Factor: x^(2/5) - 3x^(1/5) - 4 = 0
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Originally Posted by magentarita Factor: x^(2/5) - 3x^(1/5) - 4 = 0 Use a similar idea to that suggested here: http://www.mathhelpforum.com/math-he...lynomials.html
Originally Posted by magentarita Factor: x^(2/5) - 3x^(1/5) - 4 = 0 Strictly speaking, that is not a polynomial: polynomials don't have fractional powers. But it can be made into one- let and solve for y first, then solve for x.
I know how to solve this equation using the substitution method but is there another way?
Originally Posted by magentarita I know how to solve this equation using the substitution method but is there another way? You have to realise that it's a quadratic. Technically all methods for factorising involve this "substitution" - it's just whether you want to make it look a bit more pleasing to the eyes. You could just as easily write it as which becomes .
Originally Posted by Prove It You have to realise that it's a quadratic. Technically all methods for factorising involve this "substitution" - it's just whether you want to make it look a bit more pleasing to the eyes. You could just as easily write it as which becomes . How about this: (5th root of x)^2 - 3(5th root of x) - 4 = 0
Originally Posted by magentarita How about this: (5th root of x)^2 - 3(5th root of x) - 4 = 0 That's the exact same question, because .
Originally Posted by Prove It That's the exact same question, because . I understand now. Thank you so much.
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