What is the value of #a^2+b^2#?

Let a and b be real numbers such that (a^2+1)(b^2+4) = 10ab - 5. What is the value of a^2+b^2?

2 Answers

Expand the left-hand side to obtain

#4a^2 + b^2 + 4 + a^2b^2 = 10ab - 5#

Rearranging a bit, to get

#4a^2-4ab + b^2 = -(ab)^2+6ab - 9#

Finally this is equal to
#(2a-b)^2 = -(ab-3)^2#

or
#(2a-b)^2+(ab-3)^2=0#

Because the sum of two squares is zero this means that both squares are equal to zero.

Which means that #2a=b# and #ab=3#

From these equations (it's easy) you will get #a^2=3/2# and #b^2=6#

Hence #a^2+b^2=15/2#

May 6, 2017

# 15/2.#

Explanation:

Given that, #(a^2+1)(b^2+4)=10ab-5; where, a,b in RR.#

#rArr a^2b^2+b^2+4a^2+4=10ab-5.#

# rArr 4a^2+b^2+a^2b^2-10ab+9=0.#

# rArr 4a^2-4ab+b^2+a^2b^2-6ab+9=0.#

# rArr (2a-b)^2+(ab-3)^2=0, where, a,b in RR.#

# rArr 2a-b=0, and, ab-3=0, or, #

# b=2a, &, ab=3.#

# :. a(2a)=3, or, a^2=3/2.........(1).#

Also, # b=2a rArr b^2=4a^2=4*3/2=6..............(2).#

From #(1) and (2)," the reqd. value="a^2+b^2=3/2+6=15/2.#

Enjoy Maths.!