Ahaan's polynomial roots

If the function $f(x)=ax^3+bx^2+cx+d$ has roots $w$ , $x$ , and $y$ , then the function with roots $\frac{1}{w}$ , $\frac{1}{x}$ , and $\frac{1}{y}$ can be represented by $dx^3+cx^2+bx+a$ .

Since we want $g(x)$ to have a leading coefficient of $1$ , we divide by $d=42$ to get

$g(x) = x^3 + \frac{13}{42} x^2 + \frac{20}{42} x + \frac{1}{42}$

Hence, $g(1) = \frac{76}{42} = \frac{38}{21}$ , and thus $a+b = 59$ .

Given $f(x) = x^{3} + 20x^{2} + 13x + 42$
Let u, v, w be the roots of the polynomial, then according to Vieta's Formula,
$u+v+w = -20$
$uv+uw+vw = 13$
$uvw = -42$
Now, the required polynomial is $g(x) = x^{3} + rx^{2} +sx + t$ with roots $\frac{1}{u} \frac{1}{v} \frac{1}{w}$
Applying Vieta's Formula on this polynomial we have:
$\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = -r$
$\frac{vw+uw+uv}{uvw} = -r$
Using values of expressions;
$\frac{13}{-42} = \frac{-13}{42}$ = -r
$r= \frac{13}{42}$
Also for t,
$\frac{1}{v}.\frac{1}{u}.\frac{1}{w} = -t$ $\frac{1}{uvw} = -t$ Substituting value for uvw we have;
$\frac{1}{-42} = -t$
$t= \frac{1}{42}$
Also for s,
$\frac{1}{uv} + \frac{1}{uw} + \frac{1}{vw} = s$
$\frac{w+v+u}{uvw} = s$
Substituting values for u+v+w and uvw we have;
$s = \frac{-20}{-42}$
$s = \frac{20}{42}$
Now,
By substituting the values for r, s and t in polynomial g(x) we have;
$g(x) = x^{3} +\frac{13}{42}x^{2} + \frac{20}{42}x + \frac{1}{42}$
Therefore,
$g(1) = 1 + \frac{13}{42} + \frac{20}{42} +\frac{1}{42}$
$g(1) = \frac{42+13+20+1}{42} = \frac{76}{42} = \frac{38}{21}$
$=> a= 38, b=21$
$a+b = 38+21$
$\boxed{a+b = 59 Answer}$

By Vieta's Theorem in $f(x)$ yields:

• $u + v + w = -20$
• $uv + vw + wu = 13$
• $uvw = -42$

Using Vieta's Theorem in $g(x)$ also yields;

• $\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = -r$
• $\frac{1}{uv} + \frac{1}{vw} + \frac{1}{wu} = s$
• $\frac{1}{uvw} = -t$

We must solve for r, s, and t to determine the value of $g(1)$

First solving for r,

$\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = \frac{uv + vw + wu}{uvw} = \frac{13}{-42} = \frac{13}{-42} = -r$

Therefore, $r = \frac{13}{42}$

Solving for s,

$\frac{1}{uv} + \frac{1}{vw} + \frac{1}{wu} = \frac{u + v + w}{uvw} = \frac{-20}{-42} = \frac{20}{42} = s$

Lastly, solving for t, $\frac{1}{uvw} = \frac{1}{-42} = -t$

Therefore, $t = \frac{1}{42}$

We are now able to solve for $g(1)$ ,

$g(1) = 1^{3} + r ( 1^{2} ) + s ( 1 ) + t = 1 + r + s + t = 1 + \frac{13}{42} + \frac{20}{42} + \frac{1}{42} = \frac{76}{42} = \frac{38}{21}$

Therefore

$a = 38 , b = 21$

and

$a + b =$ 59

Start from the factorisation of the two polynomials:

$f(x)=(x-u)(x-v)(x-w)$

$g(x)=(x-\frac 1 u)(x-\frac 1 v)(x-\frac 1 w)$

From there, just multiply $g(1)$ by $f(0)=uvw$

$g(1)=(1-\frac 1 u)(1-\frac 1 v)(1-\frac 1 w)$

$uvw g(1)=-(1-u)(1-v)(1-w)$

$-f(0)g(1)=-f(1)$

$g(1) = \frac{76}{42}=\frac{38}{21}$

The solution is therefore $38+21=59$

Because $\frac {1} {u}$ is a root of $g(x)$ , we get:

