Math 108 T10-Review Topic 10                                                                             Page 1
                                           MATH108–REVIEWTOPIC10
                                                  Quadratic Equations
                     I.   Finding Roots of a Quadratic Equation
                            A. Factoring
                            B.    Quadratic Formula
                            C. “Taking Roots”
                    II.   Guidelines for Finding Roots of a Quadratic
                   III.   Completing the Square
                            A. Perfect Square Trinomials
                            B.    Solving Quadratics by Completing the Square
                          Answers to Exercises
      Math 108 T10-Review Topic 10                                   Page 2
      Introduction:
      Any equation that can be expressed in the form ax2 + bx + c =0,a6= 0 is called
      a quadratic equation.
      Illustration:          2
                           2x +x−6=0 quadraticin x
                               2
                           −16t +80t=0 quadratic in t.
      The values that satisfy a quadratic (or any polynomial equation) are called roots.
         I. Finding Roots of a Quadratic Equation
            There are 3 primary methods for finding roots to a quadratic. Here are
            examples and comments on each.
        A. Factoring
            Consider the equation 2x2 + x − 6 = 0. When expressed as a polynomial,
            its roots are not easily apparent. Notice what happens if we rewrite this
            expression in factored form:
                        2x2 +x−6=0⇒(2x−3)(x+2)=0
            The roots now become clear: x = 3 or x = −2. When solving a quadratic
                                         2
            equation, factored form has a distinct advantage over polynomial form. Any
            value that turns a factor into 0 will automatically make the overall product
            into 0 (and is therefore a root).
            In “algebreeze” (that strange language used by math instructors), if ab =0,
            then a =0orb =0.
            Example: Solve x(x−4) = 5.
            Warning: You can only make use of factors when their product is 0.
            If the problem read x(x−4) = 0, you would have roots of 0 and 4. No such
            conclusions about roots can be drawn from x(x−4) = 5.
            Our only recourse is to remove parentheses and put into ab = 0 form.
      Math 108 T10-Review Topic 10                                  Page 3
                Solution:
                     x(x−4)=5⇒x2−4x−5=0
                                ⇒(x−5)(x+1)=0       ab =0
                                ⇒x=5orx=−1
            Example: Solve 9x2 = x.
                Solution:
                       9x2 = x ⇒ 9x2 −x =0
                              ⇒x(9x−1)=0          ab =0
                              ⇒x=0orx=1
                                           9
            What if you attempted the same problem using the following method?
                          9x2 = x ⇒ 9x = 1  divide by x
                                ⇒x=1
                                      9
            Everyquadraticequationhas2roots. Dividing byx removesthe rootx =0.
            However, dividing by a constant does not effect roots.
                               2
            Example: Solve −16t +80t =0.
                       2           2
                   −16t +80t=0⇒t −5t=0         divide by (−16)
                                ⇒t(t−5)=0
                                ⇒t=0ort=5
            Here’s one last example of how factoring finds roots.
            Example: x3+3x2−4x−12=0.
            Even though the example is not quadratic, any factorable polynomial can
            be solved using the same principles.
                Solution:      x3 +3x2−4x−12=0
                          ⇒ x2(x+3)−4(x+3)=0 i
                                                      Grouping
                          ⇒       (x2 −4)(x+3)=0      Review Topic 4
                          ⇒ (x−2)(x+2)(x+3)=0
                          ⇒       x = ±2orx=−3
       Math 108 T10-Review Topic 10                                                Page 4
              Exercise 1: Solve for x.
               a)  x(2x+1)=15
               b)  12x2 +60x+75=0
               c)    5x +3+2= 2−6                Hint: Find LCD and clear fractions.
                   x−2 x            x −2x
               d)  x3 −9x =0
               e)  2x3 −5x2−18x+45=0                                              Answers
         B.   Quadratic Formula
              Another method for finding roots to a quadratic equation is the quadratic
              formula.
              For ax2 +bx+c =0,a 6=0,
                                          −b±√b2−4ac
                                     x=         2a        .
              Example: Solve x2 −6x−4=0.
                   Solution: With a =1,b= −6andc=−4,
                                                                    √52=√4·13
                                                                           √
                                                                        =2 13
                            p      2                  √            √
                   x = 6±     (−6) −4(1)(−4) = 6± 52 = 6±2 13
                                 2(1)                 2           2
                                                =3±√13
                   Any help you need with simplifying radicals can be found in
                   Review Topic 6.
              Comments: A quadratic has real roots when b2 −4ac ≥ 0. The discussion
              of non-real or complex roots (when b2 −4ac < 0) will be left for the course.