Questions: Question from 1.5: Complex Numbers Find all solutions of the equation and express them in the form (a+bi). (Enter your answers as a comma-separated list. Simplify your answer completely.) (2 x^2-2 x+1=0) (x=)

Solution Steps

To solve the quadratic equation \(2x^2 - 2x + 1 = 0\), we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the equation. In this case, \(a = 2\), \(b = -2\), and \(c = 1\). We will calculate the discriminant \(b^2 - 4ac\) to determine the nature of the roots. Since the discriminant is negative, the solutions will be complex numbers.

The given quadratic equation is \(2x^2 - 2x + 1 = 0\). The coefficients are:

• \(a = 2\)
• \(b = -2\)
• \(c = 1\)

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by: \[ \Delta = b^2 - 4ac \] Substituting the values, we have: \[ \Delta = (-2)^2 - 4 \times 2 \times 1 = 4 - 8 = -4 \]

Since the discriminant \(\Delta = -4\) is negative, the roots of the equation are complex numbers.

The roots of the quadratic equation are given by: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \] Substituting the values, we have: \[ x = \frac{-(-2) \pm \sqrt{-4}}{2 \times 2} = \frac{2 \pm \sqrt{-4}}{4} \]

The square root of \(-4\) is \(2i\), where \(i\) is the imaginary unit. Thus, the roots are: \[ x_1 = \frac{2 + 2i}{4} = \frac{1}{2} + \frac{1}{2}i \] \[ x_2 = \frac{2 - 2i}{4} = \frac{1}{2} - \frac{1}{2}i \]

Final Answer