SOLUTION: square root 2x-5 = 1+ square root x-3 I understand I have to square both sides. Where I lose it is here: squaring 1 + square root x-3. My textbook has: 2times square root x-3 (en

Algebra ->  Radicals -> SOLUTION: square root 2x-5 = 1+ square root x-3 I understand I have to square both sides. Where I lose it is here: squaring 1 + square root x-3. My textbook has: 2times square root x-3 (en      Log On


   

Question 29115: square root 2x-5 = 1+ square root x-3
I understand I have to square both sides. Where I lose it is here:
squaring 1 + square root x-3. My textbook has: 2times square root x-3 (end square root) + (x-3)
Thanks
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%282x-5%29+=+1+%2B+sqrt%28x-3%29+

OK, to remove at least some of the square roots here, you could square both sides, as:
+%28sqrt%282x-5%29%29%5E2+=+%281+%2B+sqrt%28x-3%29%29%5E2+

This is just like %28a%2Bb%29%5E2 - don't be freaked by the fact we have 2 terms of the right.

+%28sqrt%282x-5%29%29%5E2+=+%281+%2B+sqrt%28x-3%29%29%2A%281+%2B+sqrt%28x-3%29%29+
+2x-5+=+1+%2B+sqrt%28x-3%29+%2B+sqrt%28x-3%29+%2B+sqrt%28x-3%29sqrt%28x-3%29+
+2x-5+=+1+%2B+2sqrt%28x-3%29+%2B+%28x-3%29+
+2x-5+=+1+%2B+2sqrt%28x-3%29+%2B+x-3+
+2x-5+=+2sqrt%28x-3%29+%2B+x-2+
+x-5+=+2sqrt%28x-3%29+-+2+
+x-3+=+2sqrt%28x-3%29+
+%28x-3%29%2F2+=+sqrt%28x-3%29+

Again, square both sides:
+%28%28x-3%29%2F2%29%5E2+=+%28sqrt%28x-3%29%29%5E2+
+%28x-3%29%5E2%2F2%5E2+=+x-3+
+%28x%5E2-6x%2B9%29%2F4+=+x-3+
+x%5E2-6x%2B9+=+4%28x-3%29+
+x%5E2-6x%2B9+=+4x-12+
+x%5E2-10x%2B9+=+-12+
+x%5E2-10x%2B21+=+0+
(x-7)(x-3) = 0

so x-7=0 OR x-3=0
so x=7 or x=3

This looks rather long, but i have written out every step on a different line so you can see everything.

jon.