# Week 6——Developing and Applying the Quadratic Formula

use the way that we learned to make ax²+bx+c=0 change  to

$\frac{-b+-\sqrt{b^2-4ac}}{2a}$

Ex:

x^2+9x -2=0

use the formula that we just make

a=1，b=9，c=-2

$\frac{-9+-\sqrt{9^2-4\cdot1\cdot-2}}{2\cdot1}$ $\frac{-9+-\sqrt{81+8}}{2}$ $\frac{-9+-\sqrt{89}}{2}$

finished

so we can use this formula to solve the question easily and correctly

### Aside

Add the square root of the same coefficient

ex: 1-3√5x=-3-2√5x

-3√5x+2√5x=-3-1

-√5x=-4

Than use the way that square both sides of this to get rid of the square root

5x=16

x=16/5

Review

$\frac{2\sqrt{6}}{\sqrt{7}+\sqrt{5}}$

[Than multiply both the top and the bottom by a number , which is goodfor removing the square root of the denominator(($\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})$=3-2=1)]

=$\frac{2\sqrt{6}(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}$

=$\frac{2\sqrt{42}-2\sqrt{30}}{7-5}$

=$\frac{2\sqrt{42}-2\sqrt{30}}{2}$

=$\sqrt{42}-\sqrt{30}$

We also learned how to make it simplify

You have to simplify the number in the  root as much as you can by multiplying it by several equal numbers

EX:square root：√64=√4*4*4=4√4

cube root： ³ √81=³√3*3*3*3=3 ³√3

it makes you can better to add up some numbers with the root

see the picture

# Week 3 Absolute Value and Radicals

√25=5

The absolute value of a real number is defined as the principal square root(√25) of the square of a number.

EX:|9|=3 ,|-9|=3

√9²=√81=9=|9|=|-9|

√（-9）²=√81=9=|9|=|-9|