 2 \)
Divide both sides by 3: \( x=\frac{2}{3} \)
Case 2: \( x + 4 = 0 \)
Subtract 4 from both sides: \( x=-4 \)

Answer:

s:

• For \( 3x^2 - 2x - 5 = 0 \), \( x=-1 \) or \( x = \frac{5}{3} \)
• For \( 3x^2+10x - 8 = 0 \), \( x=-4 \) or \( x=\frac{2}{3} \)