FriedChicken Posted September 20, 2006 Share Posted September 20, 2006 Here's an incredibly simple math problem that I for the love of god cannot figure out.... help.... f(x)=(1/4)x^4 - (3/2)x^2 + 2 I need to set that to zero. I know I did this, and this is a review problem. But for the love of god I have no clue. Link to comment Share on other sites More sharing options...

woelen Posted September 20, 2006 Share Posted September 20, 2006 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. Link to comment Share on other sites More sharing options...

FriedChicken Posted September 20, 2006 Author Share Posted September 20, 2006 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? Link to comment Share on other sites More sharing options...

FriedChicken Posted September 20, 2006 Author Share Posted September 20, 2006 No wait shit I got it!!!!! I plug it into the quadratic formula!!! YES!!!! TY very much answer is (x-2)(x-4) = x=2,4 or something like that Link to comment Share on other sites More sharing options...

the tree Posted September 20, 2006 Share Posted September 20, 2006 Well, no. What woelen was suggesting was that you find the value of x^{2}, then root that afterwards. Hence the idea of x^{2}=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. Link to comment Share on other sites More sharing options...

FriedChicken Posted September 20, 2006 Author Share Posted September 20, 2006 Well, no. What woelen was suggesting was that you find the value of x^{2}, then root that afterwards. Hence the idea of x^{2}=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? Link to comment Share on other sites More sharing options...

GutZ Posted September 20, 2006 Share Posted September 20, 2006 Well, no. What woelen was suggesting was that you find the value of x^{2}, then root that afterwards. Hence the idea of x^{2}=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? Link to comment Share on other sites More sharing options...

woelen Posted September 20, 2006 Share Posted September 20, 2006 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... Link to comment Share on other sites More sharing options...

FriedChicken Posted September 20, 2006 Author Share Posted September 20, 2006 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. Link to comment Share on other sites More sharing options...

the tree Posted September 20, 2006 Share Posted September 20, 2006 Which results in an answer of √2, -2 and 2.Yes, but, you're missing one. Link to comment Share on other sites More sharing options...

FriedChicken Posted September 20, 2006 Author Share Posted September 20, 2006 Yes, but, you're missing one. I just came to the forum to mention that. I know it's not possible to have a negative square root, but it's a - outside the square root. I even typed that but for some reason didn't leave it. Link to comment Share on other sites More sharing options...

the tree Posted September 20, 2006 Share Posted September 20, 2006 Well it is possible, but it isn't the sollution to this equation. Anyways, the point is that the sollutions for this are: -2, -√2, √2, 2. Stating all sollutions clearly is really important. Link to comment Share on other sites More sharing options...

woelen Posted September 20, 2006 Share Posted September 20, 2006 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. Link to comment Share on other sites More sharing options...

the tree Posted September 21, 2006 Share Posted September 21, 2006 Oh, as an added note, I'd majorly recomend that you download a program such as "graph" to help you visualise these functions. (actually, I'd recomend that program to anyone with an interest in maths) Link to comment Share on other sites More sharing options...

FriedChicken Posted September 21, 2006 Author Share Posted September 21, 2006 Ah, well I am a mac user, and apple comes with an insane program called Grapher. It's really really really good. But for some reason here in Germany they don't use graphing calculators. Link to comment Share on other sites More sharing options...

CanadaAotS Posted September 21, 2006 Share Posted September 21, 2006 Well, we arent allowed to use graphing calculators in pre-calculus math (highschool) or calculus and stats in university. I think it has to do more with being able to store information into them for an exam though. Link to comment Share on other sites More sharing options...

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