algebra-precalculus Definition and 9 Threads

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.
  1. RChristenk

    Confusion about ##\sqrt{x^2}= \left| x \right|##

    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...
  2. RChristenk

    Circle equation coefficient conditions

    ##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}...
  3. RChristenk

    Solve ##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0##

    ##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##.
  4. RChristenk

    Solve ##\left| x+3 \right|= \left| 2x+1\right|##

    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...
  5. RChristenk

    How Do You Solve the Inequality ##x^2<4## for Negative Values of ##x##?

    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##...
  6. RChristenk

    Where Did I Go Wrong in Solving the Inequality?

    ##\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...
  7. RChristenk

    Are Complex Solutions Overlooked When Solving \(16x^4 = 81\) Directly?

    ##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...
  8. RChristenk

    Equation that is defined as an identity

    ##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...
  9. RChristenk

    Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##

    ##\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...
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