Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

There are a number of ways to solve quadratics. Our ultimate goal when we say 'solve' is referring to finding the $x$`x`-intercepts or roots of the function. That is, the solutions of a quadratic equation relate to the $x$`x`-intercepts of the graph.

First we check for easy options:

- Does the equation have to be rearranged to get to a more workable form?
- Can a common factor be removed immediately?
- Can a variable substitution be made?

Then we employ one of our solving methods:

**Algebraically Solve**- For simple binomial quadratics like $x^2=49$`x`2=49.**Factorise**- Fully factorising 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, for example $\left(x-3\right)^2=16$(`x`−3)2=16. It also tells us the vertex of the quadratic graph.**Quadratic Formula**- This method will solve any quadratic equation 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 solving practical problems:

- Use the practical context to find a relationship between the unknowns and quantities involved. Use these relationships to construct some equations.
- Solve them, often for when they equal zero, or to find the maximum or minimum.
- Use common sense to interpret the results. Some quantities such as time, height or area cannot take negative values.

Solve the following equation: $\frac{10}{x}-3x=-1$10`x`−3`x`=−1.

**Solution**:

We can remove the fraction by multiplying every term by $x$`x`.

$10-3x^2=-x$10−3`x`2=−`x`

Take all the terms to one side so that a $0$0 appears in the equation (we may be able to factorise later).

$3x^2-x-10=0$3`x`2−`x`−10=0

Using the PSF method, we need two numbers that have a product of $-30$−30 and sum of $-1$−1. The numbers that do this are $-6$−6 and $+5$+5.

$3x^2-x-10=0$3`x`2−`x`−10=0

Use the two numbers found to split $-x$−`x` into two terms.

$3x^2-6x+5x-10=0$3`x`2−6`x`+5`x`−10=0

We can now factorise in pairs: $3x^2$3`x`2 and $-6x$−6`x` have common factors. $5x$5`x` and $-10$−10 have common factors.

$3x\left(x-2\right)+5\left(x-2\right)=0$3`x`(`x`−2)+5(`x`−2)=0

A common factor of $x-2$`x`−2 can be taken out.

$\left(3x+5\right)\left(x-2\right)=0$(3`x`+5)(`x`−2)=0

From here we can see that $x=2$`x`=2, $x=\frac{-5}{3}$`x`=−53.

Solve the equation $\frac{3-2x}{x}=4x$3−2`x``x`=4`x`.

$3-2x=4x^2$3−2`x`=4`x`2 (Multiply both sides by $x$`x`)

$4x^2+2x-3=0$4`x`2+2`x`−3=0 (Rearrange to get all terms to one side of the equation)

Now, using the quadratic formula:

$x$x |
$=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |

$x$x |
$=$= | $\frac{-2\pm\sqrt{2^2-4\times4\times\left(-3\right)}}{2\times4}$−2±√22−4×4×(−3)2×4 |

$x$x |
$=$= | $\frac{-2\pm\sqrt{4+48}}{8}$−2±√4+488 |

$x$x |
$=$= | $\frac{-2\pm\sqrt{52}}{8}$−2±√528 |

$x$x |
$=$= | $\frac{-2\pm2\sqrt{13}}{8}$−2±2√138 |

$x$x |
$=$= | $\frac{-1\pm\sqrt{13}}{4}$−1±√134 |

$x$x |
$=$= | $\frac{-1}{4}\pm\frac{\sqrt{13}}{4}$−14±√134 |

So we have two distinct, irrational solutions, $x=\frac{-1+\sqrt{13}}{4}$`x`=−1+√134 and $x=\frac{-1-\sqrt{13}}{4}$`x`=−1−√134.

Sometimes we may be asked to solve an equation using a specific method, in which case that's the one we should use. But when no method is specified, how do we know which to use when we want to solve an equation of the form $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0?

In some cases it is easy to see that the equation can be factorised and we wouldn't even think of using the quadratic formula. In other cases, especially when $a$`a` is not equal to $1$1, the factorisation is not as easy to determine (using the PSF method or otherwise).

If it seems as though too much time is being spent trying to factorise the equation, we can always achieve our goals using the quadratic formula. But if the quadratic formula gives rational solutions, we should be able to factorise it!

**Solve for $p$ p:**

$5\left(p^2-3\right)-705=0$5(`p`2−3)−705=0

**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 for $x$ x by substituting in $m=4^x$m=4x.**

$4^{2x}-65\times4^x+64=0$42`x`−65×4`x`+64=0

Use a comma to separate multiple solutions.