The roots of the quadratic #2x^2+3x-1=0# are #alpha# and #beta#. Without calculating the roots, find #alpha^3+beta^3#?

1 Answer
Apr 1, 2017

Answer:

# alpha^3 + beta^3 = -45/8 #

Explanation:

Suppose the roots of the general quadratic equation:

# ax^2+bx+c = 0 #

are #alpha# and #beta# , then using the root properties we have:

# "sum of roots" \ \ \ \ \ \= alpha+beta = -b/a #
# "product of roots" = alpha beta \ \ \ \ = c /a #

So for the given quadratic with roots #alpha# and #beta#:

# 2x^2+3x-1 = 0#

we know that:

# alpha+beta = -3/2 \ \ \ # ; and # \ \ \ alpha beta = -1/2 #

Consider the binomial expression:

# \ \ \ \ \ (alpha+beta)^3 = alpha^3 + 3alpha^2 beta + 3alpha beta^2 + beta^3 #
# :. (alpha+beta)^3 = alpha^3 + beta^3+ 3alpha beta(alpha+beta) #
# :. \ alpha^3 + beta^3 = (alpha+beta)^3 - 3alpha beta(alpha+beta) #

Substituting our values of #alpha+beta# and #alpha beta# we get:

# alpha^3 + beta^3 = (-3/2)^3 - 3(-1/2)(-3/2) #
# " " = -27/8 - 9/4 #
# " " = -45/8 #