SOLUTION: Hello, In the proof of the quadractic equation, you begin by dividing both parts by "a". ax2 (2 is the square)+ bx + c = 0 (divide both sides by a) x2(2 is the square

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Question 111186: Hello,
In the proof of the quadractic equation, you begin by dividing both parts by "a".
ax2 (2 is the square)+ bx + c = 0
(divide both sides by a)
x2(2 is the square) + b/a.x + c/a = 0
(then suddenly)
x2(2 is the square) 2.(b/2.a).x = c/a = 0
Then all of a sudden, a "2" appears. Where did that come from?
Thank you for your help.
Barbara
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Deriving the quadratic formula:
ax%5E2%2Bbx%2Bc+=+0 Divide by a.
x%5E2%2B%28b%2Fa%29x%2Bc%2Fa+=+0 Subtract c/a from both sides.
x%5E2%2B%28b%2Fa%29x+=+-c%2Fa Now complete the square in x on the left side by adding the square of half the x-coefficient (%28b%2F2a%29%5E2) to both sides.
x%5E2%2B%28b%2Fa%29x%2B%28b%2F2a%29%5E2+=+%28b%2F2a%29%5E2-%28c%2Fa%29 Factor the left side.
%28x%2B%28b%2F2a%29%29%5E2+=+%28b%2F2a%29%5E2-%28c%2Fa%29 Take the square root of both sides. You'll have two answers here, + and -
x%2Bb%2F2a+=+sqrt%28%28b%2F2a%29%5E2-%28c%2Fa%29%29 Simplify the contents of the radical.
x%2Bb%2F2a+=+sqrt%28b%5E2%2F4a%5E2-c%2Fa%29 Combine the fractions under the radical over a common denominator (4a%5E2)
x%2Bb%2F2a+=+sqrt%28%28b%5E2-4ac%29%2F4a%5E2%29 Take the square root of the denominator.
x%2Bb%2F2a+=+%28sqrt%28b%5E2-4ac%29%29%2F2a Subtract b%2F2a from both sides.
x+=+%28-b%2F2a%29%2B-sqrt%28b%5E2-4ac%29%2F2a Simplify the right side.
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a