Let a and b be real numbers such that a + b = 4 and a^4 + b^4 = 16.
(a) Find all possible values of ab
(b) Find all possible values of a + b
(c) Find all possible values of a and b
+266 so we can get 2 equations here, please ask if u want to know how i got these.
\(4a + 4b = 16\)
\(a^2 + 2ab + b^2 = 16\)
so now we combine them
\(4a + 4b = a^2 + 2ab + b^2\)
\(4a + 4b - 2ab - 2b^2 = (a+b)(a-b)\)
more on how that works if you ask me.
\(4(a+b) - 2b(a+b) = (a+b)(a-b)\)
\(4-2b = a-b\)
i think thats correct..
\(a+b = 4\)
oh well.....................
thats depressing oofoof
okok for b) its just 4, literally
and for c i don't think any integers work therfore no real numbers work becuase the max number for a or b is 1, becuase 2^4 is already 16 and negatives^4 are always postive for i think c is none.
and for a) when i did the failed equations u see a 2ab when i squared the first equation,try to single that out and u will get the answer. for a
HINT. for a, do \(2ab = a^4+b^4-a^2 - b^2\)
that will be equal to \(2ab = (a^2)(a^2 - 1) + (b^2)(b^2 - 1)\)
then use the second equation to substiture the a^2 and b^2 and u will get somewhere...
