Evaluate (without a calculator) the product: f(r1+1)⋅f(r2+1)⋅f(r3+1).

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Discussion Overview

The discussion revolves around evaluating the product \( f(r_1+1) \cdot f(r_2+1) \cdot f(r_3+1) \) for the function \( f(x) = x^3 + 20x - 17 \), where \( r_1, r_2, \) and \( r_3 \) are the roots of the function. The focus is on exploring different approaches to this evaluation without the use of a calculator.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants reiterate the function \( f(x) = x^3 + 20x - 17 \) and the task of evaluating the product at the roots incremented by one.
  • One participant expresses gratitude towards another for their approach but indicates that there may be a miscalculation.
  • Another participant acknowledges the correctness of a solution provided by a peer and encourages them not to worry about previous errors.
  • An alternative solution is mentioned, though details are not provided in the posts.
  • Several participants remind others to use spoiler tags in their responses, indicating a concern for maintaining the integrity of the challenge format.

Areas of Agreement / Disagreement

The discussion contains multiple competing views and approaches to the evaluation problem, with no consensus reached on a definitive solution. Participants express differing opinions on the correctness of various methods presented.

Contextual Notes

There are indications of miscalculations and alternative solutions proposed, but specific details on these are not fully elaborated in the posts. The discussion also highlights the importance of formatting in responses.

lfdahl
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Given the function:

$$f(x) = x^3+20x-17.$$

Denote the roots of the function: $r_1,r_2$ and $r_3$.

Evaluate (without a calculator) the product: $f(r_1+1) \cdot f(r_2+1) \cdot f(r_3+1)$.
 
Last edited:
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lfdahl said:
Given the function:

$$f(x) = x^3+20x-17.$$

Denote the roots of the function: $r_1,r_2$ and $r_3$.

Evaluate (without a calculator) the product: $f(r_1+1) \cdot f(r_2+1) \cdot f(r_3+1)$.
my solution:
for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$
Using Vieta's formulas
$r_1+r_2+r_3=0---(1)$
$r_1r_2+r_2r_3+r_3r_1=20---(2)$
$r_1r_2r_3=17---(3)$
$f(r_1+1)=(r_1+1)^3+20(r_1+1)-17=\dfrac {30-42r_1}{r_1-1}---(4)$
$f(r_2+1)=(r_2+1)^3+20(r_2+1)-17=\dfrac {30-42r_2}{r_2-1}---(5)$
$f(r_3+1)=(r_3+1)^3+20(r_3+1)-17=\dfrac {30-42r_3}{r_3-1}---(6)$
$(4)\times(5)\times(6)=$
$\dfrac{27000 - 37800(r_1+r_2+r_3)+ 52920(r_1r_2+r_2r_3+r_3r_1) - 74088(r_1r_2r_3)}{(r_1r_2r_3) -(r_1r_2+r_2r_3+r_3r_1 )+ (r_1 +r_2+r_3 - 1)}$
$=\dfrac{27000+52920*20-74088*17}{17-20-1}=\dfrac{-174096}{-4}=43524$
 
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Albert said:
my solution:
for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$
Using Vieta's formulas
$r_1+r_2+r_3=0---(1)$
$r_1r_2+r_2r_3+r_3r_1=20---(2)$
$r_1r_2r_3=17---(3)$
$f(r_1+1)=(r_1+1)^3+20(r_1+1)-17=\dfrac {30-42r_1}{r_1-1}---(4)$
$f(r_2+1)=(r_2+1)^3+20(r_2+1)-17=\dfrac {30-42r_2}{r_2-1}---(5)$
$f(r_3+1)=(r_3+1)^3+20(r_3+1)-17=\dfrac {30-42r_3}{r_3-1}---(6)$
$(4)\times(5)\times(6)=$
$\dfrac{27000 - 37800(r_1+r_2+r_3)+ 52920(r_1r_2+r_2r_3+r_3r_1) - 74088(r_1r_2r_3)}{(r_1r_2r_3) -(r_1r_2+r_2r_3+r_3r_1 )+ (r_1 +r_2+r_3 - 1)}=572724$

I am unable to derive /understand (4)
 
kaliprasad said:
I am unable to derive /understand (4)
$f(x)=x^3+20x-17$
$f(r)=r^3+20r-17=0---(1)$ ($r$ is a root of $f$)
$f(r+1)=(r+1)^3+20(r+1)-17=3r^2+3r+21$
from $(1):$ $r^3-1=(r-1)(r^2+r+1)=16-20r$
$r^2+r+1=\dfrac {16-20r}{r-1}$
$\therefore 3r^2+3r+21=\dfrac {30-42r}{r-1}$
 
Albert said:
my solution:
for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$
Using Vieta's formulas
$r_1+r_2+r_3=0---(1)$
$r_1r_2+r_2r_3+r_3r_1=20---(2)$
$r_1r_2r_3=17---(3)$
$f(r_1+1)=(r_1+1)^3+20(r_1+1)-17=\dfrac {30-42r_1}{r_1-1}---(4)$
$f(r_2+1)=(r_2+1)^3+20(r_2+1)-17=\dfrac {30-42r_2}{r_2-1}---(5)$
$f(r_3+1)=(r_3+1)^3+20(r_3+1)-17=\dfrac {30-42r_3}{r_3-1}---(6)$
$(4)\times(5)\times(6)=$
$\dfrac{27000 - 37800(r_1+r_2+r_3)+ 52920(r_1r_2+r_2r_3+r_3r_1) - 74088(r_1r_2r_3)}{(r_1r_2r_3) -(r_1r_2+r_2r_3+r_3r_1 )+ (r_1 +r_2+r_3 - 1)}=572724$

Hi, Albert!Thankyou very much for your interesting approach, but ...

