+0

+2
235
1
+53

I know the quadratic formula, but I don't get why it works. Can someone post a proof of the quadratic formula.

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

Feb 15, 2019

#1
+2

a*x^2 +b*x + c = 0.

Solve for x by completing the square:

divide the equation by 'a'

x^2 + b*x/a = -c/a

compare with (x+d)^2 = (x+d)*(x+d) = x^2 + 2d*x + d^2

so we need to find 'd' such that

2d = b/a or d = b/2a

and d^2 = b^2/4a^2

so, let's add  'd^2' to both sides of the equation

x^2 + d + d^2 = -c/a +d^2

or

(x+d)^2 = -c/a + d^2

take the square root of both sides and substitute in for 'd'

x+d = (+ or -) sqrt(-c/a + d^2)

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

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

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

Feb 15, 2019

#1
+2

a*x^2 +b*x + c = 0.

Solve for x by completing the square:

divide the equation by 'a'

x^2 + b*x/a = -c/a

compare with (x+d)^2 = (x+d)*(x+d) = x^2 + 2d*x + d^2

so we need to find 'd' such that

2d = b/a or d = b/2a

and d^2 = b^2/4a^2

so, let's add  'd^2' to both sides of the equation

x^2 + d + d^2 = -c/a +d^2

or

(x+d)^2 = -c/a + d^2

take the square root of both sides and substitute in for 'd'

x+d = (+ or -) sqrt(-c/a + d^2)

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

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

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

Guest Feb 15, 2019