SOLUTION: Solve: {{{ sqrt (b-20) +4= sqrt (b) }}}

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Question 593889: Solve: +sqrt+%28b-20%29+%2B4=+sqrt+%28b%29+
Found 2 solutions by unlockmath, jsmallt9:
Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
With this +sqrt+%28b-20%29+%2B4=+sqrt+%28b%29+
we square both sides to get:
b-20 +8 sqrt (b-20) +16 = b
combine like terms and subtract b and add 4 to get:
8 sqrt (b-20)= 4
Square each side again:
64(b-20)=16
Expand out:
64b-1280=16
Add 1280 and divide by 64 to get:
b=20.25
Make sense?
RJ
www.math-unlock.com

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a procedure for solving this kind of equation:
  1. Isolate (or "solve for") one of the square roots.
  2. Square both sides of the equation. (Remember, squaring a side of an equation is multiplying it by itself. It is not squaring each individual term.)
  3. If there are still any square roots in the equation, repeat steps 1, 2 and 3.
  4. At this point, there should be NO square roots remaining. Use appropriate techniques to solve the type equation you now have.
  5. Check your answer(s)! This is not optional. Squaring both sides of an equation can introduce what are called extraneous solutions. Extraneous solutions are solutions that work in the squared equation but not in the original equation. Check your answer(s) in the original equation. Any that do not work are extraneous and must be rejected! Notes:
    • Extraneous solutions are NOT caused by errors. Even expert mathematicians must check their answers on these problems.
    • If it happens that all your answers are extraneous, it just means that there are no solutions to the equation.

Let's see this in action:
+sqrt+%28b-20%29+%2B4=+sqrt+%28b%29+
1. Isolate a square root. The square root on the right side is already by itself so we are ready to move to the next step.

2. Square both sides
+%28sqrt+%28b-20%29+%2B4%29%5E2=+%28sqrt+%28b%29%29%5E2+
The right side is just a single term an so it is simple to square. The left side, however, has two terms. So you must use FOIL on +%28sqrt+%28b-20%29+%2B4%29%2A%28sqrt+%28b-20%29+%2B4%29 or use the pattern: %28a%2Bb%29%5E2+=+a%5E2+%2B+2ab+%2B+b%5E2. I like using the pattern:
+%28sqrt+%28b-20%29%29%5E2+%2B+2%2Asqrt%28b-20%29%2A4+%2B+4%5E2=+%28sqrt+%28b%29%29%5E2+
which simplifies as follows:
+%28b-20%29+%2B8%2Asqrt%28b-20%29+%2B+16+=+b
b+%2B+8%2Asqrt%28b-20%29+-+4+=+b

3. There is still a square root so we have to repeat.

1. Isolate a square root. There is only one square root so that is what we must isolate. Adding 4 and subtracting b we get:
8%2Asqrt%28b-20%29+=+4

2. Square both sides. Both sides are single terms and so easy to square:
%288%2Asqrt%28b-20%29%29%5E2+=+%284%29%5E2
which simplifies as follows:
64(b-20) = 16
64b - 1280 = 16

3. No square roots! On to step 4...

4. Solve the equation. This is a simple first degree equation. Adding 1280:
64b = 1296
dividing by 64:
b = 1296/64 = 20 16/64 = 20 1/4

5. Check answer(s) using the original equation:
+sqrt+%28b-20%29+%2B4=+sqrt+%28b%29+
+sqrt+%2820%261%2F4-20%29+%2B4=+sqrt+%2820%261%2F4%29+
sqrt%281%2F4%29+%2B+4+=+sqrt%281296%2F64%29
1%2F2+%2B+4+=+sqrt%281296%29%2Fsqrt%2864%29
4%261%2F2+=+36%2F8
4%261%2F2+=+4%261%2F2 Check!