SOLUTION: Solve the system of equations: y-3x-2=0 x^2+y^2=4

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Question 447021: Solve the system of equations: y-3x-2=0
x^2+y^2=4
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
y-3x-2=0
y=3x+2
x2+(3x+2)2-4=0
x2+9x2+12x+4-4=0
10x2+12x=0
10x2=-12x
10x=-12
x=-12/10=-6/5
y=-8/5
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 10x%5E2%2B12x%2B0+=+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=%2812%29%5E2-4%2A10%2A0=144.

Discriminant d=144 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-12%2B-sqrt%28+144+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2812%29%2Bsqrt%28+144+%29%29%2F2%5C10+=+0
x%5B2%5D+=+%28-%2812%29-sqrt%28+144+%29%29%2F2%5C10+=+-1.2

Quadratic expression 10x%5E2%2B12x%2B0 can be factored:
10x%5E2%2B12x%2B0+=+10%28x-0%29%2A%28x--1.2%29
Again, the answer is: 0, -1.2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+10%2Ax%5E2%2B12%2Ax%2B0+%29

Throwing out 0, we get x=-6/5..