Question

Find the roots of the quadratic equation \( 2x^2 - 4x - 6 = 0 \).

• A. \( x = 1 \, \text{or} \, x = -3 \)
• B. \( x = -1 \, \text{or} \, x = 3 \)
• C. \( x = 2 \, \text{or} \, x = -1 \)
• D. \( x = 3 \, \text{or} \, x = -2 \)

Hint:

Remember: For a quadratic equation \( ax^2 + bx + c = 0 \), use the quadratic formula to find the roots. The discriminant \( b^2 - 4ac \) determines the nature of the roots.

Answer 1

The Correct Option is A

Step 1: Use the quadratic formula The quadratic formula to solve the equation \( ax^2 + bx + c = 0 \) is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the given equation \( 2x^2 - 4x - 6 = 0 \), we have: - \( a = 2 \), - \( b = -4 \), - \( c = -6 \). Step 2: Substitute the values into the quadratic formula Substitute \( a = 2 \), \( b = -4 \), and \( c = -6 \) into the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 2 \times (-6)}}{2 \times 2} \] \[ x = \frac{4 \pm \sqrt{16 + 48}}{4} \] \[ x = \frac{4 \pm \sqrt{64}}{4} \] \[ x = \frac{4 \pm 8}{4} \] Step 3: Solve for the two roots The two possible values for \( x \) are: \[ x = \frac{4 + 8}{4} = \frac{12}{4} = 3 \] and \[ x = \frac{4 - 8}{4} = \frac{-4}{4} = -1 \] Answer: Therefore, the roots of the quadratic equation are \( x = 3 \, \text{or} \, x = -1 \). So, the correct answer is option (1).