SOLUTION: 3+√z-10=√z+5 I need help solving radical equations please

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Question 543047: 3+√z-10=√z+5
I need help solving radical equations please

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given to solve:
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3+%2B+sqrt%28z-10%29+=+sqrt%28z%2B5%29
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Square both sides. For the left side this involves multiplying:
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%283+%2B+sqrt%28z-10%29%29%2A%283+%2B+sqrt%28z-10%29%29
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and you can do this by the FOIL process to get:
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3%5E2+%2B+3%2Asqrt%28z-10%29+%2B+3%2Asqrt%28z-10%29+%2B+%28sqrt%28z+-+10%29%29%5E2
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To simplify this, note that 3%5E2+=+9.
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Also note that 3%2Asqrt%28z-10%29+%2B+3%2Asqrt%28z-10%29+=+6%2Asqrt%28z-10%29
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And finally note that %28sqrt%28z+-+10%29%29%5E2+=+z+-+10
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Make these changes and the squared left side becomes:
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9+%2B+6%2Asqrt%28z-10%29+%2B+z+-+10
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Squaring the right side results in: %28sqrt%28z%2B5%29%29%5E2+=+z+%2B+5
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So when both sides of this equation are squared the result is:
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9+%2B+6%2Asqrt%28z-10%29+%2B+z+-+10+=+z+%2B+5
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Note that you have a +z on the left side and a +z on the right side. Subtracting z from both sides eliminates these two terms and you are left with:
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9+%2B+6%2Asqrt%28z-10%29+-+10+=++5
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Combine the 9 and the -10 on the left side and you have:
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6%2Asqrt%28z-10%29+-+1+=+5
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Get rid of the -1 on the left side by adding +1 to both sides to get:
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6%2Asqrt%28z-10%29+=+6
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Divide both sides by 6 and the equation is then:
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sqrt%28z-10%29+=+1
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Now square both sides and you have:
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z+-+10+=+1
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Finally eliminate the -10 on the left side by adding 10 to both sides and the answer you are looking for becomes:
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z+=+11
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You can check this out by returning to the original equation:
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3+%2B+sqrt%28z-10%29+=+sqrt%28z%2B5%29
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and substituting 11 for z to get:
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3+%2B+sqrt%2811-10%29+=+sqrt%2811%2B5%29
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This simplifies to:
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3+%2B+sqrt%281%29+=+sqrt%2816%29
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and taking the square roots results in:
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3+%2B+1+=+4
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This is obviously true, so the answer z = 11 checks.
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Hope this helps you to understand the problem.
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