Question

If \( \alpha \) and \( \beta \) are the roots of the equation \( 2x^2 - 4x + 3 = 0 \), then \( \frac{2(\alpha^4 + \beta^4) + 3(\alpha^2 + \beta^2)}{\alpha + \beta} = \)

• A. \( -1 \)
• B. \( -2 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

Vieta's formulas provide a direct relationship between the roots of a polynomial and its coefficients. For a quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \), \( \alpha + \beta = -b/a \) and \( \alpha \beta = c/a \). Algebraic identities like \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \) and \( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \) are useful for expressing higher powers of roots in terms of the sum and product of the roots.

Answer 1

The Correct Option is B

Given the quadratic equation \( 2x^2 - 4x + 3 = 0 \), for roots \( \alpha \) and \( \beta \), by Vieta's formulas: $$ \alpha + \beta = -\frac{-4}{2} = 2 $$ $$ \alpha \beta = \frac{3}{2} $$ We need to find \( \alpha^2 + \beta^2 \) and \( \alpha^4 + \beta^4 \).
$$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (2)^2 - 2\left(\frac{3}{2}\right) = 4 - 3 = 1 $$ Now, for \( \alpha^4 + \beta^4 \): $$ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 = (1)^2 - 2\left(\frac{3}{2}\right)^2 = 1 - 2\left(\frac{9}{4}\right) = 1 - \frac{9}{2} = \frac{2 - 9}{2} = -\frac{7}{2} $$ Now substitute these values into the expression: $$ \frac{2(\alpha^4 + \beta^4) + 3(\alpha^2 + \beta^2)}{\alpha + \beta} = \frac{2\left(-\frac{7}{2}\right) + 3(1)}{2} = \frac{-7 + 3}{2} = \frac{-4}{2} = -2 $$