$g(\frac {1} {u} ) = \frac {1} {u^3} + \frac {1} {u^2} + \frac {s} {u} + t = 0$

Multiplying by $42 u^3$ we get:

$42tu^3 + 42su^2 + 42ru + 42 = 0$

Because $f(u) = u^3 + 20u^2 + 13u + 42 = 0$ , the unknown coefficients can be determined by:

$42t=1\\ 42s=20\\ 42r=13$

$t=\frac {1} {42} \\ s = \frac {20} {42} \\ r = \frac {13} {42}$

$g(1) = 1 + r + s + t = 1 + \frac {13} {42} + \frac {20} {42} + \frac {1} {42} = \frac {38} {21}$

$a + b = 38 + 21 = \boxed {59}$

By Vieta's formulas, we can conclude that $u+v+w=-20$ , $uv+uw+vw=13$ and $uvw=-42$ .

By Vieta's again for $g(x)$ :

$\frac { 1 }{ u } +\frac { 1 }{ v } +\frac { 1 }{ w } =\frac { uv+uw+vw }{ uvw } \Longrightarrow \frac { 13 }{ -42 }$ $\therefore r=\frac { 13 }{ 42 }$

$\frac { 1 }{ uv } +\frac { 1 }{ uw } +\frac { 1 }{ vw } =\frac { u+v+w }{ uvw } \Longrightarrow \frac { 20 }{ 42 }$ $\therefore s=\frac { 20 }{ 42 }$

$\frac { 1 }{ uvw } =\frac { 1 }{ -42 }$ $\therefore t=\frac { 1 }{ 42 }$

So, $g(x)= { x }^{ 3 }+\frac { 13 }{ 42 } { x }^{ 2 }+\frac { 20 }{ 42 } x+\frac { 1 }{ 42 }$

When $x=1, g(x)=\frac{38}{21}=\frac{a}{b}$ , so $a+b=\boxed{59}$

put 1/x at f(x)

we get 1/x^3 + 20/x^2 + 13/x +1

• x^3

42x^3 + 13x^2 + 20x +1

divided /42

g(x) = x^3 + 13/42 x^2 + 20/42 x + 1/42

g(1) = 38/21

a+b =59

$f({x})={x^3+20x^2+13x+42}$ So ${a=1}, {b=20}, {c=13}, {d=42}$ ${u+v+w}=\frac{-b}{a}={-20}$ ${uvw}=\frac{-d}{a}={-42}$ ${uv+uw+vw}=\frac{c}{a}={13}$ ${\frac{wv}{uvw}+\frac{uw}{uvw}+\frac{uv}{uvw}=\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=\frac{-13}{42}}$ ${\frac{1}{uvw}=\frac{-1}{42}}$ ${\frac{w}{uvw}+\frac{v}{uvw}+\frac{u}{uvw}=\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=\frac{-20}{-42}=\frac{10}{21}}$ $g({x})={x^3+rx^2+sx+t}$ So ${a=1}, {b=r}, {c=s}, {d=t}$ ${\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=\frac{-b}{a}={-r}=\frac{-13}{42}}$ from which ${r}=\frac{13}{42}$ ${\frac{1}{uvw}=\frac{-d}{a}={-t}=\frac{-1}{42}}$ from which ${t}={1}{42}$ ${\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=\frac{c}{a}={s}=\frac{10}{21}}$ from which ${s}=\frac{10}{21}$ thus $g({x})={x^3+\frac{13}{42}x^2+\frac{10}{21}x+\frac{1}{42}}$ and $g({1})={{1}+\frac{13}{42}+\frac{10}{21}+\frac{1}{42}}$ which equals $g({1})=\frac{38}{21}$ but $g({1})=\frac{a}{b}=\frac{38}{21}$ Hence ${a+b}={38+21}=\boxed{59}$

I will generalize the problem:

Suppose $u$ , $v$ , $w$ are the roots of polynomial $f(x)=x^3+ax^2+bx+c,$ where $c\not=0$ and $\dfrac1u$ , $\dfrac1v$ , $\dfrac1w$ are the roots of polynomial $g(x)=x^3+rx^2+sx+t$ Find the coefficients of $g(x)$ in terms of $a,b,c$ .

