Completing the square math question.?

Question

Use completing the square to show that x-intercepts of y=ax^2+bx+c occurs at x= (negative b plus or minus the square root of b squared minus 4ac) divided by 2a.

I've seen these steps on many differens sites and my question is how did the bx/a between steps 4 and 5 disappear?

1. ax^2 + bx + c = 0
2. x^2 + bx/a + c/a = 0
3. x^2 + bx/a + c/a + b^2/4a^2 = b^2/4a^2
4. x^2 + bx/a + b^2/4a^2 = b^2/4a^2 - c/a
5. (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

4. x^2 + bx/a + b^2/4a^2 = b^2/4a^2 - c/a

5. (x + b/2a)^2=
x^2+b2/4a^2

Answer 1

It didn't disappear. Expand (x + b/2a)^2 and you will find it.

Answer 2

That is the "trick" to completing the square. Between steps 4 and 5, you do FOIL in reverse.

To check, you can do FOIL on (x + b/2a)^2 (step 5) and you will see that it equals x^2 + bx/2a + b^2/4a^2 as is in step 4.

If you dont know the term FOIL, its a trick some teachers use to explain the distributive property.

Answer 3

The bx/a disappeared because they factored (x^2 + bx/a + b^2/4a^2).

Think x^2 + 2x + 1 to (x+1)^2.

Answer 4

actually, it din't disappear. if u expand (x+b/2a)^2, you can get x^2 + bx/a + b^2/4a^2. if factorize x^2 + bx/a + b^2/4a^2, you can get (x+b/2a)^2.

Answer 5

actually, it din't disappear. if u expand (x+b/2a)^2, you can get x^2 + bx/a + b^2/4a^2. if factorize x^2 + bx/a + b^2/4a^2, you can get (x+b/2a)^2.

Answer 6

Multiply out 13². You get 169. Multiply out (t + 3)². You get t² + 6t + 9. So if you're factoring t² + 6t + 9 to get (t + 3)², where did the 6 go? Half of it is the 3.

So with x² + (b/a)x + b²/(4a²), what goes in ( )² with x is half of b/a, and that's b/2a.

Answer 7

it didn't, by factorizing (x^2 + bx/a + b^2/4a^2) you get (x + b/2a)^2

if u resemble it again:
x^2 + b^2/4a^2 + 2bx/2a = x^2 + bx/a + b^2/4a^2