Questions: Use the quadratic formula to solve the equation. Any solution is a real number. 4/7 x^2 - 5/14 x = 6/7 x =

Solution Steps

To solve the given quadratic equation using the quadratic formula, we need to first bring the equation to the standard form \(ax^2 + bx + c = 0\). Then, we can identify the coefficients \(a\), \(b\), and \(c\) and apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

1. Rewrite the equation in the standard form \(ax^2 + bx + c = 0\).
2. Identify the coefficients \(a\), \(b\), and \(c\).
3. Use the quadratic formula to find the solutions for \(x\).

We start with the equation given in the problem: \[ \frac{4}{7} x^{2} - \frac{5}{14} x = \frac{6}{7} \] Rearranging this into standard form \(ax^2 + bx + c = 0\), we have: \[ \frac{4}{7} x^{2} - \frac{5}{14} x - \frac{6}{7} = 0 \]

From the standard form, we identify the coefficients: \[ a = \frac{4}{7}, \quad b = -\frac{5}{14}, \quad c = -\frac{6}{7} \]

Next, we calculate the discriminant \(D\) using the formula: \[ D = b^2 - 4ac \] Substituting the values of \(a\), \(b\), and \(c\): \[ D = \left(-\frac{5}{14}\right)^{2} - 4 \cdot \frac{4}{7} \cdot \left(-\frac{6}{7}\right) = \frac{25}{196} + \frac{96}{196} = \frac{121}{196} \] Since \(D > 0\), there are two real solutions.

Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] we substitute \(b\), \(D\), and \(a\): \[ x = \frac{-\left(-\frac{5}{14}\right) \pm \sqrt{\frac{121}{196}}}{2 \cdot \frac{4}{7}} = \frac{\frac{5}{14} \pm \frac{11}{14}}{\frac{8}{7}} \]

Calculating the two possible values for \(x\):

1. For the positive root: \[ x_1 = \frac{\frac{5}{14} + \frac{11}{14}}{\frac{8}{7}} = \frac{\frac{16}{14}}{\frac{8}{7}} = \frac{16}{14} \cdot \frac{7}{8} = \frac{16 \cdot 7}{14 \cdot 8} = \frac{112}{112} = 1 \]
2. For the negative root: \[ x_2 = \frac{\frac{5}{14} - \frac{11}{14}}{\frac{8}{7}} = \frac{-\frac{6}{14}}{\frac{8}{7}} = -\frac{6}{14} \cdot \frac{7}{8} = -\frac{6 \cdot 7}{14 \cdot 8} = -\frac{42}{112} = -\frac{3}{8} \]

Final Answer