Question

The equation whose roots are the squares of the roots of the equation \( 2x^2 + 3x + 1 = 0 \) is:

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

Hint:

To find the equation with squared roots, just solve $ax^2 + c = -bx$, then square both sides: $a^2x^4 + 2acx^2 + c^2 = b^2x^2$. Finally, replace $x^2$ with $y$.

Answer 1

The Correct Option is D

Concept: If \( \alpha, \beta \) are roots of \( ax^2 + bx + c = 0 \), then the equation with roots \( \alpha^2, \beta^2 \) can be found by substituting \( \sqrt{x} \) for \( x \) and rationalizing the equation.

Step 1: Use the transformation method.
Let the original roots be \( \alpha, \beta \). The new roots are \( y = \alpha^2 \), so \( \alpha = \sqrt{y} \). Substitute \( x = \sqrt{y} \) into \( 2x^2 + 3x + 1 = 0 \): \[ 2(\sqrt{y})^2 + 3\sqrt{y} + 1 = 0 \] \[ 2y + 1 = -3\sqrt{y} \]

Step 2: Square both sides to remove the radical.
\[ (2y + 1)^2 = (-3\sqrt{y})^2 \] \[ 4y^2 + 4y + 1 = 9y \]

Step 3: Simplify and replace \( y \) with \( x \).
\[ 4y^2 - 5y + 1 = 0 \] The equation is \( 4x^2 - 5x + 1 = 0 \).