What is the value of a^2+b^2a2+b2?

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
Konstantinos Michailidis
May 6, 2017

Expand the left-hand side to obtain

4a^2 + b^2 + 4 + a^2b^2 = 10ab - 54a2+b2+4+a2b2=10ab−5

Rearranging a bit, to get

4a^2-4ab + b^2 = -(ab)^2+6ab - 94a2−4ab+b2=−(ab)2+6ab−9

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

or
(2a-b)^2+(ab-3)^2=0(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=b2a=b and ab=3ab=3

From these equations (it's easy) you will get a^2=3/2a2=32 and b^2=6b2=6

Hence a^2+b^2=15/2a2+b2=152

Ratnaker Mehta
May 6, 2017

15/2.152.

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.!

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