x intercepts of the equations

[musicrocks429]

1. y=x^2-3x+2

2. y=x^2-1

I used the formula -b/2(a)

for # 1 i came up w/ -1.5
for # 2 i have -.5



how do you find the roots of the equations
1. y=x^2-2x-8

2. y=x^2-4

for 1 i got 0,-8
for 2 i got 0,-4

 

[jwpaine]

Hi. My post will cover all your questions, including your last two problems.

-b/2a is used to find the vertex. Not the intercepts.

To find the X intercepts of a quadratic, you need to factor, or if the equation is not factorable, you will have to complete the square.




Now for your first equation what are the factors of x^2-3x+2? What two numbers when multiplied will equal 2, and when added will equal -3?

Following the rational roots theorem: we know that all the roots of this equation are going to two of: +/- all factors of c / a in ax^2+bx+c

so, all factors are going to be +/- 2/1 = {-2/1, -1/1, 2/1, 1/1}

Which two will work?

I'll give you the first one. The factors are (x-2) and (x-1) because when you FOIL, (x-2)(x-1) you will arrive back at your trinomial x^2-x3+1




If we set both factors of your quadratic to equal 0: (x-2) = 0 and (x-1) = 0 we will find the X intercepts to be 2 and 1