Proving quadratic formula is completing the square?

Question

I've got to the bit where I have:
x+b/2a=the square root of b squared-4ac over 4a squared

I know that x+b/2a=the square root of b squared-4ac over 2a comes next but I don't know why.

Please can someone explain and I'm sorry I can't use maths symbols on the computer. If anyone could tell me how to do that too it would be helpful.

Answer 1

x + (b/2a) = +- sqrt [(b^2 - 4ac)/(4a^2)]

x + (b/2a) = +- sqrt (b^2 - 4ac) / sqrt (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

Answer 2

Just start with ax^2 + bx + c = 0, and use the usual steps for completing-the-square to solve for x.

ax^2 + bx + c = 0
x^2 + (b/a)x + c/a = 0
x^2 + (b/a)x = -c/a
x^2 + (b/a)x + (b/2a)^2 = -c/a + (b/2a)^2
(x + (b/2a))^2 = -c/a + (b/2a)^2
(x + (b/2a))^2 = -4ac/4a^2 + (b^2 / 4a^2)
(x + (b/2a))^2 = (b^2 - 4ac) / 4a^2
x + (b/2a) = ±√(b^2 - 4ac) / 2a
x = (-b±√(b^2 - 4ac)) / 2a

That's where the quadratic formula comes from.

Answer 3

ax^2 + bx + c = 0

a*(x^2 + b/ax + c/a) = 0

complete the square on (x^2 + b/ax + c/a)

(x^2 + b/ax + c/a) = (x^2 + b/ax + (b/2a)^2) + c/a - (b/2a)^2

(x + b/2a)^2 + (c/a - (b/2a)^2) = 0

but (c/a - (b/2a)^2) = c/a - b^2/4a^2 = 1/4a^2*(4ac - b^2)

(x + b/2a)^2 =-(c/a - (b/2a)^2) = -1/4a^2*(4ac - b^2) = 1/4a^2*(b^2 - 4ac)

then take sqrt of both sides

x + b/2a = +/- 1/2a*sqrt(b^2 - 4ac)

x = -b/2a +/- 1/2a*sqrt(b^2 - 4ac) = (-b +/- sqrt(b^2 - 4ac))/2a, the quadratic formula!

Answer 4

Michael Brozinsky: The substitution z=x+b/(2a) transforms the standard quadratic equation ax^2+bx+c=0 right into a quadratic equation in z lacking a linear term. The roots for z are consequently opposites and the quadratic formulation for x follows at as quickly as

Answer 5

square root of 4a^2 in denominator is 2a
(2a*2a=4a^2)

Answer 6

ax^2 + bx + c = 0
x^2 +b/ax +c/a=0
x^2 +b/ax = -c/a
x^2 +b/ax + b^2/4a^2 =b^2/4a^2 -c/a
(x+b/2a)^2 = (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