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Originally Posted by Punch The method here is the same as before. If we multiply by 2 throughout: Again square both sides Simplify and solve for x. Remember to check for extraneous solutions
I did it like this, however won't we get a power of 3 when we expand?
Originally Posted by Punch I did it like this, however won't we get a power of 3 when we expand? Yep (x-7)^2(x+2) (x+2)(x^2-14x+49) = x^3-14x^2+49x+2x^2-28x+98 Collect like terms: x^3-12x^2+21x+98 x^3-12x^2+21x-478=0 Solve for x, very tricky: http://www.wolframalpha.com/input/?i=x^3-12x^2%2B21x-478%3D0
For general solution of the cubic, you might Google "Cardano". The history is interesting also.
Originally Posted by e^(i*pi) The method here is the same as before. If we multiply by 2 throughout: Again square both sides Simplify and solve for x. Remember to check for extraneous solutions Just a correction. here
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