Question

The point \( (t^2 + 2t + 5, 2t^2 + t - 2) \) { lies on the line} \( x + y = 2 \) { for:}

• A. All real values of \( t \)
• B. Some real values of \( t \)
• C. \( t = -3 \pm \frac{\sqrt{3}}{6} \)
• D. None of these

Hint:

When solving quadratic equations, check the discriminant. If it's negative, there are no real solutions.

Answer 1

The Correct Option is D

Step 1: The point \( (x, y) = (t^2 + 2t + 5, 2t^2 + t - 2) \) lies on the line \( x + y = 2 \). Therefore, we substitute the coordinates of the point into the equation of the line: \[ (t^2 + 2t + 5) + (2t^2 + t - 2) = 2. \] Step 2: Simplify the equation: \[ t^2 + 2t + 5 + 2t^2 + t - 2 = 2 \quad \Rightarrow \quad 3t^2 + 3t + 3 = 2. \] Step 3: Solve the equation: \[ 3t^2 + 3t + 1 = 0. \] This is a quadratic equation in \( t \). To solve for \( t \), use the discriminant: \[ \Delta = b^2 - 4ac = 3^2 - 4 \times 3 \times 1 = 9 - 12 = -3. \] Since the discriminant is negative, the quadratic equation has no real solutions. Thus, the point does not lie on the line for any real values of \( t \).