What is algebra-precalculus: Definition and 7 Discussions

##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0## ##\Leftrightarrow x^{\frac{-1}{2}}(1-2x+x^{\frac{7}{6}})=0## Then I'm at a loss as to what to do next because ##x^{\frac{7}{6}}## can't be factored here in a way to get me ##-2x##.

Both sides are in absolute values, i.e. positive, so the solution is straightforward: ##\left| x+3 \right|= \left| 2x+1\right| \Rightarrow x+3=2x+1 \Rightarrow x=2##. But the solution presents another case: ##x+3 = -(2x+1)##. How is this possible if both sides are in absolute values, i.e...

Just by inspection, the answer is obviously ##-2<x<2##. But I tried calculating it step by step and couldn't get the negative portion of the inequality. For ##x>0##, ##x^2<4 \Rightarrow x<2## . Hence ##0<x<2##. For ##x<0##, ##x^2<4 \Rightarrow x>2##. I flipped the inequality because ##x<0##...

##\dfrac{1}{x}<4## For ##x>0##: ##1<4x \Rightarrow x>\dfrac{1}{4}## For ##x<0##: ##1<4x \Rightarrow \dfrac{1}{4}<x \Rightarrow x<-\dfrac{1}{4}## But the problem is ##x<0## works in the original expression instead of just ##x<-\dfrac{1}{4}##, so from calculations alone I missed...

##x^4=\dfrac{81}{16}## ##x=\pm\dfrac{3}{2}##. But I recently realized there are complex solutions as well: ##16x^4-81=0## ##(4x^2)^2-9^2=0## ##(4x^2+9)(4x^2-9)=0## ##x^2=\dfrac{-9}{4}, x^2=\dfrac{9}{4}## ##x=\pm\dfrac{3i}{2}, x=\pm\dfrac{3}{2}## Intuitively when I see ##16x^4=81##, I see...

##5x-6=5x-6## is defined as an identity because it is true for all values of ##x##. My question is I can further simplify and arrive at ##0=0##, in which case no values of ##x## will work because the variable ##x## itself doesn't exist. Isn't this a contradiction? Or did I violate some rule...

##\sqrt{\dfrac{1}{2x^3y^5}}=\dfrac{1}{\sqrt{2\cdot x^2 \cdot x \cdot y^2 \cdot y^2 \cdot y}}=\dfrac{1}{|x|\cdot |y|\cdot |y| \cdot \sqrt{2xy}}=\dfrac{1}{|x|y^2\sqrt{2xy}}## ##\Rightarrow \dfrac{1}{|x|y^2\sqrt{2xy}} \cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{|x|y^2 \cdot...