SOLUTION: please i need help with this problem: {{{sqrt(x+4)+sqrt(x-4)=sqrt(2x+6)}}} thank you so much for your time!!

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Question 482538: please i need help with this problem:
sqrt%28x%2B4%29%2Bsqrt%28x-4%29=sqrt%282x%2B6%29
thank you so much for your time!!

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given: sqrt%28x%2B4%29%2Bsqrt%28x-4%29=sqrt%282x%2B6%29
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To solve, square both sides. For the left side this means multiplying:
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%28sqrt%28x%2B4%29%2Bsqrt%28x-4%29%29%2A%28sqrt%28x%2B4%29%2Bsqrt%28x-4%29%29
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Use the FOIL system. In the two sets of parentheses, multiply FIRST terms, then multiply OUTSIDE terms, then multiply INSIDE terms, and finally multiply LAST terms.
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Firsts: %28sqrt%28x%2B4%29%29%2A%28sqrt%28x%2B4%29%29=+x%2B4
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Outsides: %28sqrt%28x%2B4%29%29%2A%28sqrt%28x-4%29%29+=+sqrt%28x%5E2+-+16%29
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Insides: %28sqrt%28x-4%29%29%2A%28sqrt%28x%2B4%29%29+=+sqrt%28x%5E2+-+16%29
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Finally, Lasts: %28sqrt%28x-4%29%29%2A%28sqrt%28x-4%29%29=+x-4
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Add all these results together and you have:
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x%2B4%2Bsqrt%28x%5E2+-+16%29%2Bsqrt%28x%5E2+-+16%29%2Bx-4
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The two terms containing the radicals are identical. Therefore, they can be added and the overall expression becomes:
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x%2B4%2B2%2Asqrt%28x%5E2+-+16%29%2Bx-4
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Note that the +4 and the -4 cancel each other out and also the x + x sum to 2x. Therefore, the left side that we squared simplifies to:
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2x+%2B+2%2Asqrt%28x%5E2-16%29 <--- this is the left side squared
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Now let's return to the original equation and square the right side as follows:
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%28sqrt%282x%2B6%29%29%2A%28sqrt%282x%2B6%29%29=+2x%2B6 <--- this is the right side squared
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And the square of the left side must equal the square of the right side as follows:
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2x+%2B+2%2Asqrt%28x%5E2-16%29=2x%2B6
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Note that a term 2x appears on both sides. Therefore, subtract 2x from both sides and the equation reduces to:
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2%2Asqrt%28x%5E2-16%29+=+6
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Divide both sides by 2 and you get:
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sqrt%28x%5E2-16%29+=+3
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Now get rid of the radical by squaring both sides again and you have
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x%5E2-16+=+9
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Get rid of the -16 on the left side by adding 16 to both sides and you have:
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x%5E2+=+25
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And finally, solve for x by taking the square root of both sides. The result is:
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x+=+5
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And that's the answer. You can return to the original equation that you were given to solve and substitute 5 for x to get:
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sqrt%28x%2B4%29%2Bsqrt%28x-4%29=+sqrt%282x%2B6%29
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Letting x = 5 results in this becoming:
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sqrt%285%2B4%29%2Bsqrt%285-1%29=+sqrt%28%282%2A5%29%2B6%29
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and this simplifies to:
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sqrt%289%29%2Bsqrt%281%29=+sqrt%2816%29
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and finally our check shows that by taking the square roots of the three numbers:
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3%2B1+=+4
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which means that if x = 5 the original equation balances. Both sides are equal.
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Hope this helps you to understand the problem and how to work it.