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avatar+809 

Find the sum of the reciprocals of the roots of x^2-13x+4=0.

 Dec 23, 2018
 #1
avatar+3994 
+2

Hint:  Use Vieta's

 Dec 23, 2018
 #2
avatar+679 
0

Using the quadratic formula, we have the two solutions \(\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}\) and \(\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}\). The reciprocals would be \(\dfrac{1}{\dfrac{13}{2} + \dfrac{3}{2}\sqrt{17}}\)  and \(\dfrac{1}{\dfrac{13}{2} - \dfrac{3}{2}\sqrt{17}}\). Adding them together, we have \(\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}\).

 

- PM

 

wink

 Dec 23, 2018
edited by PartialMathematician  Dec 23, 2018
 #3
avatar+679 
0

But yes, a good solution would include the use of Vieta's formula. 

 

In the quadratic polynomial  \(P(x) = ax^2 + bx + c\), the roots \(x_1\) and \(x_2\) satisfy \(x_1 + x_2 = -\dfrac{b}{a}\) and \(x_1 x_2 = \dfrac{c}{a}\).

PartialMathematician  Dec 23, 2018
 #4
avatar+679 
0

Note that in this equation, \(x_1 + x_2 = 13\) and \(x_1x_2 = 4\).

PartialMathematician  Dec 23, 2018
 #6
avatar+679 
+1

So the answer is \(\left(\frac{1}{\left(\frac{13}{2}+\frac{3}{2}\sqrt{17}\right)}\right)+\left(\frac{1}{\left(\frac{13}{2}-\frac{3}{2}\sqrt{17}\right)}\right) = \boxed{\dfrac{13}{4}}\).

 

- PM

PartialMathematician  Dec 23, 2018
 #5
avatar+98107 
+3

The roots are

 

13 + 3√17                        13 - 3√17

_________     and          _________

     2                                       2

 

So

 

The reciprocal  sum is

 

      2                       2

_________  +    _________

13 + 3√17            13 - 3√17

 

 

  2 [ 13 - 3√17]  + 2 [ 13 + 2√13 ]

__________________________

        169  - 9(17)

 

 

  52              13

____  =       ___

  16               4

 

 

 

 

cool cool cool

 Dec 23, 2018
 #7
avatar+809 
+1

Thank you, everyone!

 Dec 24, 2018

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