Question

If α and β are the roots of the equation 2x² + 2px + p² = 0, where p is a non-zero real number, and α⁴ and β⁴ are the roots of x² - rx + s = 0, then the roots of 2x² - 4p²x + 4p⁴ - 2r = 0 are:
1. Real and unequal.
2. Equal and zero.
3. Imaginary.
4. Equal and non-zero.

Answer 1

Correct Answer - Option 3 : Imaginary.

Concept:

The solution to the quadratic equation Ax² + Bx + C = 0 can also be given by: \(\rm x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}\).

The quantity B² - 4AC is also called the discriminant.

• If B² - 4AC ≥ 0, the roots are real.
• If B² - 4AC = 0, the roots are real and equal.
• If B² - 4AC < 0, the roots will be complex and conjugates of each other.

The sum of both the roots of the quadratic equation Ax² + Bx + C = 0 is \(\rm -\frac{B}{A}\) and the product of the roots is \(\rm \frac{C}{A}\).

 

Calculation:

Using the expressions for the sum and the product of the roots, we have:

α + β = -p            ... (1)

αβ = \(\rm \frac{p^2}{2}\)            ... (2)

α⁴ + β⁴ = r            ... (3)

Squaring equation (1), we get:

α² + β² + 2αβ = p²

Using equation (2), we get:

⇒ α2 + β2 = 0

Squaring again, we get:

⇒ α4 + β4 + 2α2β2 = 0

Using equations (2) and (3), we get:

⇒ r = -\(\rm \frac{p^4}{2}\)            ... (4)

The discriminant of the equation 2x2 - 4p2x + 4p⁴ - 2r = 0 is:

(-4p2)² - 4(2)(4p⁴ - 2r)

= 16p⁴ - 32p4 + 16r

Using equation (4), we get:

= 16p4 - 32p4 - 8p4

= -24p⁴, which is always negative for non-zero real p.

Since the discriminant is < 0, the roots are imaginary.