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Algebra Level 3

If r 1 , r 2 , r 3 , r 4 r_{1}, r_{2}, r_{3}, r_{4} are the roots of the equation 4 x 4 − 3 x 3 − x 2 + 2 x − 6 = 0 4x^{4}-3x^{3}-x^{2}+2x-6=0 , what is the value of 1 r 1 + 1 r 2 + 1 r 3 + 1 r 4 \frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}+\frac{1}{r_4} ?

1 3 \frac{1}{3} 1 4 \frac{1}{4} 1 5 \frac{1}{5} 1 2 \frac{1}{2}

Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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2 solutions

Daniel Liu
Jun 18, 2015

If the original polynomial is P ( x ) = 4 x 4 − 3 x 3 − x 2 + 2 x − 6 P(x)=4x^4-3x^3-x^2+2x-6 with roots r 1 , r 2 , r 3 , r 4 r_1,r_2,r_3,r_4 then the polynomial Q ( x ) = x 4 P ( 1 x ) = − 6 x 4 + 2 x 3 − x 2 − 3 x + 4 Q(x)=x^4P\left(\dfrac{1}{x}\right)=-6x^4+2x^3-x^2-3x+4 has roots 1 r 1 , 1 r 2 , 1 r 3 , 1 r 4 \dfrac{1}{r_1}, \dfrac{1}{r_2}, \dfrac{1}{r_3}, \dfrac{1}{r_4} .

By vietas, we have 1 r 1 + 1 r 2 + 1 r 3 + 1 r 4 = − 2 − 6 = 1 3 \dfrac{1}{r_1}+ \dfrac{1}{r_2}+ \dfrac{1}{r_3}+ \dfrac{1}{r_4}=-\dfrac{2}{-6}=\boxed{\dfrac{1}{3}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Nihar Mahajan
Jun 18, 2015

By Vieta's formula we have :

r 1 r 2 r 3 + r 1 r 2 r 4 + r 1 r 3 r 4 + r 2 r 3 r 4 = − 2 r 1 r 2 r 3 r 4 = − 6 ⇒ 1 r 1 + 1 r 2 + 1 r 3 + 1 r 4 = r 1 r 2 r 3 + r 1 r 2 r 4 + r 1 r 3 r 4 + r 2 r 3 r 4 r 1 r 2 r 3 r 4 = − 2 − 6 = 1 3 \large{r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = -2 \\ r_1r_2r_3r_4= -6 \\ \Rightarrow \dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}+\dfrac{1}{r_4}=\dfrac{r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 }{r_1r_2r_3r_4} \\ = \dfrac{-2}{-6} = \boxed{\dfrac{1}{3}}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

C O R R E C T ! \large{CORRECT !}

This was a free 100 points

Vaibhav Prasad - 5 years, 12 months ago

It was quite easy.. well.. I used a similar approach

Rishabh Tripathi - 5 years, 12 months ago

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