Algebra-precalculus is a branch of mathematics that integrates algebraic concepts with precalculus topics such as functions, trigonometry, and matrices. It serves as a bridge between basic algebra and advanced calculus.
By definition, ##\sqrt{x^2}= \left| x \right|##.
For positive ##x##, such as ##4##, it is quite straightforward: ##\sqrt{4^2}=\sqrt{16}=4##.
For negative values, I am more confused: ##\sqrt{(-4)^2}=\sqrt{16}=4##. The answer will always be positive, even if you put in a negative value. So why...
##x^2+ax+y^2+by=-c##
##\Leftrightarrow (x+\dfrac{a}{2})^2+(y+\dfrac{b}{2})^2=-c+\dfrac{a^2}{4}+\dfrac{b^2}{4}##
The conditions under which the coefficients of this equation makes a circle:
##-c+\dfrac{a^2+b^2}{4}>0##
##\Leftrightarrow 4c < a^2+b^2##
Center of circle: ##(-\dfrac{a}{2}...
##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...