Solve the quadratic equation 2x^2 − 3x + 5 = 0. Find the roots.

Solve the quadratic equation 2x^2 − 3x + 5 = 0. Find the roots.

• A. x = (3 ± i√31)/4
• B. x = (3 ± √31)/4
• C. x = (−3 ± i√31)/4
• D. x = (3 ± i√7)/4

Answer: (A)

This question tests solving a quadratic using the quadratic formula and recognizing when the discriminant is negative, which leads to complex roots. For 2x^2 − 3x + 5 = 0, a = 2, b = −3, c = 5. The discriminant is Δ = b^2 − 4ac = (−3)^2 − 4·2·5 = 9 − 40 = −31, which is negative, so the roots are complex. Using the quadratic formula x = (−b ± sqrt(Δ)) / (2a), we get x = (3 ± sqrt(−31)) / 4 = (3 ± i√31) / 4. This is the root pair, matching the correct form with an imaginary part. The other options would either give real roots (which isn’t possible here due to the negative discriminant) or alter the sign or the radical (e.g., using √7 instead of √31), which would not satisfy the equation.

This question tests solving a quadratic using the quadratic formula and recognizing when the discriminant is negative, which leads to complex roots. For 2x^2 − 3x + 5 = 0, a = 2, b = −3, c = 5. The discriminant is Δ = b^2 − 4ac = (−3)^2 − 4·2·5 = 9 − 40 = −31, which is negative, so the roots are complex.

Using the quadratic formula x = (−b ± sqrt(Δ)) / (2a), we get x = (3 ± sqrt(−31)) / 4 = (3 ± i√31) / 4. This is the root pair, matching the correct form with an imaginary part.

The other options would either give real roots (which isn’t possible here due to the negative discriminant) or alter the sign or the radical (e.g., using √7 instead of √31), which would not satisfy the equation.