Polynomial Roots

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gauss1181

2681 posts

gauss1181

#1 h Aug 11, 2009, 5:00 PM • 8 Y

Y by premchandj, megarnie, jhu08, Rounak_iitr, BabaLama, Adventure10, Mango247, cubres

The product of two of the four roots of the quartic equation $x^4 - 18x^3 + kx^2+200x-1984=0$ is $-32$ . Determine the value of $k$ .

This post has been edited 2 times. Last edited by gauss1181, Aug 11, 2009, 10:22 PM

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abacadaea

2176 posts

abacadaea

#2 h Aug 11, 2009, 5:04 PM • 6 Y

Y by premchandj, megarnie, jhu08, Adventure10, Mango247, cubres

i think you shuold put the solution and problem in different posts...

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gauss1181

2681 posts

gauss1181

#3 h Aug 11, 2009, 5:10 PM • 6 Y

Y by megarnie, jhu08, IAmOrrin, Adventure10, Mango247, cubres

My Solution

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Ihatepie

2083 posts

Ihatepie

#4 h Aug 11, 2009, 5:24 PM • 4 Y

Y by jhu08, Adventure10, Mango247, cubres

Your solution is so much shorter than mine.

Instead of doing that adding, I substituted a+b for 18-c-d which made it take much longer.

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abacadaea

2176 posts

abacadaea

#5 h Aug 11, 2009, 8:28 PM • 3 Y

Y by jhu08, Adventure10, cubres

oh, 2 more things.
1. take out "And again, how do I add this to the resources section?"

2. change the title. it shouldnt have vieta because it hints at the solution(regardless of how obvious it may be)

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Ihatepie

2083 posts

Ihatepie

#6 h Sep 11, 2009, 3:12 AM • 11 Y

Y by evanyao10, hamup1, Wood2009, OlympusHero, jhu08, tricky.math.spider.gold.1, Adventure10, Mango247, centslordm, cubres, and 1 other user

Better solution

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tc1729

1221 posts

tc1729

#7 h Feb 15, 2012, 10:50 PM • 6 Y

Y by Wood2009, jhu08, Adventure10, Mango247, cubres, and 1 other user

Solution

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iamalive10

45 posts

iamalive10

#8 h Oct 17, 2016, 10:09 PM • 3 Y

Y by jhu08, Adventure10, cubres

Nvm this doesn't work

This post has been edited 1 time. Last edited by iamalive10, Nov 6, 2016, 12:18 AM

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AopsUser101

1750 posts

AopsUser101

#9 h Nov 26, 2019, 4:24 PM • 6 Y

Y by Wood2009, v4913, jhu08, Adventure10, Mango247, cubres

Let $a,b,c,d$ be roots of the polynomial.

By Vieta's formulas:

$a+b+c+d = 18$ $ab+ac+ad+bc+bd+cd = k$ $abc+abd+acd+bcd =-200$ $abcd =-1984.$ Without loss of generality, assume that $ab=-32$ , then by the last equation, $cd=62$ . We can re-write the equations as follows,
$(a+b)+(c+d) = 18$ $ac+ad+bc+bd + 30 = (a+b)(c+d) + 30 = k$ $-32c-32d+62d+62d =-200$ Letting $m=a+b$ and $n=c+d$ , we see that $m+n = 18$ , $-32n+62m=-200$ , so $m=4$ and $n=14$ . $k = mn + 30 = \boxed{86}$ .

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anonman

698 posts

anonman

#10 h May 11, 2020, 1:37 PM • 2 Y

Y by jhu08, cubres

I wish USAMO was this easier nowadays

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srisainandan6

2811 posts

srisainandan6

#11 h May 11, 2020, 1:38 PM • 3 Y

Y by jhu08, oralayhan, cubres

Here is a good polynomial's handout: Did you get this from YCMA?
https://youthconway.com/handouts/polynomials.pdf

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anonman

698 posts

anonman

#12 h May 11, 2020, 1:40 PM • 2 Y

Y by jhu08, cubres

srisainandan6 wrote:

Here is a good polynomial's handout: Did you get this from YCMA?
https://youthconway.com/handouts/polynomials.pdf

yup

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ATGY

2502 posts

ATGY

#13 h Jul 20, 2020, 2:16 PM • 3 Y

Y by jhu08, Mango247, cubres

Let $a$ , $b$ , $c$ and $d$ be the roots to this quartic equation.

