The quadratic equation # x^2+px+q = 0# has a complex root #2+3i#. Find #p# and #q#?

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Answer 1 · 2017-04-01T18:16:33

# p=-4 #
# q=13 #

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 #

Complex roots always appear in conjugate pairs, so if one root of the given quadratic is #alpha=2+3i# then the other root is #beta=2-3i#

# x^2+px+q = 0#

we know that:

# alpha+beta = -p/1 \ \ \ # ; and # \ \ \ alpha beta = q/1 #

And we can calculate:

# alpha + beta = 2+3i + 2-3i = 4 => p=-4 #
# alpha beta = (2+3i)(2-3i) = 4+9 = 13 => q=13 #

Answer 2 · 2017-04-01T19:16:00

#p=-4, and, q=13.#

It is known that #2+3i# is a root of the Quadr. Eqn.#:x^2+px+q=0.#

Therefore, #x=2+3i# must satisfy the eqn.

#:. (2+3i)^2+p(2+3i)+q=0.#

#:. 4+12i+9i^2+2p+3ip+q=0.#

#:. 4+12i-9+2p+3ip+q=0.#

#:. (2p+q-5)+i(12+3p)=0.#

Comparing the Real and Imaginary Parts of both sides, we get,

# 2p+q=5......(1), and, 12+3p=0#

#rArr p=-4, &, q=5-2p=5-2(-4)=13,# as Respected Steve Sir, has alreay derived.

Enjoy Maths.!