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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

Best Answer 

 #1
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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
avatar
+2
Best Answer

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

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