SOLUTION: Solve square root of h+3 + square root of h-1 =5

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Question 403598: Solve square root of h+3 + square root of h-1 =5
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28h%2B3%29+%2B+sqrt%28h-1%29+=5
Here's a procedure for solving equations where the variable is in a radicand (A radicand is the expression inside a radical.):
  1. Isolate a square root that has a variable in its radicand.
  2. Square both sides of the equation.
  3. If there is still a square root with a variable in its radicand, repeat steps 1 and 2.
  4. At this point there should no longer be a square root that has a variable in its radicand. Use appropriate Algebra to solve the equation you now have.
  5. Check your answer(s). This is not optional. Whenever you square both sides of an equation, which we have done at least once at step 2, extraneous solutions may occur. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation! Extraneous solutions can occur even if no mistakes have been made! So you must check and reject any extraneous solutions if you find any.

Let's see this in action.
1) Isolate a square root.
Subtracting sqrt%28h-1%29 from each side we get:
sqrt%28h%2B3%29++=5-sqrt%28h-1%29
2) Square both sides:
%28sqrt%28h%2B3%29%29%5E2++=%285-sqrt%28h-1%29%29%5E2
Squaring the left side is easy. Squaring the right side is easy to get wrong. Exponents do not distribute. You must use FOIL or the %28a-b%29%5E2+=+a%5E2-2ab%2Bb%5E2 pattern to square this correctly. I like using patterns:
h%2B3+=+%285%29%5E2-2%285%29%28sqrt%28h-1%29%29%2B%28sqrt%28h-1%29%29%5E2
which simplifies as follows:
h%2B3+=+25-10sqrt%28h-1%29%2B+%28h-1%29
h%2B3+=+24-10sqrt%28h-1%29%2B+h
3) We still have a square root with a variable in its radicand. So we repeat steps 1 and 2
1) Isolate a square root.
Subtracting h and 24 from each side we get:
-21+=+-10sqrt%28h-1%29
(The -10 in front of the square root is not a problem. You can divide both sides by -10 if it bothers you but you don't have to. The idea behind this step is to have one side of the equation be something you can square and eliminate the square root. The -10 will not get in the way.)
2) Square both sides.
%28-21%29%5E2+=+%28-10sqrt%28h-1%29%29%5E2
which simplifies as follows:
441 = 100(h-1)
441 = 100h-100
3) The square roots are gone so we can proceed to step 4.
4) Solve the equation
Add 100:
541 = 100h
Divide by 100:
541%2F100+=+h
5) Check you answer(s).
Always use the original equation to check:
sqrt%28h%2B3%29+%2B+sqrt%28h-1%29+=5
Checking h = 541/100:
sqrt%28%28541%2F100%29%2B3%29+%2B+sqrt%28%28541%2F100%29-1%29+=5
which simplifies as follows:
sqrt%28541%2F100%2B300%2F100%29+%2B+sqrt%28541%2F100-100%2F100%29+=5
sqrt%28841%2F100%29+%2B+sqrt%28441%2F100%29+=5
sqrt%28841%29%2Fsqrt%28100%29+%2B+sqrt%28441%29%2Fsqrt%28100%29+=5
29%2F10+%2B+21%2F10+=+5
50%2F10+=+5
5 = 5 Check!
So h+=+541%2F100 or 5.41 is the only solution. (Note: If 541%2F100 did not check then we would have to reject it. And since it was the only possible solution, there would then be no solution.)