SOLUTION: How do I solve this algebra problem? solve x(x-8)= -15

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Question 214177: How do I solve this algebra problem?
solve x(x-8)= -15
Found 3 solutions by Alan3354, jim_thompson5910, drj:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x(x-8)= -15
x%5E2+-+8x+%2B+15+=+0
(x-5)*(x-3) = 0
x=5, x=3

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x%28x-8%29=+-15 Start with the given equation


x%5E2-8x=+-15 Distribute


x%5E2-8x%2B15=0 Add 15 to both sides.


Notice we have a quadratic in the form of ax%5E2%2Bbx%2Bc where a=1, b=-8, and c=15


Let's use the quadratic formula to solve for x


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


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


x+=+%288+%2B-+sqrt%28+%28-8%29%5E2-4%281%29%2815%29+%29%29%2F%282%281%29%29 Negate -8 to get 8.


x+=+%288+%2B-+sqrt%28+64-4%281%29%2815%29+%29%29%2F%282%281%29%29 Square -8 to get 64.


x+=+%288+%2B-+sqrt%28+64-60+%29%29%2F%282%281%29%29 Multiply 4%281%29%2815%29 to get 60


x+=+%288+%2B-+sqrt%28+4+%29%29%2F%282%281%29%29 Subtract 60 from 64 to get 4


x+=+%288+%2B-+sqrt%28+4+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%288+%2B-+2%29%2F%282%29 Take the square root of 4 to get 2.


x+=+%288+%2B+2%29%2F%282%29 or x+=+%288+-+2%29%2F%282%29 Break up the expression.


x+=+%2810%29%2F%282%29 or x+=++%286%29%2F%282%29 Combine like terms.


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


So the answers are x+=+5 or x+=+3

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
How do I solve this algebra problem?
solve x%28x-8%29=+-15

Step 1. Multiply and simplify

x%28x-8%29=+-15

x%5E2-8x=+-15
Step 2. Add 15 to both sides to get a quadratic equation

x%5E2-8x%2B15=+-15%2B15

x%5E2-8x%2B15=0

Step 3. Factor as follows

x%5E2-8x%2B15=%28x-5%29%28x-3%29=0

Then x-5=0 and x-3

x=5 and x=3 ANSWER
We can also use the quadratic formula x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=1, b=-8 and c=15

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%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=%28-8%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-%28-8%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-8%29-sqrt%28+4+%29%29%2F2%5C1+=+3

Quadratic expression 1x%5E2%2B-8x%2B15 can be factored:
1x%5E2%2B-8x%2B15+=+1%28x-5%29%2A%28x-3%29
Again, the answer is: 5, 3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-8%2Ax%2B15+%29



Based on the above steps using the quadratic formula, we have x=5 and x=3...same as before.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J