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

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) (Show Source): You can put this solution on YOUR website!

Let r be a root of

This will mean 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

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 to
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.

and due to the quadratic formula.

t and u are roots of cx^2+bx+a = 0
and due to the quadratic formula.
The numerators are the same, but the denominators are now 2c instead of 2a.

We must require that and to avoid division by zero errors.

Claim: i.e. the reciprocal of r is u.

Proof:

where d = b^2-4ac is the discriminant

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

Difference of squares rule in the denominator

This proves that the reciprocal of r is u.

Similar steps will show that the reciprocal of s would be 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.

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