Question

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?

Answer 1

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`

Answer 2

`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.!