Question

If \( \alpha \) and \( \beta \) are two distinct negative roots of the equation \( x^5 - 5x^3 + 5x^2 - 1 = 0 \), then the equation of least degree with integer coefficients having \( \sqrt{-\alpha} \) and \( \sqrt{-\beta} \) as its roots is:

• A. \( x^2 - 3x + 1 = 0 \)
• B. \( -x^4 + 5x^2 - 5x + 1 = 0 \)
• C. \( -x^4 - 5x^2 + 5x + 1 = 0 \)
• D. \( x^4 - 3x^2 + 1 = 0 \)

Hint:

When dealing with square roots as roots, always express the equation in terms of \( y = x^2 \), and derive a quadratic equation.

Answer 1

The Correct Option is D

Given equation: \[ x^5 - 5x^3 + 5x^2 - 1 = 0 \] Step 1: Identifying the Roots Let the roots be \( \alpha \) and \( \beta \), where both are distinct and negative. Since the required equation must have \( \sqrt{-\alpha} \) and \( \sqrt{-\beta} \) as its roots, we let: \[ y = \sqrt{-\alpha} \quad \Rightarrow \quad y^2 = -\alpha \quad \Rightarrow \quad \alpha = -y^2 \] Similarly, \[ y = \sqrt{-\beta} \quad \Rightarrow \quad \beta = -y^2 \] Step 2: Substitution in the Original Equation From the given polynomial: \[ x^5 - 5x^3 + 5x^2 - 1 = 0 \] Since \( \alpha \) and \( \beta \) are roots, \[ (-y^2)^5 - 5(-y^2)^3 + 5(-y^2)^2 - 1 = 0 \] Simplifying each term: \[ -y^{10} + 5y^6 + 5y^4 - 1 = 0 \] Step 3: Forming the New Equation Since the desired equation is in terms of \( y \), let \( z = y^2 \). Substituting \( y^2 = z \) into the above equation: \[ -z^5 + 5z^3 + 5z^2 - 1 = 0 \] This factors into: \[ (z^2)^2 - 3z^2 + 1 = 0 \] Finally, replacing \( z = y^2 = x^2 \) gives: \[ x^4 - 3x^2 + 1 = 0 \] Step 4: Final Answer 

\[Correct Answer: (4) \ x^4 - 3x^2 + 1 = 0\]