Thread: Disguised quadratic equations
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Old May 22nd, 2008, 03:09 PM
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Quote:
Originally Posted by Mouseman View Post
I'm having troubles solving the following problem.

x^6 + 5x^3 -24 = 0

Please post solution.

Thanks in advance

Mouseman
Rewrite this as

(x^3)^2+5x^3-24=0

Now let u=x^3

giving us

u^2+5u-24=0

which implies that

u=\frac{-5\pm\sqrt{25-4(-24)(1)}}{2}=\frac{-5\pm{11}}{2}

So either

u=-8

or

u=3

So backsubbing we get

x^3=-8\Rightarrow{x=-2}

or

x^3=3\Rightarrow{x=\sqrt[3]{3}}

Note this is very useful trick...for example say you were given

solve

\sin^2(x)+2\sin(3x)-8=0

Let u=\sin(x)

do the same thign

or

e^{2x}+10e^x-11=0

rewriting this as

(e^x)^2+10e^x-11=0

let e^x=u

same trick

Remember that when you solve the u formula with the quadratic formula that you must take the inverse to get the actual answer


for example

in the sin(x) one

lets say after you sub you get

u=5

then by backsubbing you get

\sin(x)=5\Rightarrow{x=arcsin(5)}

This is a very useful tool
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