SOLUTION: Please help me solve the following problem: Solve: x^2 + 3x - 4 = 0

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Question 153536: Please help me solve the following problem:
Solve: x^2 + 3x - 4 = 0

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve: x^2 + 3x - 4 = 0
Factor to get:
(x+4)(x-1) = 0
x = -4 or x = 1
=====================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+3x+-+4
Solving this means finding the roots, or finding
the values of x that make x%5E2+%2B+3x+-+4+=+0
true.
If I just say, off the top of my head, x+=+2 for
a solution, then I'm saying 2%5E2+%2B+3%2A2+-+4+=+0 and
4+%2B+6+-+4+=+0
10+-+4+=+0
6+=+0
That's not true, so x+=+2 is not a solution
One way to solve this is by completing the square.
First add 4 to both sides
x%5E2+%2B+3x+=+4
Now take one-half of the coefficient of the x term,
square it, and add it to both sides, like this:
x%5E2+%2B+3x+%2B+%283%2F2%29%5E2+=+4+%2B+%283%2F2%29%5E2
x%5E2+%2B+3x+%2B+%289%2F4%29+=+%2816%2F4%29+%2B+%289%2F4%29
x%5E2+%2B+3x+%2B+%289%2F4%29+=+25%2F4
This is the same as
%28x+%2B+%283%2F2%29%29%5E2+=+25%2F4
Now take the square root of both sides
x+%2B+%283%2F2%29+=+5%2F2
and also
x+%2B+%283%2F2%29+=+-%285%2F2%29 (since 25%2F4 has a + and a - square root)
Subtract 3%2F2 from both sides
x+=+%285%2F2%29+-+%283%2F2%29
x+=+1
And also
x+=+-%285%2F2%29+-+%283%2F2%29
x+=+-4
So, the solutions are x+=+1 and x+=+-4
Plug these values into the equation to see if they
are roots, for instance:
%28-4%29%5E2+%2B+3%2A%28-4%29+-+4+=+0
16+-+12+-+4+=+0
16+-+16+=+0
0+=+0
So, x+=+-4 is a solution