SOLUTION: Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx^2 + bx + a = 0. Assume that b^2-4ac is greater than or e

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Question 1207458: Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx^2 + bx + a = 0. Assume that b^2-4ac is greater than or equal to 0.

Do I equate both equations? If not, how is this done?
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Let r be a root of y+=+ax%5E2%2Bbx%2Bc

This will mean ar%5E2%2Bbr%2Bc+=+0 is the case.
The input x = r leads to the output y = 0.
If the root is a real number, then it is where the curve crosses or touches the x axis which we consider it to be an x intercept.

Divide both sides by r^2
Do a bit of algebraic rearranging like so
ar%5E2%2Bbr%2Bc+=+0

%28ar%5E2%2Bbr%2Bc%29%2F%28r%5E2%29+=+0%2F%28r%5E2%29

%28ar%5E2%29%2F%28r%5E2%29%2B%28br%29%2F%28r%5E2%29%2Bc%2F%28r%5E2%29+=+0

a%2Bb%2A%281%2Fr%29%2Bc%2A%281%2F%28r%5E2%29%29+=+0

c%2A%281%2Fr%29%5E2%2Bb%2A%281%2Fr%29%2Ba+=+0
This shows that 1/r is a root of y = cx^2+bx+a

We have proven that r being a root of y = ax^2+bx+c leads to 1/r being a root of y = cx^2+bx+a

You can reverse this flow of logic to go from c%2A%281%2Fr%29%5E2%2Bb%2A%281%2Fr%29%2Ba+=+0 to ar%5E2%2Bbr%2Bc+=+0
This will show that 1/r being a root of y = cx^2+bx+a leads to r being a root of y = ax^2+bx+c.

A rephrasing of this method is found here
https://math.stackexchange.com/questions/2751986/show-that-the-solutions-from-one-quadratic-equation-are-reciprocal-to-the-soluti

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Another approach

Let r and s be roots of ax^2+bx+c = 0.

r+=+%28-b+%2B+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29 and s+=+%28-b+-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29 due to the quadratic formula.

t and u are roots of cx^2+bx+a = 0
t+=+%28-b+%2B+sqrt%28b%5E2+-+4ac%29%29%2F%282c%29 and u+=+%28-b+-+sqrt%28b%5E2+-+4ac%29%29%2F%282c%29 due to the quadratic formula.
The numerators are the same, but the denominators are now 2c instead of 2a.

We must require that a+%3C%3E+0 and c+%3C%3E+0 to avoid division by zero errors.

Claim: 1%2Fr+=+u i.e. the reciprocal of r is u.

Proof:
r+=+%28-b+%2B+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29



1%2Fr+=+%282a%29%2F%28-b+%2B+sqrt%28b%5E2+-+4ac%29%29

1%2Fr+=+%282a%29%2F%28-b+%2B+sqrt%28d%29%29 where d = b^2-4ac is the discriminant

Multiply top and bottom by (-b-sqrt(d)) to rationalize the denominator.

1%2Fr+=+%282a%28-b+-+sqrt%28d%29%29%29%2F%28b%5E2-d%29 Difference of squares rule in the denominator

1%2Fr+=+%282a%28-b+-+sqrt%28d%29%29%29%2F%28b%5E2-%28b%5E2-4ac%29%29

1%2Fr+=+%282a%28-b+-+sqrt%28d%29%29%29%2F%284ac%29

1%2Fr+=+%28-b+-+sqrt%28d%29%29%2F%282c%29

1%2Fr+=+%28-b+-+sqrt%28b%5E2-4ac%29%29%2F%282c%29

1%2Fr+=+u

This proves that the reciprocal of r is u.

Similar steps will show that the reciprocal of s would be t.
1%2Fs+=+t
I'll leave such steps for the student to do.

Example:
The roots of x^2+5x+6 = 0 would be r = -2 and s = -3
The roots of 6x^2+5x+1 = 0 are t = -1/3 and u = -1/2
This example demonstrates that conditions 1/r = u and 1/s = t are both satisfied.
I recommend exploring other examples.