SOLUTION: I can't seem to come up with the correct way to answer the problem C = √3c-8 c=(√3C-8) the numbers in parentheses is all under the s

Algebra ->  Radicals -> SOLUTION: I can't seem to come up with the correct way to answer the problem C = √3c-8 c=(√3C-8) the numbers in parentheses is all under the s      Log On


   

Question 78915This question is from textbook Study Guide and practice workbook
: I can't seem to come up with the correct way to answer the problem C = √3c-8

c=(√3C-8) the numbers in parentheses is all under the square root sign This question is from textbook Study Guide and practice workbook

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
I interpret your problem to involve solving for c in the equation:
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C+=+sqrt%283C+-+8%29
.
square both sides to get:
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C%5E2+=+3C+-+8
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get this into the standard quadratic form by subtracting 3C - 8 from both sides so that the
right side becomes zero. With this subtraction you get:
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C%5E2+-+3C+%2B+8+=+0
.
This does not factor nicely so use the quadratic formula to solve. In the quadratic
formula you know that for an equation of the form:
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ax%5E2+%2B+bx+%2B+c+=+0
.
the values of x that satisfy this equation (or in other words, the answers) are:
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x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
For your problem x = C, and by comparison with the standard form, a = 1, b = -3, and c = 8.
Substitute these values into the answer form and you get:
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C+=+%28-%28-3%29+%2B-+sqrt%28+%28-3%29%5E2-4%2A1%2A8+%29%29%2F%282%2A1%29+
.
which simplifies to:
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C+=+%283+%2B-+sqrt%289+-32%29%29%2F%282%29
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and by working on the radical you get:
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C+=+%283+%2B-+sqrt%28-23%29%29%2F2
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and by substituting i%5E2+=+-1 under the radical the two answers become:
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C+=+%283%2B-i%2Asqrt%2823%29%29%2F2
.
Hope I interpreted the problem correctly and that this clarifies the problem for you.