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  • Paul's Online Notes
    Paul's Online Notes
    Home / Algebra Trig Review / Algebra / Solving Equations, Part I
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    General Notice

    I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

    Paul
    August 27, 2020

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    Solving Equations, Part I

    Solve each of the following equations.Show All Solutions Hide All Solutions

    1. \({x^3} - 3{x^2} = {x^2} + 21x\)
      Show Solution

      To solve this equation we’ll just get everything on side of the equation, factor then use the fact that if \(ab = 0\) then either \(a = 0\) or \(b = 0\).

      \[\begin{align*}{x^3} - 3{x^2} & = {x^2} + 21x\\ {x^3} - 4{x^2} - 21x & = 0\\ x\left( {{x^2} - 4x - 21} \right) & = 0\\ x\left( {x - 7} \right)\left( {x + 3} \right) & = 0\end{align*}\]

      So, the solutions are \(x = 0\), \(x = 7\), and \(x = - 3\).

      Remember that you are being asked to solve this not simplify it! Therefore, make sure that you don’t just cancel an \(x\) out of both sides! If you cancel an \(x\) out as this will cause you to miss \(x = 0\) as one of the solutions! This is one of the more common mistakes that people make in solving equations.

    2. \(3{x^2} - 16x + 1 = 0\)
      Show Solution

      In this case the equation won’t factor so we’ll need to resort to the quadratic formula. Recall that if we have a quadratic in standard form,

      \[a{x^2} + bx + c = 0\]

      the solution is,

      \[x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

      So, the solution to this equation is

      \[\begin{align*}x & = \frac{{ - \left( { - 16} \right) \pm \sqrt {{{\left( { - 16} \right)}^2} - 4\left( 3 \right)\left( 1 \right)} }}{{2\left( 3 \right)}}\\ & = \frac{{16 \pm \sqrt {244} }}{6}\\ & = \frac{{16 \pm 2\sqrt {61} }}{6}\\ & = \frac{{8 \pm \sqrt {61} }}{3}\end{align*}\]

      Do not forget about the quadratic formula! Many of the problems that you’ll be asked to work in a Calculus class don’t require it to make the work go a little easier, but you will run across it often enough that you’ll need to make sure that you can use it when you need to. In my class I make sure that the occasional problem requires this to make sure you don’t get too locked into “nice” answers.

    3. \({x^2} - 8x + 21 = 0\)
      Show Solution

      Again, we’ll need to use the quadratic formula for this one.

      \[\begin{align*}x & = \frac{{8 \pm \sqrt {64 - 4\left( 1 \right)\left( {21} \right)} }}{2}\\ & = \frac{{8 \pm \sqrt { - 20} }}{2}\\ & = \frac{{8 \pm 2\sqrt 5 i}}{2}\\ & = 4 \pm \sqrt 5 i\end{align*}\]

      Complex numbers are a reality when solving equations, although we won’t often see them in a Calculus class, if we see them at all.

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