Derive the Quadratic Formula from the general formula form: ax^2 + bx + c, where a,b, and c are constants.

Question

The Quadratic Formula provides a way to determine if a quadratic equation has a solution. One? Two solutions? None? Used the quadratic formula XD.

To derive the Quadratic Formula, you need to apply a process called Completing the Square.

Geometrically, by completing the square on a quadratic equation, the equation becomes closest to a square.

Answer 1

ax^2 + bx + c

complete the square

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

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

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

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

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

Answer 2

As has been pointed out, the Quadratic formula itself is proved by “Completing the Square”.

Completing is often under-appreciated in that, rather than resorting to the Quadratic Formula all the time, as simply a “matter of fact", it is perfectly acceptable to not do so, and complete the square yourself, in-situ.

An example:

3f²(x) - kf(x) = g(x) # Solve for f(x) in terms of g(x)

From (a - b)² = a² - 2ab + b² = a² - (2b)a + b² ==> i.e. the “b²” term = (1/2) the 2nd term’s coefficient (2b), then squared.

Note the convenience here of having the f²(x) coefficient = 1. If otherwise, then just divide the whole equation by the offending coefficient.

3f²(x) - kf(x) = g(x)

f²(x) - (k/3)f(x) = g(x)/3

f²(x) - (k/3)f(x) + ((k/3)/2)² = g(x)/3 + ((k/3)/2)² # Half middle coefficient, then squared

f²(x) - (k/3)f(x) + (k²/6²) = g(x)/3 + (k²/6²)

[f(x) - k/6]² = g(x)/3 + k²/36

f(x) - k/6 = ±√[g(x)/3 + k²/36]

f(x) = k/6 ± √[g(x)/3 + k²/36] # ± to be determined

Now, we use this same technique on the general Quadratric:

ax² + bx + c = 0 # Isolate the ‘a’ and ‘c’, so we can work with ‘b'

x² + (b/a)x + c/a = 0

x² + (b/a)x = -c/a

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

[x + (b/2a)]² = b²/4a² - 4ac/4a² = (b² - 4ac) / 4a²

x + (b/2a) = ±√[(b² - 4ac) / 4a²] = [ ±√[(b² - 4ac) ] / 2a

x = -b/2a ±√(b² - 4ac) / 2a

x = [-b ±√(b² - 4ac)] / 2a

Another practical example, without the Quadratic formula:

x² - 6x - √13 = 0 # Question: Are you really in the mood for bringing in that √13 inside the Discriminant right away, or let it “just happen” later ? …and just get it out of the way for now ?

x² - 6x = √13

x² - 6x + 9 = 9 + √13

(x - 3)² = 9 + √13

x - 3 = ±√( 9 + √13 )

x = 3 ± √(9 + √13) # Seems simpler than the “workhorse” Quadratic formula, but at other times the work is equivalent. Practice will help you be comfortable to choose between 2 tools, rather than just 1.

Answer 3

See the videa.