By Vieta on $f$ , we know that $\begin{aligned} u+v+w&=-a,&\text{ (eqn. 1)}\\ uv+uw+vw&=b,&\text{ (eqn. 2)}\\ uvw&=-c.&\text{ (eqn. 3)} \end{aligned}$

By Vieta on $g$ , we know that $\begin{aligned} -r&=\frac1u+\frac1v+\frac1w=\frac{uv+uw+vw}{uvw},&\text{ (eqn. A)}\\ s&=\frac1{uv}+\frac1{uw}+\frac1{vw}=\frac{u+v+w}{uvw},&\text{ (eqn. B)}\\ -t&=\frac1{uvw}.&\text{ (eqn. C)} \end{aligned}$ Substituting $(\text{eqn. 2})$ and $(\text{eqn. 3})$ into $(\text{eqn. A})$ gives $r=-\frac{uv+uw+vw}{uvw}=-\frac{b}{-c}=\frac{b}c.$ Substituting $(\text{eqn. 1})$ and $(\text{eqn. 3})$ into $(\text{eqn. B})$ gives $s=\frac{u+v+w}{uvw}=\frac{-a}{-c}=\frac{a}c.$ Substituting $(\text{eqn. 3})$ into $(\text{eqn. C})$ gives $t=-\frac1{uvw}=-\frac1{-c}=\frac1c.$ Therefore, $g(x)=x^3+\left(\frac{b}c\right)x^2+\left(\frac{a}c\right)x+\frac1c.$

In this problem, we have $\begin{aligned} g(1)&=1+r+s+t\\ &=1+\frac{b}c+\frac{a}c+\frac1c\\ &=1+\frac{13}{42}+\frac{20}{42}+\frac1{42}\\ &=\frac{38}{21}\end{aligned}$ as our answer.

Motivation: Well, this is a pretty basic Vieta's formulas exercise, you know stuff about the roots and coefficients.

According to Vieta's Formula, $u+v+w = -\frac{20}{1} \ = -20$ , $uv+uw+vw = \frac{13}{1} \ = 13$ , $uvw = -\frac{42}{1} \ = -42$ . Therefore, $-\frac{r}{1}\ = \frac{1}{u}\ + \frac{1}{v}\ + \frac{1}{w}\ = \frac{uv+uw+vw}{uvw}\ = \frac{13}{-42}\$ , $r = \frac{13}{42}\$ $\frac{s}{1} = \frac{1}{uv}\ + \frac{1}{uw}\ + \frac{1}{vw}\ = \frac{u+v+w}{uvw} = \frac{-20}{-42}\$ , $s = \frac{10}{21}\$ . $-\frac{t}{1}\ = \frac{1}{uvw}\ = \frac{1}{-42}\$ , $t = \frac{1}{42}\$ . Thus, $g(1) = 1^3 + \frac{13}{42}\ (1^2) + \frac{10}{21}\ (1) + \frac{1}{42}\ = \frac{38}{21}\$ , $a+b = 38+21 = \boxed{59}$

From the polynomial, we say that
-20 = u + v + w
13 = uv + vw + wu
-42 = uvw
And...
-r = $\frac{1}{u}$ + $\frac{1}{v}$ + $\frac{1}{w}$
= $\frac{uv + vw + wu}{uvw}$ = $\frac{13}{-42}$
r = $\frac{13}{42}$
s = $\frac{1}{uv}$ + $\frac{1}{vw}$ + $\frac{1}{wu}$
= $\frac{u + v + w}{uvw}$ = $\frac{-20}{-42}$ = $\frac{10}{21}$
s = $\frac{10}{21}$
-t = $\frac{1}{uvw}$ = $\frac{1}{-42}$
t = $\frac{1}{42}$
Therefore, g(x) = $x^{3}$ + $\frac{13}{42}$ $x^{2}$ + $\frac{10}{21}$ x + $\frac{1}{42}$
then, g(1) = 1 + $\frac{13}{42}$ + $\frac{10}{21}$ + $\frac{1}{42}$ = $\frac{38}{21}$ = $\frac{a}{b}$
Thus, a + b = $\boxed{59}$

