Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

We've looked at lots of different ways to factorise quadratic equations, where we find the highest common factor between algebraic terms.

Once we can factorise, we can solve equations algebraically to find the unknown value. To do this, we generally look for the values where $x=0$`x`=0. There is a great benefit to factorising quadratics in order to solve them, and in order to understand why, we need to think about zero.

The property of $0$0 is very special. The *only* way two things that are being multiplied can have the answer of $0$0, is if one, or both of those things are $0$0 themselves.

So if we have two factors, like $a$`a` and $b$`b`, and we multiply them together so that they equal $0$0, then one of those factors ($a$`a` or $b$`b`) *must* be $0$0. A written solution to a question like this would be similar to the following:

If $a\times b=0$`a`×`b`=0 then $a=0$`a`=0 or $b=0$`b`=0. This is known as the null factor law.

Remember!

There may be more than one solution to a quadratic equation.

- Move all terms over to one side of the equation so one side equals $0$0.
- Factorise the terms.
- Solve for when $x=0$
`x`=0.

Let's look at this process with some examples.

- If $6x=0$6
`x`=0, the solution would be $x=0$`x`=0 because $6\times0=0$6×0=0. - If $x\left(x+4\right)=0$
`x`(`x`+4)=0, there are two possible solutions because either $x=0$`x`=0 or $x+4=0$`x`+4=0. So the solutions to this equation are $x=0$`x`=0 or $-4$−4.

Remember!

The product of any number and $0$0 is $0$0. That is, anything multiplied by $0$0 equals $0$0.

What are the solutions to the equation $x^2-25=0$`x`2−25=0?

**Think**: We have seen questions like this before, which we can solve by first adding $25$25 to both sides and then taking the square root to give $x=\pm5$`x`=±5. But notice that this equations is actually a difference of two squares, and so we can rewrite it first in factorised form.

**Do**: Let's first factorise the equation, then apply the null factor law to find the solutions.

$x^2-25$x2−25 |
$=$= | $0$0 |

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

Now we know from the null factor law that either $\left(x-5\right)=0$(`x`−5)=0 or $\left(x+5\right)=0$(`x`+5)=0, and so $x=5$`x`=5 or $x=-5$`x`=−5.

Solve the equation $\left(2x-8\right)^2=0$(2`x`−8)2=0.

**Think**: As with question 1 above, we could start by taking the square root of both sides of the equation and then rearrange to find $x$`x`. However, this perfect square quadratic equation is also already in a form that can make use of the null factor law.

**Do**: To make it clear, we'll rewrite the equation using the fact that $\left(\editable{}\right)^2=\editable{}\times\editable{}$()2=×.

$\left(2x-8\right)^2$(2x−8)2 |
$=$= | $0$0 |

$\left(2x-8\right)\left(2x-8\right)$(2x−8)(2x−8) |
$=$= | $0$0 |

What does this last line mean? The null factor law tells us that either $\left(2x-8\right)=0$(2`x`−8)=0 or $\left(2x-8\right)=0$(2`x`−8)=0. But these are the same factor! This is actually an example of a double root, and we find the answer by considering the single factor $2x-8=0$2`x`−8=0 by itself, which gives $x=4$`x`=4.

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 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.

Solve $3y-15y^2=0$3`y`−15`y`2=0

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