SOLUTION: Suppose a and x satisfy x^2 + (a-(1/a))x - 1 = 0. Solve for x in terms of a. *That's a quadratic in x

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Suppose a and x satisfy x^2 + (a-(1/a))x - 1 = 0. Solve for x in terms of a. *That's a quadratic in x      Log On


   



Question 1086139: Suppose a and x satisfy x^2 + (a-(1/a))x - 1 = 0. Solve for x in terms of a.
*That's a quadratic in x

Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use a substitution for ease,
C=a-1%2Fa
Complete the square,
x%5E2%2BCx-1=0
%28x%5E2%2BCx%2B%28C%2F2%29%5E2%29=1%2B%28C%2F2%29%5E2
%28x%2BC%2F2%29%5E2=1%2BC%5E2%2F4
x%2BC%2F2=0+%2B-+sqrt%28%284%2BC%5E2%29%2F4%29
x=-C%2F2+%2B-+sqrt%284%2BC%5E2%29%2F2
So then,
C%2F2=a%2F2-1%2F%282a%29
C%2F2=a%5E2%2F%282a%29-1%2F%282a%29
C%2F2=%28a%5E2-1%29%2F%282a%29
and
C=%28a%5E2-1%29%2Fa
C%5E2=%28a%5E2-1%29%5E2%2Fa%5E2
4%2BC%5E2=4a%5E2%2Fa%5E2%2B%28a%5E2-1%29%5E2%2Fa%5E2
4%2BC%5E2=%284a%5E2%2B%28a%5E2-1%29%5E2%29%2Fa%5E2
4%2BC%5E2=%284a%5E2%2Ba%5E4-2a%5E2%2B1%29%2Fa%5E2
4%2BC%5E2=%28a%5E4%2B2a%5E2%2B1%29%2Fa%5E2
4%2BC%5E2=%28a%5E2%2B1%29%5E2%2Fa%5E2
So then substituting,
2x=-C+%2B-+sqrt%284%2BC%5E2%29
2x=%281-a%5E2%29%2Fa+%2B-+sqrt%28%28a%5E2%2B1%29%5E2%2Fa%5E2%29
2x=%281-a%5E2%29%2Fa+%2B-+%28a%5E2%2B1%29%2Fa
2xa=1-a%5E2+%2B-+%28a%5E2%2B1%29
"Positive" solution:
2xa=1-a%5E2+%2B+a%5E2%2B1
2xa=2
highlight%28x=1%2Fa%29
"Negative" solution:
2xa=1-a%5E2+-+a%5E2-1%29
2xa=-2a%5E2
highlight%28x=-a%29

Answer by ikleyn(52782) About Me  (Show Source):
You can put this solution on YOUR website!
.
I walked in the park and thought on this problem.

I got two solutions. Each is as short as 2 - 4 lines.



Solution 1   (the Vieta's theorem;   4-lines solution)

The Vieta's theorem says:  if  p and q are the roots of a quadratic equation  x%5E2+%2B+ux+%2B+v = 0  then  u = -(p+q)  and  v = pq.

The opposite is also TRUE:  if  u = -(p+q)  and  v = pq  then p and q  are the roots of the quadratic equation    x%5E2+%2B+ux+%2B+v = 0.


Now look into your equation and notice that  the numbers  -a  and  %281%2Fa%29  give  -%28a+-+%281%2Fa%29%29 when summed up and -1 when multiplied.


Hence,  the numbers "-a" and  1%2Fa are the roots of your equation.


Solution 2   (Factoring.   2-lines solution)

Factor your polynomial:  x%5E2+%2B+%28a-%281%2Fa%29%29x+-+1+ = %28x%2Ba%29%2A%28x-1%2Fa%29.


Hence, the roots are "-a" and 1%2Fa.