SOLUTION: Solve each of the following quadratic equations by completing the square. {{{2x^2+x+11=0}}} My answer was:x={{{-1/4+(2sqrt(87)/4)*i}}},x={{{-1/4-(2sqrt(87)/4)*i}}}. Is this cor

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve each of the following quadratic equations by completing the square. {{{2x^2+x+11=0}}} My answer was:x={{{-1/4+(2sqrt(87)/4)*i}}},x={{{-1/4-(2sqrt(87)/4)*i}}}. Is this cor      Log On


   

Question 75017This question is from textbook Beginning Algebra
: Solve each of the following quadratic equations by completing the square.
2x%5E2%2Bx%2B11=0
My answer was:x=-1%2F4%2B%282sqrt%2887%29%2F4%29%2Ai,x=-1%2F4-%282sqrt%2887%29%2F4%29%2Ai. Is this correct? Thanks. This question is from textbook Beginning Algebra

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure where the 2 that multiplies sqrt%2887%29 came from. Let's go through the
problem, and maybe you can catch it.
.
Start with:
.
2x%5E2%2Bx%2B11=0
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You want the coefficient of the x%5E2 term to be 1, so divide all terms on both sides
of the given equation by 2. This results in:
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x%5E2%2B%281%2F2%29%2Ax%2B11%2F2=0
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Take half of the multiplier of the x term. So take half of 1%2F2 and you get 1%2F4.
Square that and you get 1%2F16. Add and subtract 1%2F16 to the equation and you have:
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x%5E2%2B%281%2F2%29%2Ax%2B%281%2F16%29+-+%281%2F16%29+%2B+11%2F2=0
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The first three terms are now a perfect square. Write them as the square:
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%28x+%2B+1%2F4%29%5E2+-+%281%2F16%29+%2B+%2811%2F2%29+=+0
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Combine the -1%2F16 and 11%2F2 by putting the 11%2F2 over the common denominator
of 16. Do this by multiplying 11%2F2 by 8%2F8 to get 88%2F16. Then recognize
that -1%2F16 plus 88%2F16 is 87%2F16. Substitute this into the equation and
it becomes:
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%28x+%2B+1%2F4%29%5E2+%2B+87%2F16+=0
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Subtract 87%2F16 from both sides and you have:
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%28x%2B1%2F4%29%5E2+=+-87%2F16
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Take the square root of both sides to get:
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x+%2B+1%2F4+=+sqrt%28-87%2F16%29
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Subtract 1%2F4 from both sides and you have:
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x+=+-1%2F4+%2B+sqrt%28-87%2F16%29
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But the square root of -87%2F16 simplifies to sqrt%2887%29%2Ai%2Fsqrt%2816%29 which further
simplifies to %28%28sqrt%2887%29%29%2F4%29%2Ai and %28%28-sqrt%2887%29%29%2F4%29%2Ai
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Substituting these two results results in the two answers:
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x+=+-1%2F4+%2B+%28sqrt%2887%29%2F4%29%2Ai and x+=+-1%2F4+-+%28sqrt%2887%29%2F4%29%2Ai
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In this way, you don't get that multiplier of 2 that you have in your answer.
.
I suspect that you might have forgotten to make the coefficient of the x%5E2 term equal
to 1. Early on you want to divide all the terms on both sides of the quadratic equation by
the coefficient of the x%5E2 term if that multiplier is anything other than 1 ...
but I'm not sure whether or not that was the problem..
.
The rest of it was OK and it appears that, with the exception of that one glitch, you
worked everything correctly.
.
Hope this clarifies things for you.