Where the "quadratic formula" come from?

neela | Student

The quadratic formula gives us the solution to the equation of the form ax^2+bx+c = 0. The solution to the equation is x1 andx2 given by:

x1 = {-b+(b^2-4ac)^(1/2)}/2.

x2 = {-b-(b^2-4ac)^(1/2)}/2.


ax^2+bx+c= 0.

=> ax^2+bx = -c.

We add  (b^2+2a to both sides.

=> ax^2+bx+b^2/4a = b^2/4a-1.

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

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

=> x1+b/2a =  +{(b^2-4ac)^(1/2)}/(2a), or x2+b/2a = -{(b^2-4ac)^(1/2)}/(2a).

=> x1 = {b+(b^2-4ac)^(1/2)}/(2a) or

x2 = {b-(b^2-4ac)^(1/2)}/(2a)}.

Yo can find for the origin of the solution of quadratic equation in any history of mathematics of India, Babylon, China etc.

giorgiana1976 | Student

Let's recall the form of the quadratic equation:

ax^2 + bx + c = 0

We'll factorize by a:

a(x^2 + bx/a + c/a) = 0

We'll group the first 2 terms and we'll complete the square:

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

We'll combine the first 3 terms that represent the perfect square (x +b/2a)^2.

We'll add the amount b^2/4a - c/a both sides:

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

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

We'll divide by a:

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

We'll apply square root both sides, to determine x:

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

We'll subtract b/2a:

x = [-b +/- sqrt(b^2 - 4c)]/2a, the quadratic formula

Remark that the square root exists if the expression under the square root is positive or, at least, zero.

The expression under the square root is called the discriminant of equation and it is identified by the greek letter delta.

If deltais positive, the equation has 2 real distinct roots.

If delta = 0 => the equation has 2 real equal roots.

If deltais negative, the equation has no real roots, but it has imaginary roots.