# Thread: Solving for x

1. ## Solving for x

1.) Solve for x :
xlnx−3+lnx−3x=0

Take ln x as a common factor in the first and third terms, making
(x + 1)ln x
(x + 1)ln x - 3(x+1) = 0
(x+1)(ln x - 3) = 0
Therefore x + 1 = 0 or ln x - 3 = 0
Therefore x = -1 or x = e^3

2)Solve for x :
-2√2(√3 sin x + cos x)^(-2) * (√3 cos x - sin x) = 0 for 0<x<2pi

(√3 sin x + cos x)^2
√3 cos x - sin x = 0
which is equivalent to
tan x = √3
x = pi/3 or 4pi/3

3) Solve for x :
x^3 - x^{5/2} - 6x^2 = 0

i dont know how to do this one ><

The text in read is what i have tried, but something is wrong. Can anyone help me find out what i did wrong?

2. On number one, substitute the two answers you got back into the original equation (note that only one works for real numbers). Your answer to number two looks right to me. As for number three, factor out $\displaystyle x^2$:
$\displaystyle x^3-x^{\frac{5}{2}}-6x^2=0$
$\displaystyle (x-\sqrt{x}-6)x^2=0$
So either x=0 or $\displaystyle x-\sqrt{x}-6=0$. Note that as $\displaystyle x=(\sqrt{x})^2$, this is quadratic in $\displaystyle \sqrt{x}$. If we let $\displaystyle y=\sqrt{x}$, this becomes $\displaystyle y^2-y-6=0$, which is easily factorizable; note that $\displaystyle y=\sqrt{x}$ must be positive, so only one of the roots of that quadratic will square to give an x that solves the original equation.

--Kevin C.