SOLUTION: How do I find the x-intercepts for the graph of this equation? y=x^2+3x-4

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Question 295750: How do I find the x-intercepts for the graph of this equation?
y=x^2+3x-4
Found 2 solutions by Alan3354, JBarnum:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find the x-intercepts for the graph of this equation?
y=x^2+3x-4
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At any x-intercept, y = 0.
x%5E2+%2B+3x+-+4+=+0
I would factor it, that's the simplest way.

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
its a quadratic equation set y=0 and solve for x.
find 2 numbers that multiply to get -4 and add to get +3
-1 and +4
so 0=%28x-1%29%28x%2B4%29
x=1x=-4 these are the 2 intercepts
if you cant remember this method then use the quadratic method like the solver below
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%2B-4+=+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=%283%29%5E2-4%2A1%2A-4=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+1
x%5B2%5D+=+%28-%283%29-sqrt%28+25+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B3x%2B-4 can be factored:
1x%5E2%2B3x%2B-4+=+1%28x-1%29%2A%28x--4%29
Again, the answer is: 1, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B-4+%29