Numbers and Reciprocals

Algebra Level 4

Three positive real numbers $p,q,r$ satisfy

$\begin{aligned} \frac { 1 }{ p } +q &=& 2\\ \frac { 1 }{ q } +r &=& 3\\ \frac { 1 }{ r } +p &=& 5\\ \end{aligned}$

Find the sum of all possible values of $p$ .

The answer is 5.2.

2 solutions

Brian Charlesworth
Jan 30, 2015

From the second equation we have that $q = \dfrac{1}{3 - r}.$

From the third equation we have that $r = \dfrac{1}{5 - p}.$

Thus the first equation becomes

$\dfrac{1}{p} + \dfrac{1}{3 - \frac{1}{5 - p}} = 2$

$\Longrightarrow \dfrac{1}{p} + \dfrac{5 - p}{14 - 3p} = 2$

$\Longrightarrow (14 - 3p) + p(5 - p) = 2p(14 - 3p)$

$\Longrightarrow 5p^{2} - 26p + 14 = 0 \Longrightarrow p = \dfrac{26 \pm \sqrt{396}}{10} = \dfrac{13 \pm 3\sqrt{11}}{5}.$

Thus there are two possible positive values for $p$ , and they sum to

$2*\dfrac{13}{5} = \dfrac{26}{5} = \boxed{5.2}$ .

After you get quadratic equation, you can just do: sum of roots = -b/a = 26/5 = 5.2 (no need to find roots).

Bhaskar Sukulbrahman - 6 years, 4 months ago

You still need to check that the roots are real valued, and not complex valued. That requires calculating the discriminant, which is the extra step that was taken.

Calvin Lin Staff - 6 years, 4 months ago

And also that both the roots are positive!

Prasun Biswas - 6 years, 4 months ago

The question asks for the sum of the roots. If both roots are not real, then the sum cannot be defined (even to be zero).

Janardhanan Sivaramakrishnan - 6 years, 4 months ago

Niranjan Khanderia
Feb 15, 2015

$From~first~and ~third~equations..\\\dfrac{1}{q}=\dfrac{p}{2p-1}..........(1)~~~~~~~~~~~~~~~r=\dfrac{1}{5-p}........(2) \\Substituting~(1)~and~(2)~in~the~second~equation:-\\\dfrac{p}{2p-1}+\dfrac{1}{5-p}=3\\ \implies~\dfrac{5p-p^2+2p-1}{-2p^2+11p-5}=3~~\therefore~-p^2+7p-1=-6p^2+33p-15\\That ~is~5p^2-26p+14=0....\\\therefore ~ Sum~ of~ values ~of ~p =\dfrac{26}{5}\\=\boxed{ \Large \color{#D61F06}{5.2}}$

Numbers and Reciprocals

The answer is 5.2.

2 solutions

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