SOLUTION: X squared + y squared = 4 Y minus X = 1 We have to solve the system it says... But my teacher says that all we have to solve for is X and Y because it's a longer problem.

Algebra ->  Finance -> SOLUTION: X squared + y squared = 4 Y minus X = 1 We have to solve the system it says... But my teacher says that all we have to solve for is X and Y because it's a longer problem.       Log On


   

Question 1031015: X squared + y squared = 4
Y minus X = 1
We have to solve the system it says... But my teacher says that all we have to solve for is X and Y because it's a longer problem.
Found 2 solutions by Alan3354, fractalier:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
X^2 + y^2 = 4
Y - X = 1 --> y = x + 1
-----------
X^2 + y^2 = 4
Sub for y
x^2 + (x+1)^2 = 4
2x^2 + 2x + 1 = 4
2x^2 + 2x - 3 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B2x%2B-3+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282%29%5E2-4%2A2%2A-3=28.

Discriminant d=28 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+28+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+28+%29%29%2F2%5C2+=+0.822875655532295
x%5B2%5D+=+%28-%282%29-sqrt%28+28+%29%29%2F2%5C2+=+-1.8228756555323

Quadratic expression 2x%5E2%2B2x%2B-3 can be factored:
2x%5E2%2B2x%2B-3+=+%28x-0.822875655532295%29%2A%28x--1.8228756555323%29
Again, the answer is: 0.822875655532295, -1.8228756555323. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B2%2Ax%2B-3+%29

=================
x = -1/2 - sqrt(7)/2
y = +1/2 - sqrt(7)/2
======================
x = -1/2 + sqrt(7)/2
y = +1/2 + sqrt(7)/2

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, from
x%5E2+%2B+y%5E2+=+4 and y+-+x+=+1
we solve the second one and substitute into the first one...we get
y = x + 1 and then
x^2 + (x+1)^2 = 4
x^2 + x^2 + 2x + 1 = 4
2x^2 + 2x - 3 = 0
This cannot be factored so we use the quadratic formula...
x = (-2 + sqrt(4 + 24))/4 = (-2 + sqrt(28))/4 = (-1 + sqrt(7))/2
and
x = (-2 - sqrt(4 + 24))/4 = (-2 - sqrt(28))/4 = (-1 - sqrt(7))/2
Plug in and solve for y, as in y = x + 1...