- there must be a miscalculation somewhere. :( The correct end value is $43524$.
 
lfdahl said:
Hi, Albert!Thankyou very much for your interesting approach, but ...

- there must be a miscalculation somewhere. :( The correct end value is $43524$.
yes a miscalculation found the answer is 43524
the solution has been edited
what a shame ! my poor calculation
 
Last edited by a moderator:
Thankyou, Albert! - for your correct solution!
Please don´t bother because of the previous small miscalculation.
You are indeed a master in solving challenges and puzzles!:cool:

An alternative solution:

First we note, that $f(r_1+1) = 3r_1^2+3r_1+21$, since $r_1^3+20r_1-17 = 0$.

The quadratic polynomium has the roots:

\[3\left ( r_1^2+r_1+7 \right ) =3 \left ( \frac{-1+i3\sqrt{3}}{2}-r_1 \right )\cdot \left (\frac{-1-i3\sqrt{3}}{2}-r_1 \right ) = 3(\alpha -r_1)(\beta -r_1).\]

Now,
\[f(\alpha )=(\alpha-r_1)( \alpha-r_2)( \alpha-r_3 )\\\\
f(\beta )=(\beta-r_1)( \beta-r_2)( \beta-r_3 ) \\\\ \Rightarrow 3^3\cdot f(\alpha )\cdot f(\beta )=3(\alpha -r_1)(\beta -r_1) \cdot 3(\alpha -r_2)(\beta -r_2) \cdot 3(\alpha -r_3)(\beta -r_3)
\\\\=f(r_1+1)f(r_2+1)f(r_3+1)\]

\[f(\alpha )= -17+i 21\sqrt{3}, \: \: f(\beta )= -17-i 21\sqrt{3}\]

Thus, we get:

$ f(r_1+1)f(r_2+1)f(r_3+1)= 27f(\alpha ) f(\beta )= 27(17^2+3\cdot 21^2) = 43524$.
 
Last edited:
lfdahl said:
Thankyou, Albert! - for your correct solution!
Please don´t bother because of the previous small miscalculation.
You are indeed a master in solving challenges and puzzles!:cool:

An alternative solution:

First we note, that $f(r_1+1) = 3r_1^2+3r_1+21$, since $r_1^3+20r_1-17 = 0$.

The quadratic polynomium has the roots:

\[3\left ( r_1^2+r_1+7 \right ) =3 \left ( \frac{-1+i3\sqrt{3}}{2}-r_1 \right )\cdot \left (\frac{-1-i3\sqrt{3}}{2}-r_1 \right ) = 3(\alpha -r_1)(\beta -r_1).\]

Now,
\[f(\alpha )=(\alpha-r_1)( \alpha-r_2)( \alpha-r_3 )\\\\
f(\beta )=(\beta-r_1)( \beta-r_2)( \beta-r_3 ) \\\\ \Rightarrow 3^3\cdot f(\alpha )\cdot f(\beta )=3(\alpha -r_1)(\beta -r_1) \cdot 3(\alpha -r_2)(\beta -r_2) \cdot 3(\alpha -r_3)(\beta -r_3)
\\\\=f(r_1+1)f(r_2+1)f(r_3+1)\]

\[f(\alpha )= -17+i 21\sqrt{3}, \: \: f(\beta )= -17-i 21\sqrt{3}\]

Thus, we get:

$ f(r_1+1)f(r_2+1)f(r_3+1)= 27f(\alpha ) f(\beta )= 27(17^2+3\cdot 21^2) = 43524$.
to find the value of $f(\alpha)=-17+21\sqrt 3 i$ and $f(\beta)=-17-21\sqrt 3i$ seemed time-consuming
do you have a better way to get them ?
 
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Albert said:
to find the value of $f(\alpha)=-17+21\sqrt 3 i$ and $f(\beta)=-17-21\sqrt 3i$ seemed time-consuming
do you have a better way to get them ?

I´m afraid, no. Maybe, someone in the forum knows a less laborious way to follow?
The conversion to polar notation doesn´t make things better:

\[\frac{-1+i3\sqrt{3}}{2} \approx \sqrt{7}\cdot e^{i100.893^{\circ}}\]
:confused:
 
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  • #10
Hi guys. Just a friendly reminder to use spoiler tags. Thanks. :D
 
  • #11
greg1313 said:
Hi guys. Just a friendly reminder to use spoiler tags. Thanks. :D

Thankyou, greg1313, for the friendly reminder! :o
I will take precaution in future challenge & puzzle correspondence.
 

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