Proof of a classic?

Question

Prove the qudratic formula, that is:
ax^2+bx+c
x= [-b (+/-) (b^(2) - 4ac)^(1/2)]/2a

Answer 1

ax² + bx + c = 0

In order to solve this, divide by a on both sides:

ax² + bx + c = 0
= (ax² + bx + c)/a = 0/a
= x² + (bx)/a + c/a = 0

Subtract c/a from both sides:

x² + (bx)/a + c/a = 0
= x² + (bx)/a + c/a - c/a = 0 - c/a
= x² + (bx)/a = -c/a

Now, complete the square on the left side. In order to do this
add b²/4a² to both sides:

x² + (bx)/a = -c/a
= x² + (bx)/a + b²/4a² = -c/a + b²/4a²

Now factor the left side of the equation:

x² + (bx)/a + b²/4a² = -c/a + b²/4a²
= (x + b/2a)² = b²/4a² - c/a

Now square root both sides of the equation:

(x + b/2a)² = b²/4a² - c/a
= √((x + b/2a)²) = + or -√(b²/4a² - c/a)
= x + b/2a = + or -√(b²/4a² - c/a)

Now subtract c/a from b²/4a² by multiplying the denominator and numerator of c/a by 4a.

x + b/2a = + or -√(b²/4a² - c/a)
= x + b/2a = + or - √((b² - 4ac)/4a²)

After that, simplify the square root on the right side of the equation by taking the square root of 4a²:

x + b/2a = + or - √((b² - 4ac)/4a²)
= x + b/2a = + or - √(b² - 4ac)/2a

Now subtract b/2a from both sides to obtain the value of x:

x + b/2a = + or - √(b² - 4ac)/2a
= x + b/2a - b/2a = + or - √(b² - 4ac)/2a - b/2a
= x = (-b + or - √(b² - 4ac))/(2a)

This is the proof of this formula.

Hope this helps!

Answer 2

ax^2 + bx + c = 0

Take the first two terms
ax^2 + bx .... convert this to the form U^2 - T
(SQRT(a)x + b(1/(2SQRT(a)))^2 - b^2/4a

Take the last term of this and combine it with c in the form of a V^2.
- b^2/4a + c = - (SQRT(b^2/4a - c))^2

Put the whole equation back together:
(SQRT(a)x + b(1/(2SQRT(a)))^2 - (SQRT(b^2/4a - c))^2 = 0 [EQN-1]

You now have the form U^2 - V^2 = 0
This factors into (U - V)(U + V) = 0

U - V = 0 and U = V so:
SQRT(a)x + b(1/(2SQRT(a)) = SQRT(b^2/4a - c)
Divide by SQRT(a)
x + b/2a = (1/2a)SQRT(b^2 - 4ac)
x = (-b + SQRT(b^2 - 4ac))/2a

For detail on that last one:
(1/ SQRT(a))SQRT(b^2/4a - c)

SQRT(b^2/4a - c) = SQRT((b^2 - 4ac)/4a)
Take the (1/4a) outside the SQRT) to get (1/(2SQRT(a))
And then:
(1/ SQRT(a))SQRT(b^2/4a - c) = (1/2a)SQRT(b^2 - 4a)

The other solution from EQN-1 gives
SQRT(a)x + b(1/(2SQRT(a)) = -SQRT(b^2/4a - c)
It just changes the sign on the square root term. So this then gives the other solution.

Answer 3

ax^2 + bx + c = 0
ax^2 + bx = -c

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

Complete the square
x^2 + (b/a)x + (b^2)/(4a^2) = (b^2 - 4ac)/(4a^2)
(x + b/(2a))^2 = (b^2 - 4ac)/(2a)

Take the square root on both sides
(x + b/(2a)) = ±sqrt(b^2 - 4ac)/(2a)
x= [-b ± (b^(2) - 4ac)^(1/2)]/2a

Answer 4

OK, lemme see if I can remember this....

ax² + bx + c = 0 for a ≠ 0. Then:
x² + (b/a)x + c/a = 0
x² + (b/a)x + b²/4a² + c/a = b²/4a²
x² + (b/a)x + b²/4a² = b²/4a² - c/a
(x + b/2a)² = b²/4a² - c/a
(x + b/2a)² = b²/4a² - 4ac/4a²
(x + b/2a)² = (b² - 4ac)/4a²
x + b/2a = ± √((b² - 4ac)/4a²)
x + b/2a = ± √(b² - 4ac)/2a
x = -b/2a ± √(b² - 4ac)/2a
x = (-b ± √(b² - 4ac))/2a

...and done

Answer 5

It is the quadratic formula, and I would rather not prove it, sorry.

Answer 6

uhhhhhh..... ok?

Answer 7

wat?