- #1
WORLD-HEN
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I just found this really old book. In it, I found a way of solving quadratic equations using calculus. I've never seen this method in any other book. Ok, here's the method :
The discriminant of the quadratic formula i.e sqrt(b^2 - 4ac) is equal to the first derivative of the original quadratic expression. I am not sure how this is derived because the author doesn't give a derivation.
Theres also one simple relation between the first derivative of an expression and the factors of it in the form (x-a)(x-b)(x-c)... which is : The sum of the factors is equal to the first derivative. This is easy to prove but i have never seen this in any other calculus book or algebra book. Heres the proof :
Lets use a quadratic expression for simplicity,y, and its factors are (x+a)(x+b)
(d/dx)y = (x+a)((d/dx)*(x+b)) + (x+b)((d/dx)*(x+a)
= x+a + x+b
Using this, you can find the factors from the derivative or viceversa.
I think the first method I described can also be derived in a similar way, but the author didnt do it in the book.
The discriminant of the quadratic formula i.e sqrt(b^2 - 4ac) is equal to the first derivative of the original quadratic expression. I am not sure how this is derived because the author doesn't give a derivation.
Theres also one simple relation between the first derivative of an expression and the factors of it in the form (x-a)(x-b)(x-c)... which is : The sum of the factors is equal to the first derivative. This is easy to prove but i have never seen this in any other calculus book or algebra book. Heres the proof :
Lets use a quadratic expression for simplicity,y, and its factors are (x+a)(x+b)
(d/dx)y = (x+a)((d/dx)*(x+b)) + (x+b)((d/dx)*(x+a)
= x+a + x+b
Using this, you can find the factors from the derivative or viceversa.
I think the first method I described can also be derived in a similar way, but the author didnt do it in the book.