I'm having troubles solving the following problem.
x^6 + 5x^3 -24 = 0
Please post solution.
Thanks in advance
Mouseman
Rewrite this as
$\displaystyle (x^3)^2+5x^3-24=0$
Now let $\displaystyle u=x^3$
giving us
$\displaystyle u^2+5u-24=0$
which implies that
$\displaystyle u=\frac{-5\pm\sqrt{25-4(-24)(1)}}{2}=\frac{-5\pm{11}}{2}$
So either
$\displaystyle u=-8$
or
$\displaystyle u=3$
So backsubbing we get
$\displaystyle x^3=-8\Rightarrow{x=-2}$
or
$\displaystyle x^3=3\Rightarrow{x=\sqrt[3]{3}}$
Note this is very useful trick...for example say you were given
solve
$\displaystyle \sin^2(x)+2\sin(3x)-8=0$
Let $\displaystyle u=\sin(x)$
do the same thign
or
$\displaystyle e^{2x}+10e^x-11=0$
rewriting this as
$\displaystyle (e^x)^2+10e^x-11=0$
let $\displaystyle e^x=u$
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
$\displaystyle u=5$
then by backsubbing you get
$\displaystyle \sin(x)=5\Rightarrow{x=arcsin(5)}$
This is a very useful tool