Question

The roots of the equation \( 6x^2 - x - 2 = 0 \) are

• A. \( \frac{2}{3}, -\frac{1}{2} \)
• B. \( -\frac{2}{3}, \frac{1}{2} \)
• C. \( -\frac{2}{3}, -\frac{1}{2} \)
• D. \( \frac{2}{3}, \frac{1}{2} \)

Hint:

Remember the quadratic formula: For \( ax^2 + bx + c = 0 \), the roots are \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Answer 1

The Correct Option is A

Step 1: Identify the coefficients.
The given quadratic equation is \( 6x^2 - x - 2 = 0 \), where \( a = 6 \), \( b = -1 \), and \( c = -2 \). Step 2: Apply the quadratic formula.
The roots are given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-2)}}{2(6)} \] \[ x = \frac{1 \pm \sqrt{1 + 48}}{12} \] \[ x = \frac{1 \pm \sqrt{49}}{12} \] \[ x = \frac{1 \pm 7}{12} \] Step 3: Calculate the two roots.
\[ x_1 = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3} \] \[ x_2 = \frac{1 - 7}{12} = \frac{-6}{12} = -\frac{1}{2} \] The roots are \( \frac{2}{3} \) and \( -\frac{1}{2} \).