Quadratic Equations

A quadratic equation in x is an equation that can be written in the general form   ax^{2}+bx +c=0  where a, b and c are real numbers, with a \neq 0.

A quadratic equation in x is also known as a second-degree polynomial equation in x.

There are four ways to solve a quadratic equation.

1.  Factoring  –   If \; \: a*b=0,\; then \: \: a=0\: \: or\: \: b=0\: .

ExamplesAlways rewrite the equation so the LS = 0.

            a)\; 2x^{2}=3-5x

               2x^{2}+5x-3=0

              \left ( 2x-1 \right )\left ( x+3 \right )=0

             2x-1=0\:\: or\:\: x+3=0

                   2x=1\: \:            x=-3

                  x=\frac{1}{2}                                      Therefore   x=\left \{ -3,\frac{1}{2} \right \}

               b)\: \: x^{2}+4x=12

                    x^{2}+4x-12=0

                   \left ( x+6 \right )\left ( x-2 \right )=0

                      x=-6\: \: or\: \: x=2              Therefore    x=\left \{ -6,2 \right \}

                 c)\: \: \frac{3}{4}x^{2}+8x+20=0

                     \frac{1}{4}\left ( 3x^{2}+32x+80 \right )=0    

                     \frac{1}{4}\left ( 3x+20 \right )\left ( x+4 \right )=0  

                          x=-\frac{20}{3}\: \: or\: \: -4               Therefore   x=\left \{ -\frac{20}{3},-4 \right \}      

2.  Square Root Principle –  If\: \: u^{2}=c,\: \: where c> 0,\: \: then\: \: u=\pm \sqrt{c}\: .   

ExamplesIsolate the perfect square on the LS of the equation and the positive constant c on the RS of the equation. 

                 a)\: \: x^{2}=49

                      x=\pm 7

                  b)\: \: 3x^{2}=81

                       x^{2}=27                 

                       x=\pm \sqrt{27}

                       x=\pm 3\sqrt{3}

                   c)\: \: \left ( x-12 \right )^{2}=16     

                         x-12=\pm \sqrt{16}           

                              x=12\pm 4

                              x=12+4\: \: or\: \: 12-4

                              x=16\: \: or\: \: 8

                        d)\: \: \left ( 2x-1 \right )^{2}=18

                                                2x-1=\pm \sqrt{18}

                                                          2x=1\pm 3\sqrt{2}

                                                            x=\frac{1+3\sqrt{2}}{2}\: \: or\: \: x=\frac{1-3\sqrt{2}}{2}

3.  Completing the Square –       If \: \: x^{2}+bx=c,\: \: then,

                                                    x^{2}+bx+\left ( \frac{b}{2} \right )^{2}=c+\left ( \frac{b}{2} \right )^{2}

                                                       \left ( x+\frac{b}{2} \right )^{2}=c+\frac{b^{2}}{4}

Examples:       a)\: \: x^{2}-2x=0

                              x^{2}-2x+1=0+1

                               \left ( x-1 \right )^{2}=1

                                    x-1=\pm 1

                                         x=1\pm 1

                                        x=2\: \: or\: \: 0

                       b)\: \: x^{2}+6x+2=0

                           x^{2}+6x=-2

                           x^{2}+6x+3^{2}=-2+9

                           \left ( x+3 \right )^{2}=7

                            x+3=\pm \sqrt{7}

                                 x=-3\pm \sqrt{7}

                                 x=-3+\sqrt{7}\: \: or\: \: -3-\sqrt{7}

** Once you finish completing the square then you use the square root principle to find the roots of the equation.

4.  Quadratic Formula –  If \: \: ax^{2}+bx+c=0,\: \: then\: \: x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}

Examples:          a)\: \: 2x^{2}+x-1=0                   

                            a=2,\: \: b=1\: \: c=-1      x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} 

                                                              x=\frac{-1\pm \sqrt{1-4\left ( 2 \right )\left ( -1 \right )}}{2\left ( 2 \right )}

                                                            x=\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm \sqrt{9}}{4}=\frac{-1\pm 3}{4}

                                                             x=\frac{-4}{4}\: \: or\: \:\frac{2}{4}=-1\: \: or\: \: \frac{1}{2}

The above example can be solved using the factoring method as well.                     

                  a)\: \: 2x^{2}+x-1=0

                       \left ( 2x-1 \right )\left ( x+1 \right )=0

                       2x=1\: \: or\: \: x=-1

                       x=\frac{1}{2}\: \: or\: \: -1

The question can be solved using the Completing the Square method along with the Sauare Root Principle.

                a)\: \: 2x^{2}+x-1=0

                    2x^{2}+x=1

                   2\left ( x^{2}+\frac{1}{2}x \right )=1

                      \left ( x^{2}+\frac{1}{2} x\right )=\frac{1}{2}

                       \left ( x^{2} +\frac{1}{2}x+\left ( \frac{\frac{1}{2}}{2}\right )^{2}\right )=\frac{1}{2}+\left ( \frac{1}{4} \right )^{2}

                       \left ( x+\frac{1}{4} \right )^{2}=\frac{1}{2}+\frac{1}{16}=\frac{9}{16}

                        x+\frac{1}{4}=\pm \sqrt{\frac{9}{16}}

                          x=-\frac{1}{4}\pm \frac{3}{4}

                         x=\frac{2}{4}\: \: or\: \: -\frac{4}{4}

                         x=\frac{1}{2}\: \: or\: \: -1

When solving Quadratic Equations, unless the directions tell you which method to use, you can use whatever method you like.  If you can factor the trinomial, that is often the easiest method to use.  If it is difficult to tell what the factors are or if there are large coefficients, often using the quadratic formula is the best method to use.  It is always wise to practice all four methods as much as possible.  The more you do the better your skill becomes.  Remember Practice Makes Perfect.  The goal is always to become successful in math!

 

Filed under: Pre-requisites for Pre-Calculus

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