SOLUTION: Not understanding this: Quadratic Coefficients: a = 1, b = 8, c = 15 Looking for "Solutions" x1=__________; x2 =__________ Can you explain?

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Question 148758: Not understanding this:
Quadratic Coefficients: a = 1, b = 8, c = 15 Looking for "Solutions" x1=__________; x2 =__________
Can you explain?
Found 3 solutions by jim_thompson5910, checkley77, nerdybill:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Simply use the coefficients in the Quadratic Formula:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%288%29+%2B-+sqrt%28+%288%29%5E2-4%281%29%2815%29+%29%29%2F%282%281%29%29 Plug in a=1, b=8, and c=15


x+=+%28-8+%2B-+sqrt%28+64-4%281%29%2815%29+%29%29%2F%282%281%29%29 Square 8 to get 64.


x+=+%28-8+%2B-+sqrt%28+64-60+%29%29%2F%282%281%29%29 Multiply 4%281%29%2815%29 to get 60


x+=+%28-8+%2B-+sqrt%28+4+%29%29%2F%282%281%29%29 Subtract 60 from 64 to get 4


x+=+%28-8+%2B-+sqrt%28+4+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-8+%2B-+2%29%2F%282%29 Take the square root of 4 to get 2.


x+=+%28-8+%2B+2%29%2F%282%29 or x+=+%28-8+-+2%29%2F%282%29 Break up the expression.


x+=+%28-6%29%2F%282%29 or x+=++%28-10%29%2F%282%29 Combine like terms.


x+=+-3 or x+=+-5 Simplify.


So our answers are x%5B1%5D+=+-3 or x%5B2%5D+=+-5 (note: the order does not matter)


Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+8x+15=0 (if a,b & c are positive integers)
(x+5)(x+3)=0
x+5=0
x=-5 answer.
x+3=0
x=-3 answer.

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
A "quadratic" equation comes in the standard form of:
ax^2 + bx + c = 0
where a,b, and c are the coefficients.
.
What they are asking you to do is plug it into the "quadratic formula" and solve for solutions of 'x'.
.
In your problem, the coefficients are:
a = 1, b = 8, c = 15
.
solutions: -3 and -5
.
Details below:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B8x%2B15+=+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=%288%29%5E2-4%2A1%2A15=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-8%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%288%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+-3
x%5B2%5D+=+%28-%288%29-sqrt%28+4+%29%29%2F2%5C1+=+-5

Quadratic expression 1x%5E2%2B8x%2B15 can be factored:
1x%5E2%2B8x%2B15+=+1%28x--3%29%2A%28x--5%29
Again, the answer is: -3, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B8%2Ax%2B15+%29