reversing coefficients in a quadratic

apple2357 said:
I realized if you used the quadratic formula for both equations

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


This conveniently simplifies to 1.
So i think it will work for complex solutions too.
Click to expand...
Jomo said:
Now that is interesting. Does anyone see that this is obvious? Can you please explain why?
Click to expand...
I wouldn't call it immediately obvious, but it is very nice.

The direct way to get the 1 quickly is to see that the numerators are complements, so the product is (b^2 - (b^2 - 4ac))/(2a*2c) = 1.

Looking at the quadratic formula in light of the question, we see that a and c appear only in the discriminant (where interchanging them has no effect) and in the denominator. So one root of the reversed equation is a/c times the "other" root of the original.

Given that the product of the two roots h and k of a quadratic is c/a, the product of a root of the original, say h, and the "other" root of the reversed equation, (a/c)k, will be h * (a/c)k = (a/c)(hk) = (a/c)(c/a) = 1.

What Halls said in post #2 is still the classic approach (and applies to any degree), but this is, again, very nice.
 
Dr.Peterson said:
I wouldn't call it immediately obvious, but it is very nice.

The direct way to get the 1 quickly is to see that the numerators are complements, so the product is (b^2 - (b^2 - 4ac))/(2a*2c) = 1.

Looking at the quadratic formula in light of the question, we see that a and c appear only in the discriminant (where interchanging them has no effect) and in the denominator. So one root of the reversed equation is a/c times the "other" root of the original.

Given that the product of the two roots h and k of a quadratic is c/a, the product of a root of the original, say h, and the "other" root of the reversed equation, (a/c)k, will be h * (a/c)k = (a/c)(hk) = (a/c)(c/a) = 1.

What Halls said in post #2 is still the classic approach (and applies to any degree), but this is, again, very nice.
Click to expand...
Got it. Thanks.
 
You must log in or register to reply here.
Top