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Home / Algebra Trig Review / Algebra / Solving Equations, Part I
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Section 1.10 : Solving Equations, Part I

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

  1. \({x^3} - 3{x^2} = {x^2} + 21x\)
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    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. \(3{x^2} - 16x + 1 = 0\)
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    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. \({x^2} - 8x + 21 = 0\)
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    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.