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 

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

Answer 2

We begin with the equation \(x^5 - 5x^3 + 5x^2 - 1 = 0\) which has \(\alpha\) and \(\beta\) as two distinct negative roots. We are tasked with finding the equation whose roots are \(\sqrt{-\alpha}\) and \(\sqrt{-\beta}\).
Given that \(\alpha < 0\) and \(\beta < 0\), we have \(-\alpha > 0\) and \(-\beta > 0\), making \(\sqrt{-\alpha}\) and \(\sqrt{-\beta}\) real numbers.
By assuming \(\sqrt{-\alpha}=p\) and \(\sqrt{-\beta}=q\), the transformations give us \(-\alpha=p^2\) and \(-\beta=q^2\).
To deduce the polynomial in terms of \(p\), use the relationships:

• The sum \(p^2+q^2=\alpha+\beta\) by the nature of quadratic roots.
• The product \(p^2q^2=(-\alpha)(-\beta)=\alpha\beta\).

We seek an equation in terms of \(p^2\) and \(q^2\). Recognizing that \(p^2\) and \(q^2\) each satisfy: \[(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta\] we adjust this for \((-x)\) since \(\alpha\) and \(\beta\) are negative: \[(x^2-\alpha)(x^2-\beta)=x^4-(\alpha+\beta)x^2+\alpha\beta\].
Given the known polynomial's construction, compute: \(\alpha+\beta=-3\) and \(\alpha\beta=1\), inserting these into the polynomial equation results in: \[x^4-3x^2+1=0\].
The derived polynomial \(x^4-3x^2+1=0\) corresponds to the least degree equation with integer coefficients having roots \(\sqrt{-\alpha}\) and \(\sqrt{-\beta}\).