Question

If \( \alpha, \beta \) are the roots of the equation \( x^2 + bx + c = 0 \) satisfying the conditions \( \alpha+\beta=5 \) and \( \alpha^3+\beta^3=60 \), then \( 3c+2 = \)

• A. \( 2b \)
• B. \( 3b \)
• C. \( -3b \)
• D. \( -2b \)

Hint:

For problems involving roots of quadratic equations, remember Vieta's formulas relating the coefficients to the sum and product of the roots: for \( ax^2+bx+c=0 \), the sum of roots is \( -b/a \) and the product of roots is \( c/a \). Also, know common algebraic identities like \( \alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) \) or \( \alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2) \). These identities simplify calculations significantly.

Answer 1

The Correct Option is C

Given the quadratic equation \( x^2 + bx + c = 0 \), and \( \alpha, \beta \) are its roots. From Vieta's formulas, we have: 1. Sum of the roots: \( \alpha + \beta = -b \) 2. Product of the roots: \( \alpha \beta = c \) We are given the conditions: % Option (i) \( \alpha + \beta = 5 \) % Option (ii) \( \alpha^3 + \beta^3 = 60 \) From condition (i) and Vieta's formula for the sum of roots: \( 5 = -b \implies b = -5 \) Now, let's use the identity for the sum of cubes: \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \) We can rewrite \( \alpha^2 + \beta^2 \) as \( (\alpha + \beta)^2 - 2\alpha\beta \). So, \( \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) \) Substitute the given values into this identity: \( 60 = (5)((5)^2 - 3c) \) \( 60 = 5(25 - 3c) \) Divide both sides by 5: \( \frac{60}{5} = 25 - 3c \) \( 12 = 25 - 3c \) Now, solve for \( c \): \( 3c = 25 - 12 \) \( 3c = 13 \) \( c = \frac{13}{3} \) We need to find the value of \( 3c+2 \). Substitute the value of \( c \): \( 3c+2 = 3\left(\frac{13}{3}\right) + 2 \) \( = 13 + 2 \) \( = 15 \) Now, let's check the options in terms of \( b \). We found \( b = -5 \). % Option (A) \( 2b = 2(-5) = -10 \) % Option (B) \( 3b = 3(-5) = -15 \) % Option (C) \( -3b = -3(-5) = 15 \) % Option (D) \( -2b = -2(-5) = 10 \) Comparing our calculated value \( 15 \) with the options, it matches option (C).