document.write( "Question 101751This question is from textbook Precalculus with Limits
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\n" ); document.write( "\n" ); document.write( "(2z+3)^(2/3) + (2z+3)^(1/3) = 6\r
\n" ); document.write( "\n" ); document.write( "We've tried cubing each side which gets rather complicated, and we've tried factoring out a (2z+3)^(1/3), but we're not making any progress. Dad is helping, but he's having a tough time remembering complex radical problems!
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Algebra.Com's Answer #73986 by bucky(2189)\"\" \"About 
You can put this solution on YOUR website!
Interesting problem. You can make the problem a little easier to work by making the following
\n" ); document.write( "substitution: let \"A+=+%282z+%2B+3%29%5E%281%2F3%29\"
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\n" ); document.write( "When you make that substitution the problem becomes:
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\n" ); document.write( "\"A%5E2+%2B+A+=+6\"
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\n" ); document.write( "Get this into the standard quadratic form by subtracting 6 from both sides. When you do that
\n" ); document.write( "the equation becomes:
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\n" ); document.write( "\"A%5E2+%2B+A+-+6+=+0\"
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\n" ); document.write( "This equation can be solved by factoring. When factored you get:
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\n" ); document.write( "\"%28A+%2B+3%29%2A%28A+-+2%29=+0\"
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\n" ); document.write( "This can be solved for A by setting each of the factors equal to zero, which is one technique
\n" ); document.write( "for solving a quadratic equation that can be factored. Setting each of the factors equal
\n" ); document.write( "to zero results in:
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\n" ); document.write( "\"A+%2B+3+=+0\" which, after subtracting 3 from both sides, gives \"A+=+-3\"
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\n" ); document.write( "and in:
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\n" ); document.write( "\"A+-+2+=+0\" which, after adding 2 to both sides, results in \"A+=+2\"
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\n" ); document.write( "So there are two answers for A. And at this time we can return to the original definition
\n" ); document.write( "of A and substituting \"%282z+%2B+3%29%5E%281%2F3%29\" into the two answers that we got for A. When
\n" ); document.write( "we do we get:
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\n" ); document.write( "\"%282z+%2B+3%29%5E%281%2F3%29+=+-3\"
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\n" ); document.write( "Cube both sides of this to get:
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\n" ); document.write( "\"2z+%2B+3+=+-27\"
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\n" ); document.write( "Subtract 3 from both sides and you get:
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\n" ); document.write( "\"2z+=+-30\"
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\n" ); document.write( "and solve for z by dividing both sides by 2 to find that:
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\n" ); document.write( "\"z+=+-15\"
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\n" ); document.write( "You can next use that same process on the second answer that you got for A, which was
\n" ); document.write( "A = 2. Substitute \"%282z+%2B+3%29%5E%281%2F3%29\" for A and you have:
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\n" ); document.write( "\"%282z+%2B+3%29%5E%281%2F3%29+=+2\"
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\n" ); document.write( "Cube both sides and you have:
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\n" ); document.write( "\"2z+%2B+3+=+8\"
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\n" ); document.write( "Subtract 3 from both sides to get:
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\n" ); document.write( "\"2z+=+5\"
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\n" ); document.write( "Divide both sides by 2 and you get:
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\n" ); document.write( "\"z+=+5%2F2\"
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\n" ); document.write( "So you have two answers for z ... \"z+=+-15\" and \"z+=+5%2F2\"
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\n" ); document.write( "Hope this helps you to see how you can get answers for z.
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