SOLUTION: Solve: x2(squared)+2x-35=0

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Question 99342: Solve: x2(squared)+2x-35=0
Found 2 solutions by timmy1729, jim_thompson5910:
Answer by timmy1729(23) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-35+=+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%2A1%2A-35=144.

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

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+144+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%282%29-sqrt%28+144+%29%29%2F2%5C1+=+-7

Quadratic expression 1x%5E2%2B2x%2B-35 can be factored:
1x%5E2%2B2x%2B-35+=+1%28x-5%29%2A%28x--7%29
Again, the answer is: 5, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-35+%29

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:


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-35=0 ( notice a=1, b=2, and c=-35)

x+=+%28-2+%2B-+sqrt%28+%282%29%5E2-4%2A1%2A-35+%29%29%2F%282%2A1%29 Plug in a=1, b=2, and c=-35



x+=+%28-2+%2B-+sqrt%28+4-4%2A1%2A-35+%29%29%2F%282%2A1%29 Square 2 to get 4



x+=+%28-2+%2B-+sqrt%28+4%2B140+%29%29%2F%282%2A1%29 Multiply -4%2A-35%2A1 to get 140



x+=+%28-2+%2B-+sqrt%28+144+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-2+%2B-+12%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-+12%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%28-2+%2B+12%29%2F2 or x+=+%28-2+-+12%29%2F2

Lets look at the first part:

x=%28-2+%2B+12%29%2F2

x=10%2F2 Add the terms in the numerator
x=5 Divide

So one answer is
x=5



Now lets look at the second part:

x=%28-2+-+12%29%2F2

x=-14%2F2 Subtract the terms in the numerator
x=-7 Divide

So another answer is
x=-7

So our solutions are:
x=5 or x=-7

Notice when we graph x%5E2%2B2%2Ax-35, we get:

+graph%28+500%2C+500%2C+-17%2C+15%2C+-17%2C+15%2C1%2Ax%5E2%2B2%2Ax%2B-35%29+

and we can see that the roots are x=5 and x=-7. This verifies our answer