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avatar+262 

 Jul 13, 2015

Best Answer 

 #1
avatar+584 
+13

In your second step,the denominator did not change; you wrote 3 instead of 2 as the numerator in the first frist fraction.And there is something else wrong in your work.

You can do it in this way,

  2(x+3)23(x29)=0

plus 3/(x^2-9) both side

2(x+3)2=3(x29)

bot side times (x+3),

2(x+3)=3(x3)

2×x6=3×x+9

x=15

check: 2/(-15+3)^2-3/(x^2-9)=1/72-1/72=0 and x=-15 not equal to 3 or -3.

another way:

2(x+3)23(x29)=0

2[(x+3)×(x+3)]3((x+3)×(x3))=0

2×(x3)[(x+3)×(x+3)×(x3)]3×(x+3)[(x+3)×(x+3)×(x3)]=0

both sides of the equation multiply by (x+3)^2(x-3) (x not equal 3 or -3)

2×(x3)3×(x+3)=0 

2×x63×x9=0

x15=0

x=15

 Jul 13, 2015
 #1
avatar+584 
+13
Best Answer

In your second step,the denominator did not change; you wrote 3 instead of 2 as the numerator in the first frist fraction.And there is something else wrong in your work.

You can do it in this way,

  2(x+3)23(x29)=0

plus 3/(x^2-9) both side

2(x+3)2=3(x29)

bot side times (x+3),

2(x+3)=3(x3)

2×x6=3×x+9

x=15

check: 2/(-15+3)^2-3/(x^2-9)=1/72-1/72=0 and x=-15 not equal to 3 or -3.

another way:

2(x+3)23(x29)=0

2[(x+3)×(x+3)]3((x+3)×(x3))=0

2×(x3)[(x+3)×(x+3)×(x3)]3×(x+3)[(x+3)×(x+3)×(x3)]=0

both sides of the equation multiply by (x+3)^2(x-3) (x not equal 3 or -3)

2×(x3)3×(x+3)=0 

2×x63×x9=0

x15=0

x=15

fiora Jul 13, 2015
 #2
avatar+262 
+8

thank you!dont understand how i could do such a stupid mistake

 Jul 13, 2015
 #3
avatar+2973 
+8

Hey things happen.

nobodys perfect but a few people (but that is a all out opinion, so really nobodys perfect)

 Jul 13, 2015

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