Question

extra practice problems:
solve each using the quadratic formula.

1. ( x^2 + 12x + 20 = 0 )
2. ( 4x^2 + 8x + 3 = 0 )
3. ( 3x^2 + 2x - 8 = 0 )

Explanation:

Problem 1: \( x^2 + 12x + 20 = 0 \)

Step 1: Identify \( a \), \( b \), \( c \)

For \( ax^2 + bx + c = 0 \), here \( a = 1 \), \( b = 12 \), \( c = 20 \).

Step 2: Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substitute values: \( x = \frac{-12 \pm \sqrt{12^2 - 4(1)(20)}}{2(1)} \)

Step 3: Calculate discriminant

\( 12^2 - 4(1)(20) = 144 - 80 = 64 \)

Step 4: Simplify square root and solve

\( \sqrt{64} = 8 \), so \( x = \frac{-12 \pm 8}{2} \)
Two solutions:

• \( x = \frac{-12 + 8}{2} = \frac{-4}{2} = -2 \)
• \( x = \frac{-12 - 8}{2} = \frac{-20}{2} = -10 \)

Step 1: Identify \( a \), \( b \), \( c \)

Here \( a = 4 \), \( b = 8 \), \( c = 3 \).

Step 2: Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substitute: \( x = \frac{-8 \pm \sqrt{8^2 - 4(4)(3)}}{2(4)} \)

Step 3: Calculate discriminant

\( 8^2 - 4(4)(3) = 64 - 48 = 16 \)

Step 4: Simplify square root and solve

\( \sqrt{16} = 4 \), so \( x = \frac{-8 \pm 4}{8} \)
Two solutions:

• \( x = \frac{-8 + 4}{8} = \frac{-4}{8} = -\frac{1}{2} \)
• \( x = \frac{-8 - 4}{8} = \frac{-12}{8} = -\frac{3}{2} \)

Step 1: Identify \( a \), \( b \), \( c \)

Here \( a = 3 \), \( b = 2 \), \( c = -8 \).

Step 2: Quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substitute: \( x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-8)}}{2(3)} \)

Step 3: Calculate discriminant

\( 2^2 - 4(3)(-8) = 4 + 96 = 100 \)

Step 4: Simplify square root and solve

\( \sqrt{100} = 10 \), so \( x = \frac{-2 \pm 10}{6} \)
Two solutions:

• \( x = \frac{-2 + 10}{6} = \frac{8}{6} = \frac{4}{3} \)
• \( x = \frac{-2 - 10}{6} = \frac{-12}{6} = -2 \)

Answer:

\( x = -2 \) or \( x = -10 \)

Problem 2: \( 4x^2 + 8x + 3 = 0 \)