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Solve for x without a calculator.

2x^(square root of -78)*33x+-(587/369)x*(square root of -982)+36698/4895-5884*102504=(square root of x)*698/56x*589x*42x+69/(square root of -90) :wacko:

Remember: No calculator.

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Solve for x without a calculator.

2x^(square root of -78)*33x+-(587/369)x*(square root of -982)+36698/4895-5884*102504=(square root of x)*698/56x*589x*42x+69/(square root of -90) :wacko:

Remember: No calculator.

I'm still trying to decide whether you're looking for me

it's not possible because you're doing square roots of negative numbers

or if I should just go ahead and ask whether that's a 698/(56x) or (698/56)x and use complex numbers if needed.

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I'm still trying to decide whether you're looking for me
it's not possible because you're doing square roots of negative numbers
or if I should just go ahead and ask whether that's a 698/(56x) or (698/56)x and use complex numbers if needed.

Complex numbers are needed.

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Not possible for two reasons I see.

1. Your doing squareroots of negatives which are imaginary numbers.

2. You can't have radicals in the denominator of a fraction.

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Not possible for two reasons I see.

1. Your doing squareroots of negatives which are imaginary numbers.

2. You can't have radicals in the denominator of a fraction.

Just to clarify, the slashes are divide.

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Solve for x without a calculator.

2x^(square root of -78)*33x+-(587/369)x*(square root of -982)+36698/4895-5884*102504=(square root of x)*698/56x*589x*42x+69/(square root of -90) :wacko:

Remember: No calculator.

Just as a matter of clarification.....please fully parenthesize the equation (at least with regard to division)...

2x^(square root of -78)*33x+-(587/369)x*(square root of -982)+36698/(4895-5884*102504)=(square root of x)*698/(56x*589x*42x+69/(square root of -90)) ?

or

2x^(square root of -78)*33x+-(587/369)x*(square root of -982)+(36698/4895)-5884*102504=(square root of x)*(698/56x)*589x*42x+69/(square root of -90) ?

And as a response to DemonWolf44,

"Not possible for two reasons I see."

1. Your doing squareroots of negatives which are imaginary numbers.

yes...the answer may or may not be a complex number.....that doesn't mean it isn't possible.

2. You can't have radicals in the denominator of a fraction.

Sure you can, and you can move them if needed. (eg, 1/sqrt(2) = sqrt(2)/(sqrt(2)*sqrt(2)) = sqrt(2)/2 )

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To tell the truth, i just threw down a bunch af random numbers to see if it can be solved. I belive it can be solved if you look at it carefully. Also, this equation does not consist of fractions. / is also another symbol for divide. I will be more careful to make future puzzles/problems clearer and easier to understand.

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To tell the truth, i just threw down a bunch af random numbers to see if it can be solved. I belive it can be solved if you look at it carefully. Also, this equation does not consist of fractions. / is also another symbol for divide. I will be more careful to make future puzzles/problems clearer and easier to understand.

Haha. Did I read that correctly?? The problem that YOU put down MIGHT have an answer but YOU'RE NOT SURE?! Hahahaha. Wow...I'd post a picture of a short bus but I have to go play some tennis.

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2. You can't have radicals in the denominator of a fraction.

Sure you can, and you can move them if needed. (eg, 1/sqrt(2) = sqrt(2)/(sqrt(2)*sqrt(2)) = sqrt(2)/2 )

Or another example... What about 1/sqrt(4)? :huh:

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Or another example... What about 1/sqrt(4)? :huh:

Isn't the radical in the denominator thing just one of those formalities?? I can't remember terminology, but if you do encounter a 1/sqrt(4), then just multiply it by sqrt(4)/sqrt(4) (conjugate might be term??). That way it looks prettier to those who don't like radicals in denominator. 1/sqrt(4) = sqrt(4)/2.

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Isn't the radical in the denominator thing just one of those formalities?? I can't remember terminology, but if you do encounter a 1/sqrt(4), then just multiply it by sqrt(4)/sqrt(4) (conjugate might be term??). That way it looks prettier to those who don't like radicals in denominator. 1/sqrt(4) = sqrt(4)/2.

Except it would be

1/sqrt(4) = sqrt(4)/4

Since

sqrt(4) = 2

thus,

1/2 = 2/2 (in your showing)

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To tell the truth, i just threw down a bunch af random numbers to see if it can be solved. I belive it can be solved if you look at it carefully. Also, this equation does not consist of fractions. / is also another symbol for divide. I will be more careful to make future puzzles/problems clearer and easier to understand.

When did fractions become different than dividing?

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Except it would be

1/sqrt(4) = sqrt(4)/4

Since

sqrt(4) = 2

thus,

1/2 = 2/2 (in your showing)

Yeah so I screwed that one up when I typed it because my multitasking ability SUCKS.

1/sqrt 4 = 1/2

multiply 1/sqrt (4) times sqrt (4)/sqrt (4) = sqrt (4)/4.

Therefore, 1/2 = 1/2

Obviously the 2 in the denominator was supposed to be a 4. I guess everyone else realized that, though, but glad you probably feel better for doing such an "advanced proof." :rolleyes:

Also, I'm taking up collections to hire "giterdone" a tutor. My 3rd grade cousin made an A in math last semester, so I was thinking of hiring him.

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