Question

If the sum of two roots of the equation \( x^4 + 2x^3 - 7x^2 - 8x + 12 = 0 \) is zero, then the sum of the squares of the other two roots is

• A. \( 5 \)
• B. \( 10 \)
• C. \( 13 \)
• D. \( 25 \)

Hint:

When a polynomial has roots with a specific relationship (like sum being zero), use Vieta's formulas. If \( \alpha + \beta = 0 \), then this immediately simplifies the sum of roots relation. This information can then be used to simplify other Vieta's relations, eventually leading to a system of equations to solve for the unknown terms. In the case of \( x^4 \), if two roots sum to zero (\( \alpha, -\alpha \)), then \( (x-\alpha)(x+\alpha) = x^2-\alpha^2 \) is a factor of the polynomial, allowing you to reduce the polynomial to a quadratic.

Answer 1

The Correct Option is B

Let the four roots of the equation \( x^4 + 2x^3 - 7x^2 - 8x + 12 = 0 \) be \( \alpha, \beta, \gamma, \delta \). According to Vieta's formulas for a quartic equation \( a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 = 0 \): 1. Sum of roots: \( \sum \alpha = \alpha+\beta+\gamma+\delta = -\frac{a_3}{a_4} \) 2. Sum of products of roots taken two at a time: \( \sum \alpha\beta = \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta = \frac{a_2}{a_4} \) 3. Sum of products of roots taken three at a time: \( \sum \alpha\beta\gamma = \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = -\frac{a_1}{a_4} \) 4. Product of roots: \( \alpha\beta\gamma\delta = \frac{a_0}{a_4} \) For the given equation \( x^4 + 2x^3 - 7x^2 - 8x + 12 = 0 \): \( a_4 = 1, a_3 = 2, a_2 = -7, a_1 = -8, a_0 = 12 \). So, we have: 1. \( \alpha+\beta+\gamma+\delta = -2 \) 2. \( \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta = -7 \) 3. \( \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta = 8 \) 4. \( \alpha\beta\gamma\delta = 12 \) We are given that the sum of two roots is zero. Let \( \alpha + \beta = 0 \). This implies \( \beta = -\alpha \). Now, let's use this information in Vieta's formulas: From (1): \( (\alpha+\beta) + \gamma+\delta = -2 \) Since \( \alpha+\beta = 0 \), we have \( 0 + \gamma+\delta = -2 \implies \gamma+\delta = -2 \). From (2): \( \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = -7 \) Substitute \( \beta = -\alpha \): \( \alpha(-\alpha) + \alpha\gamma + \alpha\delta + (-\alpha)\gamma + (-\alpha)\delta + \gamma\delta = -7 \) \( -\alpha^2 + \alpha\gamma + \alpha\delta - \alpha\gamma - \alpha\delta + \gamma\delta = -7 \) \( -\alpha^2 + \gamma\delta = -7 \) From (3): \( \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = 8 \) Substitute \( \beta = -\alpha \): \( \alpha(-\alpha)\gamma + \alpha(-\alpha)\delta + \alpha\gamma\delta + (-\alpha)\gamma\delta = 8 \) \( -\alpha^2\gamma - \alpha^2\delta + \alpha\gamma\delta - \alpha\gamma\delta = 8 \) \( -\alpha^2(\gamma+\delta) = 8 \) We know \( \gamma+\delta = -2 \). \( -\alpha^2(-2) = 8 \) \( 2\alpha^2 = 8 \) \( \alpha^2 = 4 \) Now use \( \alpha^2 = 4 \) in \( -\alpha^2 + \gamma\delta = -7 \): \( -4 + \gamma\delta = -7 \) \( \gamma\delta = -7 + 4 \) \( \gamma\delta = -3 \) The problem asks for the sum of the squares of the other two roots, which are \( \gamma \) and \( \delta \). We need to find \( \gamma^2 + \delta^2 \). We know that \( \gamma^2 + \delta^2 = (\gamma+\delta)^2 - 2\gamma\delta \). Substitute the values we found: \( \gamma+\delta = -2 \) and \( \gamma\delta = -3 \). \( \gamma^2 + \delta^2 = (-2)^2 - 2(-3) \) \( = 4 + 6 \) \( = 10 \) To verify, we can find the roots. Since \( \alpha^2 = 4 \), \( \alpha = \pm 2 \). Let \( \alpha = 2 \), then \( \beta = -2 \). The equation is \( x^4 + 2x^3 - 7x^2 - 8x + 12 = 0 \). Since \( x=2 \) and \( x=-2 \) are roots, \( (x-2)(x+2) = x^2-4 \) is a factor. We can perform polynomial division: \( (x^4 + 2x^3 - 7x^2 - 8x + 12) \div (x^2 - 4) \) The other factor is \( x^2 + 2x - 3 = 0 \). The roots of this quadratic are the other two roots, \( \gamma \) and \( \delta \). \( x^2 + 2x - 3 = 0 \) \( (x+3)(x-1) = 0 \) So, the other two roots are \( \gamma = 1 \) and \( \delta = -3 \) (or vice versa). Let's check \( \gamma+\delta = 1+(-3) = -2 \). This matches. Let's check \( \gamma\delta = 1(-3) = -3 \). This matches. The sum of squares of these roots is \( \gamma^2 + \delta^2 = (1)^2 + (-3)^2 = 1 + 9 = 10 \).