Use completing the square to show that the x-intercepts of y+ax^2+bx+c occur at the value x= (negative b plus or minus the square root of b squared minus 4ac) divided by 2a.
If you can't type it out can you find a site where it shows how to do this step by step? Thanks in advance.
2006-11-27 15:43:36 · 3 answers · asked by culture_killer 3 in Science & Mathematics ➔ Mathematics
Ah, we can fit it in one of these boxes:
ax^2 + bx + c = 0
x^2 + bx/a + c/a = 0
x^2 + bx/a + c/a + b^2/4a^2 = b^2/4a^2
x^2 + bx/a + b^2/4a^2 = b^2/4a^2 - c/a
(x + b/2a)^2 = b^2/4a^2 - c/a
(x + b/2a)^2 = (b^2 - 4ac)/4a^2
x + b/2a = +-sqrt((b^2 - 4ac)/4a^2)
x + b/2a = +-sqrt(b^2 - 4ac)/2a
x = - b/2a +-sqrt(b^2 - 4ac)/2a
x = (-b +-sqrt(b^2 - 4ac))/2a
2006-11-27 16:00:48 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I think that you want y=ax^2+bx+c
Setting y=0 and solving for x gives the x-intercepts.
The point of this exercise is to have you essentially derive the quadratic formula.
ax^2+bx+c=0
x^2+bx/a+c/a = 0
To complete the square add
and subtract [b/(2a)]^2 (this is chosen so that the linear term is the twice the cross product. This technique of adding and subtracting the same thing is used in higher math)
back to it:
x^2+bx/a+c/a +[b/(2a)]^2-[b/(2a)]^2 = 0
Now you've made a perfect square:
(x-b/2a)^2 +c/a-[b/(2a)]^2 = 0
I'll let you do the rest. Move the last two terms to the right, take square root, and solve for x, note that it is in the process of taking the square root that the + or - sneaks in.
2006-11-27 23:55:08 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋
I believe the web site below has your answer
2006-11-27 23:50:16 · answer #3 · answered by sep_n 3 · 0⤊ 0⤋