SOLUTION: Hi. I have 3 problems that I am having difficulty with and I was wondering if you could help. 1) 4x^2 = 12x 2) 6x^2 = 6x 3) 3x^2 + 6 = 2x^2 +18 Thanks!

Algebra ->  Radicals -> SOLUTION: Hi. I have 3 problems that I am having difficulty with and I was wondering if you could help. 1) 4x^2 = 12x 2) 6x^2 = 6x 3) 3x^2 + 6 = 2x^2 +18 Thanks!      Log On


   

Question 897660: Hi. I have 3 problems that I am having difficulty with and I was wondering if you could help.
1) 4x^2 = 12x
2) 6x^2 = 6x
3) 3x^2 + 6 = 2x^2 +18

Thanks!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(1)
+4x%5E2+=+12x+
+4x%5E2+-+12x+=+0+
+4x%2A%28+x+-+3+%29+=+0+
For this to be true, either
+x+=+0+ or +x+=+3+
----------------------
(2)
+6x%5E2+=+6x+
+6x%5E2+-+6x+=+0+
+6x%2A%28+x+-+1+%29+=+0+
+x+=+0+ and
+x+=+1+
--------------
(3)
+3x%5E2+%2B+6+=+2x%5E2+%2B+18+
Subtract +2x%5E2+ from both sides
+x%5E2+%2B+6+=+18+
Subtract +6+ from both sides
+x%5E2+=+12+
+x+=+sqrt%2812%29+
+x+=+2%2Asqrt%283%29+
and, also
+x+=+-sqrt%2812%29+
+x+=+-2%2Asqrt%283%29+