SOLUTION: Solve: 8/(x+2)+8/(x-2)=3 ok I have been working and working on this and it comes up (8x+16)+ (8x--16)=3 then iwent 24+-8 = 3 but something is wrong we are working on quadratic an

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Question 384935: Solve: 8/(x+2)+8/(x-2)=3 ok I have been working and working on this and it comes up (8x+16)+ (8x--16)=3 then iwent 24+-8 = 3 but something is wrong we are working on quadratic and rooting. Please help!!
Here's another one I'm stuck with,Solve by factoring and using the principle of zero products:
2x^2-11x+15=0
So I did this 2(x-2)(x+2)-11+15 then add 11 on both sides which would be 2x2=4 and 15+11 =4 -11+11=0 But does not make any since.
Found 2 solutions by tara0066, ewatrrr:
Answer by tara0066(31)   (Show Source): You can put this solution on YOUR website!
Going to be tricky to explain on this one.
2x^2-11x+15=0 (Multiply the a and c values=30, then find two multiples that =30 but the sum is -11
These numbers are -6 and -5.)
2x^2 -6x-5x+15=0 (Replaced the -11x with those two numbers.)
2x(x-3)-5(x-3)=0 (Pull out what is common in the first two and then do the same for the last two. You
should have the same left over in both sets.)
(2x-5)(x-3)=0 (Write the repeated set only once and the factors you pulled out of both sets as the other.
2x-5=0 (add 5 to both sides)
2x=5 (divide by 2)
x=5/2
x-3=0 (add 3 to both sides)
x=3

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi, 

solving        

 Multiplying ALL terms by (x+2)(x-2)

 8(x-2) + 8(x+2) = 3[(x+2)(x-2)]

 8x - 16 + 8x + 16 = 3(x^2 - 4)

     16x = 3x^2 - 12

  3x^2 -16x - 12 = 0





  

     x = 6

     x = -4/6 or - 2/3



2x^2-11x+15=0 

factoring

(2x -5)(x -3)=0 Note:SUM of the inner product(-5x) and the outer product(-6x) = -11x

Using the principle of zero products:

(2x -5)=0

  x = 5/2

(x -3)=0

  x = 3 

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