SOLUTION: What are the roots of y=x^2+2x-3

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Question 98414: What are the roots of y=x^2+2x-3
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's use the quadratic formula to find the roots:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve x%5E2%2B2%2Ax-3=0 ( notice a=1, b=2, and c=-3)

x+=+%28-2+%2B-+sqrt%28+%282%29%5E2-4%2A1%2A-3+%29%29%2F%282%2A1%29 Plug in a=1, b=2, and c=-3



x+=+%28-2+%2B-+sqrt%28+4-4%2A1%2A-3+%29%29%2F%282%2A1%29 Square 2 to get 4



x+=+%28-2+%2B-+sqrt%28+4%2B12+%29%29%2F%282%2A1%29 Multiply -4%2A-3%2A1 to get 12



x+=+%28-2+%2B-+sqrt%28+16+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-2+%2B-+4%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-2+%2B-+4%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%28-2+%2B+4%29%2F2 or x+=+%28-2+-+4%29%2F2

Lets look at the first part:

x=%28-2+%2B+4%29%2F2

x=2%2F2 Add the terms in the numerator
x=1 Divide

So one answer is
x=1



Now lets look at the second part:

x=%28-2+-+4%29%2F2

x=-6%2F2 Subtract the terms in the numerator
x=-3 Divide

So another answer is
x=-3

So our solutions are:
x=1 or x=-3

Notice when we graph x%5E2%2B2%2Ax-3, we get:

+graph%28+500%2C+500%2C+-13%2C+11%2C+-13%2C+11%2C1%2Ax%5E2%2B2%2Ax%2B-3%29+

and we can see that the roots are x=1 and x=-3. This verifies our answer