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Simple math review problem.


FriedChicken

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I assume you know how to solve quadratic equations of the form

 

a*z² + b*z + c = 0

 

Look at your equation. It can be rewritten as follows:

 

(1/4)*(x²)² - (3/2)*(x²) + 2 = 0

 

Now take x² = z. I hope this hint helps. Please try to give the solution of the equation over here in this thread. Then I am certainly willing to help you with that solution, or check the correctness of your solution.

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Yes I understand how to solve equations using the quadratic form...

Isn't it something like x^2 + x + 2 = (x+1)^2 = (x+1)(x+1) ?

or something like x^2 - 2x + 1 = (x-1)(x+2) ?

 

I can understand that, but throwing in fractions throws me off.

 

...

 

Unless I divide everything by 1/4, which would give me (x^2)^2 - 6(x^2) + 8 = f(x)

But what multiplies to -6 and gives me 8?

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Well, no. What woelen was suggesting was that you find the value of x2, then root that afterwards. Hence the idea of x2=z, so the equation that you will be looking at will be [math]\frac{1}{4}z^2 - \frac{6}{4}z + 2 = 0[/math]

 

And once you've looked at that, and worked out a value for z, then you must work out a value for x.

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Well, no. What woelen was suggesting was that you find the value of x2, then root that afterwards. Hence the idea of x2=z, so the equation that you will be looking at will be [math]\frac{1}{4}z^2 - \frac{6}{4}z + 2 = 0[/math]

 

And once you've looked at that, and worked out a value for z, then you must work out a value for x.

 

Crap... so is that a wrong answer what I posted above?

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Well, no. What woelen was suggesting was that you find the value of x2, then root that afterwards. Hence the idea of x2=z, so the equation that you will be looking at will be [math]\frac{1}{4}z^2 - \frac{6}{4}z + 2 = 0[/math]

 

And once you've looked at that, and worked out a value for z, then you must work out a value for x.

 

wow I didn't know you could do that. Interesting. It's been so long for me. I need to get back into math again.

 

why 6/4 though? just to simplify?

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Your answer is almost correct. You should solve the following equation:

(x²-2)(x²-4) = 0 instead of (x-2)(x-4) = 0

 

Now, show me all steps you take, starting from your original one. So, you first multiply by 4 to get rid of the fractions, then you substitute x² by z, giving you an equation in z.

You solve that equation in z, and then you perform a final step to solve the equation in x. Keep in mind that there are 4 different solutions for x.

 

Now you have all steps mentioned. Now you should use them in the right order to solve your problem. So, please write down all steps and let's see if you got the idea. I do not simply post answers here, I really want you to understand how to solve these things... ;)

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I do not simply post answers here, I really want you to understand how to solve these things... ;)

 

That's awesome, thanks.

 

Ok, this is what I did:

starts off:

f(x) = 1/4(x^2)^2 - (3/2)x^2 + 2

f(x) = (x^2)^2 - 6x^2 + 8

 

Then quadratic formula

 

That gives me 4 and 2

 

Which results in (x^2 - 2) and (x^2 - 4), which results in an answer of sq. rt. of 2, -2 and 2.

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FriedChicken, all of your steps look good and I have the idea that you understand this now. The only missing thing indeed was the omission of -sqrt(2) as the tree pointed out.

 

Just as an exercise for you, try other equations of this type from your textbook or exercise book, but I'm quite confident you can solve this kind of things now.

 

One added comment. You can use the formula for solving quadratic equations also directly for fractional coefficients. Just plug in the values of the coefficients a, b and c as I gave in my first post into the formula for quadratic equations and you'll get the answer. But if that is too cumbersome for you, the multiplication with an integer to get rid of fractional coefficients also is OK.

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