## last bit of a quadradic... solving 4x^2 + x -1/4 = 0

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
balkenator
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### last bit of a quadradic... solving 4x^2 + x -1/4 = 0

I'm having trouble figuring out how to simplify the last bit of this quadradic equation: 4x^2 + x -1/4 = 0

I've gotten down to x + 1/4 = + or - the square root of 1/32 (sorry i'm new on this board and don't really know how to write it out)... is the answer already simplfied (it doesn't seem like it) or can I go further? I would not remember how to simplfy the square root of 1/32.

Thanks

stapel_eliz
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Would it be correct to assume that the instructions for this equations specify that you must solve by completing the square, rather than any other method?

If so, then:

. . . . .$4x^2\, +\, x\, =\, \frac{1}{4}$

. . . . .$4(x^2\, +\, \frac{1}{4}x)\, =\, \frac{1}{4}$

. . . . .$x^2\, +\, \frac{1}{4}x\, =\, \frac{1}{16}$

. . . . .$x^2\, +\, \frac{1}{4}x\, +\, \frac{1}{64}\, =\, \frac{4}{64}\, +\, \frac{1}{64}$

. . . . .$\left(x\, +\, \frac{1}{8}\right)^2\, =\, \frac{5}{64}$

. . . . .$x\, +\, \frac{1}{8}\, =\, \pm \frac{\sqrt{5}}{8}$

...and so forth.

balkenator
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### Re: last bit of a quadradic... solving 4x^2 + x -1/4 = 0

OMG I was stuck in a rut- i totally see it now... thank you so much!