Question

solve each of the quadratic equations.
3x = 0.5x²
× x = -6 or x = 0
x = -4 or x = 3
x = -2 or x = 1.5
✔️ x = 0 or x = 6
complete
0 = 5x² - 2x + 6
x= (1±3i)/2
x= (1±√11)/5
x= (1±i√29)/5
done

Explanation:

Step1: Solve \(3x = 0.5x^2\)

Rearrange the equation to standard quadratic form: \(0.5x^2 - 3x = 0\). Multiply both sides by 2 to eliminate the decimal: \(x^2 - 6x = 0\). Factor out \(x\): \(x(x - 6) = 0\). Set each factor to zero: \(x = 0\) or \(x - 6 = 0 \implies x = 6\). So the solution is \(x = 0\) or \(x = 6\), which matches the checked option.

Step2: Solve \(0 = 5x^2 - 2x + 6\) (or \(5x^2 - 2x + 6 = 0\))

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -2\), \(c = 6\). Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(5)(6) = 4 - 120 = -116\). Since the discriminant is negative, we have complex roots. Simplify \(\sqrt{-116} = \sqrt{4 \times (-29)} = 2i\sqrt{29}\). Then \(x = \frac{-(-2) \pm 2i\sqrt{29}}{2(5)} = \frac{2 \pm 2i\sqrt{29}}{10} = \frac{1 \pm i\sqrt{29}}{5}\). Wait, wait, let's recalculate discriminant: Wait, \(b = -2\), so \(b^2 = 4\), \(4ac = 4*5*6 = 120\), so \(4 - 120 = -116\)? Wait no, \(b\) is \(-2\), so \(-b = 2\). Wait, the formula is \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), so \(a = 5\), \(b = -2\), \(c = 6\). So \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 456}}{25} = \frac{2 \pm \sqrt{4 - 120}}{10} = \frac{2 \pm \sqrt{-116}}{10}\). Simplify \(\sqrt{-116} = \sqrt{4(-29)} = 2\sqrt{29}i\), so \(x = \frac{2 \pm 2\sqrt{29}i}{10} = \frac{1 \pm \sqrt{29}i}{5}\), which is the third option: \(x = \frac{1 \pm i\sqrt{29}}{5}\).

Answer:

For \(3x = 0.5x^2\): \(x = 0\) or \(x = 6\) (the checked option).
For \(0 = 5x^2 - 2x + 6\): \(x = \frac{1 \pm i\sqrt{29}}{5}\) (the third option, \(x = \frac{1 \pm i\sqrt{29}}{5}\)).