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
done
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 \(ax^2+bx + c = 0\). Subtract \(3x\) from both sides: \(0.5x^2-3x = 0\). Multiply both sides by 2 to eliminate the decimal: \(x^2 - 6x=0\). Factor out \(x\): \(x(x - 6)=0\). By the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So \(x=0\) or \(x - 6=0\), which gives \(x = 0\) or \(x=6\).

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

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) for a quadratic equation \(ax^2+bx + c = 0\). Here, \(a = 5\), \(b=-2\), and \(c = 6\). First, calculate the discriminant \(\Delta=b^2-4ac=(-2)^2-4\times5\times6=4 - 120=- 116\). Since \(\Delta=-116=4\times(-29)\), \(\sqrt{\Delta}=\sqrt{-116}=2i\sqrt{29}\). Then \(x=\frac{-(-2)\pm2i\sqrt{29}}{2\times5}=\frac{2\pm2i\sqrt{29}}{10}=\frac{1\pm i\sqrt{29}}{5}\).

Answer:

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