Quadratic Relations

Lesson

There are a number of ways to solve quadratics. Remember that when we say solve we are actually finding the $x$`x`-intercepts or roots of the equation.

We have seen:

**Algebraically Solve**- for simple binomial quadratics like $x^2=49$`x`2=49.**Factor**- fully factoring a quadratic means we can then use the null factor law: if $a\times b=0$`a`×`b`=0 then either $a=0$`a`=0 or $b=0$`b`=0.**Completing the square**- this method gets us to a point where we can then solve algebraically. It also tells us the vertex of the quadratic.**Quadratic Formula**- this method will solve any quadratic function of the form $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0, but it is not always the easiest to deal with algebraically, sometimes the other methods are a better choice.

When looking to solve a quadratic, check for easy options:

- Can we remove a common factor immediately?
- Can we solve it straight away algebraically?
- Can we factor it easily?

If these first two options haven't worked then we can either complete the square or use the quadratic formula.

Let's have a look at these questions.

Solve for $x$`x`:

$x^2=17x+60$`x`2=17`x`+60

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

Solve for $x$`x`, expressing your answer in exact form.

$\left(x-5\right)^2-4=8$(`x`−5)2−4=8

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

Solve for the unknown:

$-8x+x^2=-6-x-x^2$−8`x`+`x`2=−6−`x`−`x`2

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

Solve the following equation:

$x-\frac{45}{x}=4$`x`−45`x`=4

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

Solve quadratic equations that have real roots, using a variety of methods