Disguised quadratic equations

Mouseman · #1 $Old$ May 22nd, 2008, 10:19 AM

I'm having troubles solving the following problem.

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

Please post solution.

Thanks in advance

Mouseman

sean.1986 · #2 $Old$ May 22nd, 2008, 10:26 AM

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

(x^3 + 8)(x^3 - 3) = 0

x = cubert(-8) or x = cubert(3)

masters · #3 $Old$ May 22nd, 2008, 10:34 AM

Mathstud28 · #4 $Old$ May 22nd, 2008, 03:09 PM

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