SOLUTION: Suppose that ab = 7 and a^2b + ab^2 + a+b = 80. What is a^2+b^2?

Question 1036921: Suppose that ab = 7 and a^2b + ab^2 + a+b = 80. What is a^2+b^2?
Found 3 solutions by Boreal, MathTherapy, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
a=7/b
(49/b^2)*b+(7/b)*b^2+(7/b)+b=80
(49/b)+7b+(7/b)+b=80
multiply through by b
49+7b^2+7+b^2=80b
collect terms on left side
8b^2-80b+56=0
divide by 8
b^2-10b+7=0
b=(1/2)(10+/-sqrt (100-28)); sqrt 72=6 sqrt(2)
b=(1/2)(10+/-6 sqrt(2))
b=5+/-3sqrt(2)
a=7/5+3sqrt(2). that is 7(5-3 sqrt(2)/25-18) or 7(5-3 sqrt(2))/7= 5-3sqrt(2), rationalizing the denominator by using the conjugate
a=7/5-3 sqrt(2), and by the same method, get 7(5+3 sqrt(2))/7=5+ 3 sqrt(2)
One term is 5+3 sqrt(2) and the other is 5-3 sqrt(2).
square the first and get 43+30 sqrt (2); square the second and get 43-30 sqrt (2). Their sum is 86.
ab is 7, because they are conjugates.
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Alternative way.
a^2b+a=a(ab+1)
ab^2+b=b(ab+1)
Their sum is (a+b)(ab+1)=80
But (ab+1)=8
so a+b=10
7/b+ b=10
b^2-10b+7=0 and solve for b as above.

Answer by MathTherapy(10559)   (Show Source): You can put this solution on YOUR website!

Suppose that ab = 7 and a^2b + ab^2 + a+b = 80. What is a^2+b^2?

ab = 7 ------- eq (i)
---- eq (ii)
ab(a + b) + 1(a + b) = 80 ------- Factoring polynomial on left
(ab + 1)(a + b) = 80
(7 + 1)(a + b) = 80 ------ Substituting 7 for ab
8(a + b) = 8(10) ------- Factoring out GCF, 8
a + b = 10 ------- eq (iii)
------ FOILing binomial
------- Substituting 10 for a + b, and 7 for ab

Answer by ikleyn(52932)   (Show Source): You can put this solution on YOUR website!
.
Suppose that ab = 7 and a^2b + ab^2 + a+b = 80. What is a^2+b^2?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

= ab*(a+b) + (a+b) = (replace ab by 7) = 7(a+b) + (a+b) = 8(a+b). Now, since = 80, we have 8(a+b) = 80, which implies a+b = 10. Next, = = = 100 - 14 = 86. Answer. = 86.

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