SOLUTION: hi please help me with this question: i think it's under quadratic equation im not sure. Show that the real solution of the equation ax2+bx+c=0 are the reciprocals of the real s

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Question 750840: hi please help me with this question: i think it's under quadratic equation im not sure.
Show that the real solution of the equation ax2+bx+c=0 are the reciprocals of the real solutions of the equation cx2+bx+a=0. Assume that b2-4ac≥0
Thank You
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
 ax²+bx+c=0

The solutions are

%28-b+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ and %28-b+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+ 


 cy²+by+a=0

The solutions are

%28-b+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+ and %28-b+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+

They are reciprocals then their product must be 1:

Let's multiply the solution of the first with a positive radical times
the solution of the second with a negative radical:

%28%28-b+%2B+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29%29+%22%22%2A%22%22%28%28-b+-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Ac%29+%29%29 





The two factors in the numerators are conjugates so FOILing them
causes outers and inners to cancel, so we get:

%28%28-b%29%5E2+-+%28sqrt%28b%5E2-4ac%29%29%5E2%29%2F%284ac%29+ =

%28b%5E2-%28b%5E2-4ac%29%29%2F%284ac%29 =

%28b%5E2-b%5E2%2B4ac%29%2F%284ac%29 =

%284ac%29%2F%284ac%29 =

1

so they are reciprocals.  Multiplying the other pair
gives the same results.

Edwin