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3.4 hmwk #1-9 solve the equation using the quadratic formula. (see exam…

Question

3.4 hmwk #1-9
solve the equation using the quadratic formula. (see examples 1, 2,

  1. ( x^2 - 4x + 3 = 0 )
  2. ( x^2 + 6x + 15 = 0 )

Explanation:

Response
Problem 1: \( x^2 - 4x + 3 = 0 \)

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

For a quadratic equation \( ax^2 + bx + c = 0 \), here \( a = 1 \), \( b = -4 \), \( c = 3 \).

Step 2: Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a \), \( b \), \( c \):
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} \)

Step 3: Simplify discriminant

Calculate discriminant \( D = b^2 - 4ac = 16 - 12 = 4 \).

Step 4: Solve for \( x \)

\( x = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2} \).

  • For \( + \): \( x = \frac{4 + 2}{2} = 3 \)
  • For \( - \): \( x = \frac{4 - 2}{2} = 1 \)

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

For \( ax^2 + bx + c = 0 \), \( a = 1 \), \( b = 6 \), \( c = 15 \).

Step 2: Quadratic Formula

Use \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute values:
\( x = \frac{-6 \pm \sqrt{6^2 - 4(1)(15)}}{2(1)} \)

Step 3: Simplify discriminant

Discriminant \( D = 36 - 60 = -24 \).

Step 4: Solve (complex numbers)

Since \( D < 0 \), solutions are complex: \( x = \frac{-6 \pm \sqrt{-24}}{2} = \frac{-6 \pm 2i\sqrt{6}}{2} = -3 \pm i\sqrt{6} \)

Answer:

\( x = 1 \) or \( x = 3 \)

Problem 2: \( x^2 + 6x + 15 = 0 \)