Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

There are many techniques for factorising quadratics, these are covered in the factorisation section.

We can factorise by:

- Using the highest common factor
- Using knowledge of the difference of two squares or perfect squares
- Grouping in pairs

There is a great benefit to factorising quadratics in order to solve them. Remember that once we factorise the quadratic, we can make use of the null factor law to find the values of $x$`x` that solve the equation.

Let's have a look at some examples using different factorising techniques. Note that the method of solving once factorised is always the same.

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

**Think**: This quadratic equation is already in factorised form, so we can go straight to using the null factor law to find the solutions.

**Do**: Using the null factor law, either

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

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

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

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

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 factorise 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 |

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 both of these functions are monic quadratics, meaning 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 factorised. 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.

**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 $x^2+6x-55=0$`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.