SOLUTION: I need help solving: x(x+2)(x+4)=315 i know the answer, i just have to figure out how to get there other than trial and error.

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Question 78021This question is from textbook algebra 1
: I need help solving:
x(x+2)(x+4)=315
i know the answer, i just have to figure out how to get there other than trial and error. This question is from textbook algebra 1

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
+x%28x%2B2%29%28x%2B4%29=315
x%5E3%2B6x%5E2%2B8x=315 Foil

x%5E3%2B6x%5E2%2B8x-315=0 Subtract 315 from both sides

Use the calculator's root feature to find one factor. This will allow us to factor the quick way (the other route involves a lot of luck and patience).
%28x-5%29%28x%5E2%2B11x%2B63%29=0
So we have one answer of x=5. Lets use the quadratic formula to find the rest:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B11x%2B63+=+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=%2811%29%5E2-4%2A1%2A63=-131.

The discriminant -131 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -131 is + or - sqrt%28+131%29+=+11.4455231422596.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B11%2Ax%2B63+%29


So our only real answer is x=5