#1

my math homework blows, i've got it all done except two problems any hel anyone can offer would be greatly appreciated here they are:

1) (x+1)(2x-3) = x+2 this one's answer should be in radical form

2) 3/x-2 + 4/x = 1

find the values of x

1) (x+1)(2x-3) = x+2 this one's answer should be in radical form

2) 3/x-2 + 4/x = 1

find the values of x

#2

ha ha u have math homework on the weekend. thats too bad. u know what rip dat shit up u aint ever gone use that later in life

#3

1:

2x{2}+2x-3x-3=0

2x(x+1)-3(x+1)=0

(2x-3)(x+1)=0

x=3/2 or x=-1

for 2 is it:

3/(x-2)

or

(3/x)-2

2x{2}+2x-3x-3=0

2x(x+1)-3(x+1)=0

(2x-3)(x+1)=0

x=3/2 or x=-1

for 2 is it:

3/(x-2)

or

(3/x)-2

#4

1 do the radicals yourself:

2x squared + 3x + 2x + 3x = x + 2

2x squared + 7x= 2

2, I don't have the symbols for on my computer

EDIT: Someone got to it before me. XD That always happens

2x squared + 3x + 2x + 3x = x + 2

2x squared + 7x= 2

2, I don't have the symbols for on my computer

EDIT: Someone got to it before me. XD That always happens

#5

Defy: LOL, i usualy get beat too so i rush it, and then when i post it i check for mistakes

#6

For number one expand bracket:-

2x^2 + 2x -3x -3 = x + 2

2x^2 - 2x - 5 = 0

For number two, multiply everyone by x(x-2) so it is:-

3x + 4(x-2) = x(x-2)

3x + 4x - 8 = X^2 - 2x

7x + 8 = x^2 - 2x

x^2 - 9x - 8 = 0

For both use the quadratic equation. Have fun

2x^2 + 2x -3x -3 = x + 2

2x^2 - 2x - 5 = 0

For number two, multiply everyone by x(x-2) so it is:-

3x + 4(x-2) = x(x-2)

3x + 4x - 8 = X^2 - 2x

7x + 8 = x^2 - 2x

x^2 - 9x - 8 = 0

For both use the quadratic equation. Have fun

#7

thanks

*Last edited by eviledge87 at Jan 6, 2007,*

#8

3/(x-2) + 4/x = 1x

3x/(x-2) +4=1x

3x+4(x-2)=1x(x-2)

3x+4x-8=x^2-2x

-x^2+9x-8=0

x^2-9x+8=0

blah blah blah

3x/(x-2) +4=1x

3x+4(x-2)=1x(x-2)

3x+4x-8=x^2-2x

-x^2+9x-8=0

x^2-9x+8=0

blah blah blah