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Alegebraic equation halp

Name: Anonymous 2008-10-15 21:30

I'm stuck on one particular problem:

n - 2/n = 23/5

Now I cancel out the denominators by 5n:

5n^2 - 10 = 23n

Then go further:

5n^2 - 23n - 10 = 0

I'm stuck now as I can't find a way to factor this out as. I'm not asking for anyone to solve it, but just where I need to go to solve it.

Name: 4tran 2008-10-15 21:53

(5n+2)(n-5)=0
If you can't factor, then quadratic equation/complete the square.

Name: Anonymous 2008-10-15 21:56

(x-5 )( 5x+2 )

x-5         5x+2=0  =  -2/5

(x=5)(x=-2/5)


i wish i could explain how i did it, i backward foiled it in my head .

or basically multiply   (5)(-10)   -50   and the coefficient of x=  -23           find the number who's product equals to -50  and equals the sum to -23
 

those numbers are 2 and -25

so now we have these number   take out  5x  and put in  5x^2+2x-25x-10=0

now factor     5x^2+2x   and   25x-10
                5x+2=0          x=5

sorry im bad at explaining

Name: Anonymous 2008-10-15 22:06

Thanks guys.

Name: Anonymous 2008-10-15 23:17

ax^2+bx+c

5x^2 - 23x - 10
5x^2 - 25x + 2x - 10
5x^2 - 25x | + 2x - 10
5x(x-5) | + 2x - 10
5x(x-5) | + 2(x-5)
(x-5)(5x+2)
(x-5)(5x+2) = 0
x = 5, 5/2

Sorry if I didn't explain it well enough

The breakers (  |  ) are simply for visual management, they are not actually part of the equation

Name: Anonymous 2008-10-15 23:18

Samefag as last, meant to put x=5, -2/5

Name: Anonymous 2008-10-17 5:44

use the fuckin' quadratic formula.

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