Question

solve each of the quadratic equations.

(3x = 0.5x^2)

• (x = -6) or (x = 0)
• (x = -4) or (x = 3)
• (x = -2) or (x = 1.5)
• (x = 0) or (x = 6)

(0 = 5x^2 - 2x + 6)

• (x = \frac{1 pm 3i}{2})
• (x = \frac{1 pm sqrt{11}}{5})
• (x = \frac{1 pm isqrt{29}}{5})

Explanation:

⚡ Using what you learned: factoring and solving quadratic equations · quadratic formula and its applications

Step 1: Solve the first equation

$$ 3x = 0.5x^2 $$

$$ 0.5x^2 - 3x = 0 $$

Multiply by 2 to clear the decimal:

$$ x^2 - 6x = 0 $$

Factor out \(x\):

$$ x(x - 6) = 0 $$

$$ x = 0 \quad \text{or} \quad x = 6 $$

Step 2: Solve the second equation

$$ 5x^2 - 2x + 6 = 0 $$

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 5\), \(b = -2\), and \(c = 6\):

$$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(5)(6)}}{2(5)} $$

$$ x = \frac{2 \pm \sqrt{4 - 120}}{10} $$

$$ x = \frac{2 \pm \sqrt{-116}}{10} $$

Simplify the radical using \(i = \sqrt{-1}\):

$$ \sqrt{-116} = \sqrt{-1 \cdot 4 \cdot 29} = 2i\sqrt{29} $$

$$ x = \frac{2 \pm 2i\sqrt{29}}{10} $$

Divide numerator and denominator by 2:

$$ x = \frac{1 \pm i\sqrt{29}}{5} $$

Answer:

• First equation (\(3x = 0.5x^2\)):

\(x = 0 \text{ or } x = 6\)

• Second equation (\(0 = 5x^2 - 2x + 6\)):

\(x = \frac{1 \pm i\sqrt{29}}{5}\)