Questions: Solve the equation by applying the quadratic formula. 5/12 x^2 - 1/2 x + 1/4 = 0

Solution Steps

To solve the quadratic equation using the quadratic formula, we need to identify the coefficients \(a\), \(b\), and \(c\) from the equation \(ax^2 + bx + c = 0\). Then, we apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] to find the roots of the equation.

The given quadratic equation is: \[ \frac{5}{12} x^{2} - \frac{1}{2} x + \frac{1}{4} = 0 \]

We need to identify the coefficients \(a\), \(b\), and \(c\) from the standard form of a quadratic equation \(ax^2 + bx + c = 0\).

Here: \[ a = \frac{5}{12}, \quad b = -\frac{1}{2}, \quad c = \frac{1}{4} \]

The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The discriminant \(\Delta\) is: \[ \Delta = b^2 - 4ac \]

Substitute the values of \(a\), \(b\), and \(c\): \[ \Delta = \left(-\frac{1}{2}\right)^2 - 4 \left(\frac{5}{12}\right) \left(\frac{1}{4}\right) \]

Calculate each term: \[ \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ 4 \left(\frac{5}{12}\right) \left(\frac{1}{4}\right) = \frac{5}{12} \cdot \frac{1}{4} \cdot 4 = \frac{5}{12} \]

So, \[ \Delta = \frac{1}{4} - \frac{5}{12} \]

To subtract these fractions, find a common denominator: \[ \frac{1}{4} = \frac{3}{12} \]

Thus, \[ \Delta = \frac{3}{12} - \frac{5}{12} = -\frac{2}{12} = -\frac{1}{6} \]

Since the discriminant \(\Delta\) is negative, the quadratic equation has two complex roots.

Substitute \(a\), \(b\), and \(\Delta\) into the quadratic formula: \[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]

Since \(\Delta = -\frac{1}{6}\), we have: \[ \sqrt{\Delta} = \sqrt{-\frac{1}{6}} = i \sqrt{\frac{1}{6}} = \frac{i}{\sqrt{6}} = \frac{i \sqrt{6}}{6} \]

Now, substitute \(b\) and \(a\): \[ x = \frac{\frac{1}{2} \pm \frac{i \sqrt{6}}{6}}{\frac{10}{12}} = \frac{\frac{1}{2} \pm \frac{i \sqrt{6}}{6}}{\frac{5}{6}} \]

Simplify the fraction: \[ x = \frac{1}{2} \cdot \frac{6}{5} \pm \frac{i \sqrt{6}}{6} \cdot \frac{6}{5} = \frac{3}{5} \pm \frac{i \sqrt{6}}{5} \]

Final Answer