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# recriprocal

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

Dec 23, 2018

#1
+2

Hint:  Use Vieta's

Dec 23, 2018
#2
+1

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 Dec 23, 2018
edited by PartialMathematician  Dec 23, 2018
#3
+1

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
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Note that in this equation, $$x_1 + x_2 = 13$$ and $$x_1x_2 = 4$$.

PartialMathematician  Dec 23, 2018
#6
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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
+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   Dec 23, 2018
#7
+1

Thank you, everyone!

Dec 24, 2018