Section 1.10 : Solving Equations, Part I

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

1. ${x^3} - 3{x^2} = {x^2} + 21x$
   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.