Finding+the+x-intercepts++Set+3

–To find the x-intercept of a quadratic equation you have to set y to equal zero.

For instance: y = x²+10x+21 0= x² +10x+21 0=(x+7)(x+3) x=-7,-3 therefore those are the x-intercepts.

Now try to solve the following three problems:

y=x²+12x+27

y = 6x²-5x -25

y=-9 x^4+6x^2+35 (-3x2+7)(3x2+5)

Answer 0= x^2+12x+27 0 = (x+9)(x+3) x= -9,-3

0 = 6x²-5x -25 0 = (2x -5)(3x+5) x= 5/2, -5/3

0 =-9 x^4+6x^2+35 0= (-3x2+7)(3x2+5) -3x2=-7 3x2 = 7 x2= 7/3 x = 7/3^1/2

3x2=-5 x2 = -5/3 cant solve negative x= 7/3^1/2

2. Jesse McCarthy Find the x intercepts of the following

y=x²+5x+6

y=2x²+18x+40

y=3x²-7x+2

Answers: From Josh Fisher

A. 0=x²+5x+6 (x+2)(x+3) --> **x= -2, -3**

B. 0=2x²+18x+40 (2x+8)(x+5) --> 2x= -8 **x= -4 or -5**

C. 0=3x²-7x+2 (3x - 1)(x - 2) --> x = 1/3 or 2