+584 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)2−3(x2−9)=0
plus 3/(x^2-9) both side
2(x+3)2=3(x2−9)
bot side times (x+3),
2(x+3)=3(x−3)
2×x−6=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)2−3(x2−9)=0
2[(x+3)×(x+3)]−3((x+3)×(x−3))=0
2×(x−3)[(x+3)×(x+3)×(x−3)]−3×(x+3)[(x+3)×(x+3)×(x−3)]=0
both sides of the equation multiply by (x+3)^2(x-3) (x not equal 3 or -3)
2×(x−3)−3×(x+3)=0
2×x−6−3×x−9=0
−x−15=0
x=−15
+584 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)2−3(x2−9)=0
plus 3/(x^2-9) both side
2(x+3)2=3(x2−9)
bot side times (x+3),
2(x+3)=3(x−3)
2×x−6=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)2−3(x2−9)=0
2[(x+3)×(x+3)]−3((x+3)×(x−3))=0
2×(x−3)[(x+3)×(x+3)×(x−3)]−3×(x+3)[(x+3)×(x+3)×(x−3)]=0
both sides of the equation multiply by (x+3)^2(x-3) (x not equal 3 or -3)
2×(x−3)−3×(x+3)=0
2×x−6−3×x−9=0
−x−15=0
x=−15
+2973 Hey things happen.
nobodys perfect but a few people (but that is a all out opinion, so really nobodys perfect)
