Question

if a^2+b^2=170 and ab=77 fine (a+b)^2. please teach me a quick method on how to solve this. thanks!

Answer 1

as we know that (a+b)^2 = (a^2) + (b^2) + 2ab
now it is given that a^2 + b^2 = 170
and ab = 77
(a+b)^2 = 170 + 2(77)
= 170 + 154
= 324

Answer 2

got it phoebeco

Answer 3

well thanks for awarding me with a medal

Answer 4

there are no "quick" methods; there are sensible methods at best

Answer 5

seeing that they give you 2 equations with the same variables in them we can equate them to find out the as and bs that they have in common

Answer 6

ab=77; we can solve this for either a or b

a = 77/b or b = 77/a pick one of these options and use it as a value to replace in the other equation; ill use a = 77/b

a^2 + b^2 = 170

(77/b)^2 + b^2 = 170 ; now we have 1 variable to worry about and can try to solve for b

Answer 7

5929/ b^2 + b^2 = 170 ; *b^2

5929 + b^4 = 170b^2

b^4 -170b^2+5929 = 0 ; this is a quadratic in disguise, to see it
let x = b^2

x^2 -170x+5929 = 0; solve for x and equate those solutions to x = b^2

Answer 8

i get x=49 and x=121 from that... sooo

x = b^2
49 = b^2
49 = (+-7)^2

x = b^2
121 = b^2
121 = (+-11)^2

try out all these options to weed out the extra ones

Answer 9

a * b = 77
-7 * -11 = 77

a^2 + b^2 = 170
49 + 121 = 170 ; that works out


a * b = 77
7 * 11 = 77

a^2 + b^2 = 170
49 + 121 = 170 ; that works out as well; :)

Answer 10

(7+11)^2

18^2

18
18
---
144
18
----
324 .... same conclusion :)

Answer 11

that was a perfect answer amistre 64