# How do you solve using the completing the square method  x^2-14x+45=0?

May 2, 2016

$x = 9$ OR $x = 5$

#### Explanation:

Completing the square is a method based on the product of the square of a binomial: ${\left(x + m\right)}^{2} = {x}^{2} + 2 x m + {m}^{2}$
This is in the form of a quadratic equation : $a {x}^{2} \pm b x + c$

(x + 3)^2 = x² + 6x + 9
(x - 6)^2 = x² - 12x + 36

Notice that
$a = 1$
the first and last terms are always perfect squares.
There is a specific relationship between $b \mathmr{and} c$

(b÷2)^2 gives the value of c. (half of $b$, then square the answer.)

If you have a trinomial in this form it can be written as ${\left(x \pm . .\right)}^{2}$

Complete the following square: ${x}^{2} + 10 x + \ldots \ldots .$
Do you see that the missing value is 25?

${x}^{2} + 10 x + 25$ can be written as ${\left(x + 5\right)}^{2}$

In x² - 14x + 45 = 0, 45 is not the correct value for c.

Move 45 to the other side: x² - 14x ....... = -45

Add the required value TO BOTH SIDES
x² - 14x + color(red)49 = -45 + color(red)49

now: ${\left(x - 7\right)}^{2} = 4$ ................... [7 is half of 14 or$\sqrt{49}$]

$x - 7$ = $\pm 2$ .........................find the square root of both sides

$x = 2 + 7$ OR $x = - 2 + 7$

$x = 9$ OR $x = 5$