x^3+ y^3 = 91. x^2- xy + y2 = 13. Find the values of x and y.
Answers
Answer:
Step-by-step explanation:
The first thing anyone would try is x=3,y=4 , which checks. It will be useful to have one root so we can turn our cubic equation into a quadratic one.
Proceeding like Mr. Pradeep, let s=x+y . I use s because my t s look like + signs. Of course my s looks like a 5 ... anyway.
s2=(x+y)2=x2+2xy+y2=25+2xy
2xy=s2−25
s3=(x+y)3=x3+3x2y+3y2x+y3=91+3xy(x+y)=91+3sxy
0=−2s3+3s(2xy)+182= −2s3+3s(s2−25)+182=s3−75s+182
Since 3,4 is a solution, s=3+4=7 must be a root of this equation, so we can factor out (s−7). Doing the long division, we get
s3−75s+182=(s−7)(s2+7s−26)=0
The three roots are s=7 and s=−7±317√2 . The latter two are real and real ugly. I can see why Mr. Pradeep resorted to decimal approximations.
Now that we have s , we can go back and solve for x and y . Let's do it symbolically first.
We have s=x+y , so y=s−x , and 2xy=s2−25 .
2x(s−x)=s2−25
0=2x2−2sx+(s2−25)
x=14(2s±(2s)2−4(−2)(s2−25)−−−−−−−−−−−−−−−−−−√)=12(s±50−s2−−−−−−√)
So for s=7 we have x=12(7±50−72−−−−−−√)=12(7±1) . x=3 or x=4 . Note that when one is x the other is y .
Now we're left with the messier parts. I'm busy now so I will come back to those.
10 days later and I'm back. This is messy, but let's get those other roots. We have
s=−7±317√2 and x=12(s±50−s2−−−−−−√)
That's enough to calculate the answers, but I was after the closed form. It doesn't simplify down much, but we can make progress.
(2s)2=49+9⋅17−±2⋅2117−−√=202−±4217−−√
200−4s2=−2±4217−−√
We're going to have to take the square root of that. Let's first focus on the s with the plus sign, which gives a positive 200−4s2 .
250−s2−−−−−−√=200−4s2−−−−−−−−√=−2+4217−−√−−−−−−−−−−√
x=12(s±50−s2−−−−−−√)
x= 14(−7+317−−√±−2+4217−−√−−−−−−−−−−√)
Simple. These give two more real solutions, x≈4.613 and x≈−1.928. Again, each is the other's y .
What about that complex solution? That's from s=−7−317√2
x= 14(−7−317−−√±i2+4217−−√−−−−−−−−−√)
Those are two more complex roots, x≈−4.842±3.309i
OK, we found six roots. They come as three x,y pairs that you can swap due to the symmetry of the problem.