Since $u, v, w$ are the roots of $f(x)$ then:

1. $(x-u)(x-v)(x-w)=0$ or $x^3-(u+v+w)x^2+(uv+uw+vw)x-uvw=0$

By substituting $1$ to the right coeffiecient of $f(x)$ we can get:

a. $u+v+w=-20$

b. $uv+vw+uw=13$

c. $uvw=42$

Using the same method for $g(x)$ , we can derive:

1. $\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=-r$

2. $\frac{1}{uv}+\frac{1}{vw}+\frac{1}{uw}=s$

3. $\frac{1}{uvw}=-t$

By solving $1, 2, 3$ using $a, b, c$ we can get:

1. $r = - \frac{uv+uw+vw}{uvw} = \frac{13}{42}$

2. $s = \frac{u+v+w}{uvw} = \frac{20}{42}$

3. $t =- \frac{1}{uvw} = \frac{1}{42}$

Then $g(1) = 1 + \frac{13}{42} + \frac{20}{42} + \frac{1}{42} = \frac{76}{42} = \frac{38}{21}$

Since $38$ and $21$ are coprime positive integers, then $a+b =\boxed{59}$

From Vieta's Formuals we know that:

$u + v + w = \frac{-b}{a} =-20$

$uv + vw + uw = \frac{c}{a} = 13$

$wuv = \frac{-d}{a} = -42$

Using the same ideas for $g(x)$ :

$\frac{1}{u}+\frac{1}{v} + \frac{1}{w} = \frac{vu+uw+uv}{uvw} = -\frac{13}{42} = -r \rightarrow r = \frac{13}{42}$

$(\frac{1}{u}*\frac{1}{v})+(\frac{1}{w}*\frac{1}{u}) + (\frac{1}{w}*\frac{1}{v}) = \frac{v+u+w}{wuv} = \frac{20}{42} = s \rightarrow s = \frac{20}{42}$

$\frac{1}{u}*\frac{1}{v}*\frac{1}{w}=\frac{1}{uvw}=-\frac{1}{42} = -d \rightarrow d = \frac{1}{42}$

$g(x) = x^{3} + \frac{13}{42}x^{2} + \frac{20}{42}x + \frac{1}{42}$

$g(1) = 1 + \frac{34}{42} = \frac{76}{42} = \frac{38}{21}\Longrightarrow 38 + 21 = \boxed{59}$

Using Vi $e$ ta's relations,we get,

For $f(x)$ , $u+v+w=-20$ , $uv+uw+vw=13$ & $uvw=-42.$

For $g(x)$ , $\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=-r$ $= \frac{uv+vw+uw}{uvw}$ $=> \frac{13}{42}=r.$

and $\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=s$ $=\frac{u+v+w}{uvw}=>20/42=s.$

Also, $\frac{1}{uvw}=-t=>t=\frac{1}{42}.$

Therefore, $g(1)=1+r+s+t$ $=1+\frac{13}{42}+\frac{20}{42}+\frac{1}{42}$ $= \frac{38}{21}$ , which gives required answer as: $38+21=$ $\boxed{59}$

From "de vieta's formula", u + v+ w = - 20, uv + vw + wu = 13, uvw=-42 And again \frac {1}{u} + \frac {1}{v} + \frac {1}{w} = - r \Rightarrow \frac {uv + vw + wu}{uvw} = -r \Rightarrow r= \frac {13}{42}

\frac {1}{uv} + \frac{1}{vw} + \frac{1}{wu} = s \Rightarrow \frac {u + v + w}{uvw}=s \Rightarrow s = \frac {10}{21}

\frac {1}{uvw} = - t \Rightarrow r = \frac {1}{42}

So g(1) = 1 + r + s + t =1 + \frac {13}{42} + \frac {10}{21} + \frac {1}{42} = \frac {38}{21} Our ans is \boxed{59}