10. Quadratics

Lesson

We have previously seen two ways to solve a quadratic equation (i.e. an equation of the form $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0):

- By using square roots and algebraic manipulation
- By factoring

Next, we'll discuss a third way to solve quadratic equations. That is, by using the quadratic formula.

If $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0, then:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$`x`=−`b`±√`b`2−4`a``c`2`a`

The advantage of using the quadratic formula is that it always works (unlike factoring) and it always follows the exact same process. The only downside to using it is that it gives no insight into the problem you are trying to solve, hence why teachers will often suggest other methods when solving quadratic equations.

Nevertheless, to see why using the quadratic formula is so efficient, refer to the following demonstration. It shows how any quadratic equation can be solved no matter what the values of $a$`a`, $b$`b` and $c$`c` are. It also allows you to see how the answer is affected when each of the values change. (To change the values, just move the sliders.)

Note that you will need to simplify the expression fully to the solutions.

The quadratic formula might seem quite complex when you first come across it, but it can be broken down into smaller parts.

- The ± allows for the possibility of two solutions.
- The $b^2-4ac$
`b`2−4`a``c`under the square root sign is important as it will tell us how many solutions there are. This is known as the discriminant.

Find the $x$`x`-intercepts (zeros) of the graph $y=2x^2-3x+1$`y`=2`x`2−3`x`+1 using the quadratic formula.

**Think**: The $x$`x`-intercepts are found by substituting $y=0$`y`=0. This results in a quadratic equation $2x^2-3x+1=0$2`x`2−3`x`+1=0. Identify $a$`a`,$b$`b` and $c$`c` for the formula.

$a=2$`a`=2 and $b=-3$`b`=−3 and $c=1$`c`=1

**Do**: Once we have identified these values we can substitute them into the quadratic formula:

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

$=$= | $\frac{-\left(-3\right)\pm\sqrt{\left(-3\right)^2-4\times2\times1}}{2\times2}$−(−3)±√(−3)2−4×2×12×2 | |

$=$= | $\frac{3\pm\sqrt{9-8}}{4}$3±√9−84 | |

$=$= | $\frac{3\pm\sqrt{1}}{4}$3±√14 | |

$=$= | $\frac{3\pm1}{4}$3±14 |

The last line gives $x=\frac{3+1}{4}$`x`=3+14 and $x=\frac{3-1}{4}$`x`=3−14, which gives $x=\frac{4}{4}$`x`=44 and $x=\frac{2}{4}$`x`=24. So, for the quadratic $y=2x^2-3x+1$`y`=2`x`2−3`x`+1 the zeros or solutions are $x=1$`x`=1 and $x=\frac{1}{2}$`x`=12.

Solve the equation $x^2-5x+6=0$`x`2−5`x`+6=0 by using the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$`x`=−`b`±√`b`2−4`a``c`2`a`.

Write each solution on the same line, separated by a comma.

Solve the following equation: $-6-13x+5x^2=0$−6−13`x`+5`x`2=0.

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

Solve the following equation by using the quadratic formula.

$x\left(x-4\right)=5$`x`(`x`−4)=5

Solve quadratic equations in one variable.