SOLUTION: What is Quadratic inform method how do you solve this: (x^2+2x)^2-5(x^2+2x)+6=0 and x^1/2-11x^1/4+30 How do you this Solve Equation 6x^3/5-24=0

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Question 3581: What is Quadratic inform method
how do you solve this:
(x^2+2x)^2-5(x^2+2x)+6=0
and
x^1/2-11x^1/4+30

How do you this Solve Equation
6x^3/5-24=0
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
no idea what the "quadratic inform" method is, but you solve these as follows:

Given %28x%5E2%2B2x%29%5E2-5%28x%5E2%2B2x%29%2B6=0, this has a quadratic form, which you can see better if you let y=x%5E2%2B2x, because we can re-write the equation now as y%5E2-5y%2B6+=+0. This is now better to see and we can solve it by factorising:
(y-6)(y+1) = 0
so, y-6 = 0 OR y+1 = 0

so y=6 or y=-1

Hence x%5E2%2B2x+=+6 or x%5E2%2B2x+=+-1 or, rewritten:

x%5E2%2B2x-6+=+0 or x%5E2%2B2x%2B1+=+0

Now, solve these 2 equations to give the values of x. The first one looks to be a "formula" job, but the second can be factorised. You do these.

Second example...Let y = x%5E%281%2F4%29...we then get y%5E2-11y%2B30+=+0. This can be factorised and hence solved quickly. Then go back and find the values of x.

Third... 6x%5E3%2F5-24=0 becomes 6x%5E%283%2F5%29=24 then x%5E%283%2F5%29+=+4

now x%5E%283%2F5%29 means either %28x%5E3%29%5E%281%2F5%29 or %28x%5E%281%2F5%29%29%5E3. I shall use the first of these:

%28x%5E3%29%5E%281%2F5%29+=+4
x%5E3+=+4%5E5
x%5E3+=+1024
so x = cube root of 1024 (roughly 10.079).

jon.