Question

1. $2x^2 + 3x + 3 = 0$
2. $3x^2 - 3x + 1 = 0$

Explanation:

Problem 28: \(2x^2 + 3x + 3 = 0\)

Step 1: Identify coefficients

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

Step 2: Calculate discriminant \(\Delta\)

Discriminant formula: \(\Delta = b^2 - 4ac\)
Substitute values: \(\Delta = 3^2 - 4\times2\times3 = 9 - 24 = -15\)

Step 3: Find roots (complex numbers)

Quadratic formula: \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\)
Since \(\Delta = -15\), \(\sqrt{\Delta} = \sqrt{-15} = i\sqrt{15}\) (where \(i\) is imaginary unit)
Substitute into formula: \(x = \frac{-3 \pm i\sqrt{15}}{2\times2} = \frac{-3 \pm i\sqrt{15}}{4}\)

Problem 30: \(3x^2 - 3x + 1 = 0\)

Step 1: Identify coefficients

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

Step 2: Calculate discriminant \(\Delta\)

Discriminant formula: \(\Delta = b^2 - 4ac\)
Substitute values: \(\Delta = (-3)^2 - 4\times3\times1 = 9 - 12 = -3\)

Step 3: Find roots (complex numbers)

Quadratic formula: \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\)
Since \(\Delta = -3\), \(\sqrt{\Delta} = \sqrt{-3} = i\sqrt{3}\) (where \(i\) is imaginary unit)
Substitute into formula: \(x = \frac{-(-3) \pm i\sqrt{3}}{2\times3} = \frac{3 \pm i\sqrt{3}}{6}\)

Answer:

• For \(2x^2 + 3x + 3 = 0\): \(x = \frac{-3 + i\sqrt{15}}{4}\) or \(x = \frac{-3 - i\sqrt{15}}{4}\)
• For \(3x^2 - 3x + 1 = 0\): \(x = \frac{3 + i\sqrt{3}}{6}\) or \(x = \frac{3 - i\sqrt{3}}{6}\)