2. Complex Numbers & Quadratics

Lesson

There is a great benefit to factoring quadratics in order to solve them. Remember, the zeros of a function are the $x$`x`-values of the $x$`x`-intercepts of the function. Therefore, when the quadratic is set equal to zero, the factors of the function can be set equal to zero. By doing this, we can solve each factor for the $x$`x`-intercept. We use the zero product property when solving.

Let's have a look at some examples using different factoring techniques.

Solve $\left(x-2\right)\left(x+5\right)=0$(`x`−2)(`x`+5)=0.

**Think**: This quadratic equation is already in its factored form, so we can go straight to finding the solutions.

**Do**: Set each factor equal to zero:

$\left(x-2\right)$(x−2) |
$=$= | $0$0 | or | $\left(x+5\right)$(x+5) |
$=$= | $0$0 |

$x$x |
$=$= | $2$2 | $x$x |
$=$= | $-5$−5 |

The solutions to the equation $\left(x-2\right)\left(x+5\right)=0$(`x`−2)(`x`+5)=0 are $x=2$`x`=2 or $x=-5$`x`=−5.

Solve $2x^2+12x=0$2`x`2+12`x`=0.

**Think**: Notice that both these terms have a common factor of $2x$2`x`, so we can factor this one using common factors.

**Do **:

$2x^2+12x$2x2+12x |
$=$= | $0$0 |

$2x\left(x+6\right)$2x(x+6) |
$=$= | $0$0 |

So either

$2x$2x |
$=$= | $0$0 | or | $\left(x+6\right)$(x+6) |
$=$= | $0$0 |

$x$x |
$=$= | $0$0 | $x$x |
$=$= | $-6$−6 |

The solutions to the equation $x^2+12x=0$`x`2+12`x`=0 are $x=0$`x`=0 or $x=-6$`x`=−6.

Find the $x$`x`-intercepts for the following quadratics

a) $y=x^2+6x+8$`y`=`x`2+6`x`+8

b) $y=-x^2+3x-2$`y`=−`x`2+3`x`−2

**Think**: Notice that the coefficient of $x^2$`x`2 is either $1$1 or $-1$−1. The $x$`x`-intercepts are the roots of the quadratic, where the value of the quadratic is equal to zero. So, we set the quadratic equal to $0$0 then solve, then solve for $x$`x`.

**PART a)**

We start with $y=x^2+6x+8$`y`=`x`2+6`x`+8 and let $x^2+6x+8=0$`x`2+6`x`+8=0.

**Think**: Check to see if this can be factored. Are there two numbers that multiply to give $8$8 and add to give $6$6?

**Do**: The numbers that multiply to give $8$8 are $2$2 and $4$4, and these also add to give $6$6.

$x^2+6x+8$x2+6x+8 |
$=$= | $0$0 |

$\left(x+2\right)\left(x+4\right)$(x+2)(x+4) |
$=$= | $0$0 |

So either

$\left(x+2\right)$(x+2) |
$=$= | $0$0 | or | $\left(x+4\right)$(x+4) |
$=$= | $0$0 |

$x$x |
$=$= | $-2$−2 | $x$x |
$=$= | $-4$−4 |

These are the solutions to the equation $x^2+6x+8=0$`x`2+6`x`+8=0 and represent the $x$`x`-intercepts of the function $y=x^2+6x+8$`y`=`x`2+6`x`+8, which are $\left(-2,0\right)$(−2,0) and $\left(-4,0\right)$(−4,0).

**PART b)**

Start with $y=-x^2+3x-2$`y`=−`x`2+3`x`−2 and let $-x^2+3x-2=0$−`x`2+3`x`−2=0.

**Think**: Negative signs can sometimes be tricky. We can remove them by factoring out $-1$−1 as a common factor first.

$-x^2+3x-2$−x2+3x−2 |
$=$= | $0$0 |

$-1\left(x^2-3x+2\right)$−1(x2−3x+2) |
$=$= | $0$0 |

Now, we need to find two numbers that multiply to give $2$2, and add to give $-3$−3. These numbers will be $-2$−2 and $-1$−1.

**Do**:

$-1\left(x^2-3x+2\right)$−1(x2−3x+2) |
$=$= | $0$0 |

$-1\left(x-2\right)\left(x-1\right)$−1(x−2)(x−1) |
$=$= | $0$0 |

Here we have $3$3 things being multiplied to give $0$0, so either

$-1=0$−1=0 (which can't happen), or $\left(x-2\right)=0$(`x`−2)=0, or $\left(x-1\right)=0$(`x`−1)=0.

So $x=2$`x`=2 and $x=1$`x`=1 are the $x$`x`-intercepts of the function $y=-x^2+3x-2$`y`=−`x`2+3`x`−2.

Solve $x\left(x+6\right)=0$`x`(`x`+6)=0 for $x$`x`.

Write all solutions on the same line, separated by commas.

Solve $x^2+6x-55=0$`x`2+6`x`−55=0 for $x$`x`.

Write all solutions on the same line, separated by commas.

Solve for the two possible values of $x$`x`:

$\left(17x-19\right)\left(11x-5\right)=0$(17`x`−19)(11`x`−5)=0

Write all solutions in fraction form, on the same line separated by commas.