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Simple Algebra problem

coder1

Senior member
Some of you may know that I have been teaching myself alebra again after being away rom it for 10 years. Recently I came across this problem:

solve for X:

x^2-8x-20=0

I see some of the answers as being in a 8,2 format.

What does this mean, that x has more than one answer? Can anyone explain this and walk me through the process of solving this problem.

Thanks again for your help.

 
x^2-8x-20=0
(x-10)(x+2)=0
x-10= 0... x+2=0
x= 10 ..... x= -2

x= 10 or -2

The reason the question has two answers is because problems like this can be factored down and then both of the factors set equal to zero.

I think that's right.


: ) Amanda
 
use the quadratic formula.

(-b +/- sqrt(b^2-4ac))2a

Where a, b, and c are the coeeficients of your polynomial.

In this problem x = (8 +/- sqrt((-8)^2-4*1*-20))2

x = (8 +/- 12)/2

X = 10 or -2
 
Originally posted by: coder1
Some of you may know that I have been teaching myself alebra again after being away rom it for 10 years. Recently I came across this problem:

solve for X:

x^2-8x-20=0

I see some of the answers as being in a 8,2 format.

What does this mean, that x has more than one answer? Can anyone explain this and walk me through the process of solving this problem.

Thanks again for your help.

yes, x has 2 possible values because it is an equation of degree 2 (highest power x is raised to)

you can use the quadratic equation, but you can eyeball this one... just sort of guess... what factors of -20 would add up to -8. and viola! you come up with 10 and -2.

you can check

(x-10)(x+2) = x^2 -8x -20
 
x^2-8x-20=0
can be re-written as:
(x-10)(x+2)=0
by reverse FOIL
and therefore for the equation to work
x must be 10 or x must be -2

x=10,-2



-MC
 
I know that you can break this binomial down to (x-10)(x+2)=0, then I forget how to get to the answer after that.


edit: other guys answered much better than I did. Maybe I need to reteach myself some algebra.
 
Originally posted by: CPA
I know that you can break this binomial down to (x-10)(x+2)=0, then I forget how to get to the answer after that.


edit: other guys answered much better than I did. Maybe I need to reteach myself some algebra.

from there, you basically know that the product of two terms is 0, so one of the terms must be 0 (since x-10 and x+2 can't be equal). so you have two equations for two possible solutions...

x-10 = 0
x+2 = 0

these yield

x = 10
x = -2
 
wow, I wish I had been doing simple problems like that while I was taking my calculus final today..ERR
 
here

If you enter the equation you have x*x - 8x - 20
and hit graph you can see how there are 2 answers to the equation
where a value of x can result in a answer of 0, ie the crossings on the x axis.

for defactoring it just find 2 numbers where their sum is -8 and their product will be -20 ie. -2 and 10
if you can't find any just use the quadratic equation.
 
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