WLOG, let $a$ and $b$ be the roots to this quartic whose product is $-32$ .

By Vieta's Formula, we know that:
$\implies a + b + c + d = 18$ $\implies ab + ac + ad + bc + bd + cd = k$ $\implies abc + abd + bcd + acd = -200$ $\implies abcd = -1984$ Note that:
$\implies \frac{abcd}{ab} = 62 = cd$ $\implies ab + ac + ad + bc + bd + cd = -32 + ac + ad + bc + bd + 62 = k$
$\implies ac + ad + bc + bd = k - 30$ We can factor $ac + ad + bc + bd$ as $(a + b)(c + d)$ .
$\implies (a + b)(c + d) = k - 30$ Note that:
$\implies abc + abd + bcd + acd = ab(c + d) + cd(a + b) = -200$ We know that $a + b + c + d = 18$ so $a + b = 18 - (c + d)$ . We also figured $ab = -32$ and $cd = 62$ .
$\implies -32(18 - (c + d)) + 62(c + d) = -200$ $\implies -576 + 32c + 32d + 62c + 62d = -200$ $\implies 94c + 94d = 376$ $\implies 94(c + d) = 376$ $\implies c + d = 4$ $\implies a + b = 14$ (or vice versa).

Hence:
$\implies (14)(4) = k - 30$ $\implies 56 = k - 30$ $\implies k = \boxed{86}$ $\blacksquare$

credits to my blog

This post has been edited 1 time. Last edited by ATGY, Jul 20, 2020, 2:16 PM

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OlympusHero

17020 posts

OlympusHero

#14 h Jul 20, 2020, 3:21 PM • 3 Y

Y by Wood2009, jhu08, cubres

There is a much easier way (in my opinion)

Solution

Shoot this is the same as #6 :blush:

This post has been edited 3 times. Last edited by OlympusHero, Dec 2, 2020, 8:40 PM

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mathstats

123 posts

mathstats

#15 h Jul 20, 2020, 3:24 PM • 2 Y

Y by jhu08, cubres

@above i think you meant 86 but nice solution

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jasperE3

11234 posts

jasperE3

#39 h Aug 24, 2021, 12:42 AM • 1 Y

Y by cubres

My solution from WOOT:

Let the roots of the quartic be $a,b,c,d$ with $ab=-32$ . By Vieta's, we have:
$\begin{cases}a+b+c+d=18\\ab+ac+ad+bc+bd+cd=k\\abc+bcd+cda+dab=-200\\abcd=-1984\end{cases}$ Using $ab=-32$ , this is:
$\begin{cases}a+b+c+d=18\\ac+ad+bc+bd+cd=k+32\\bcd+cda-32c-32d=-200\\cd=62\end{cases}$ $\Rightarrow\begin{cases}a+b+c+d=18\\(a+b)(c+d)=k-30\\62a+62b-32c-32d=-200\\cd=62\end{cases}$ Let $x=a+b$ and $y=c+d$ . We have $x+y=18$ and $62x-32y=-200$ . Solving, $x=4$ and $y=14$ . Substituting into $xy=k-30$ , we will find that $\boxed{k=86}$ .

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Tafi_ak

309 posts

Tafi_ak

#40 h Oct 18, 2022, 7:50 PM • 4 Y

Y by Mango247, Mango247, Mango247, cubres

The answer is $86$ . By Vieta we have
$\begin{align*} \sum r_i&=18\tag{$1$} \\ \sum r_ir_j&=k\tag{$2$} \\ \sum r_ir_jr_k&=-200\tag{$3$} \\ r_1r_2r_3r_4&=-1984\tag{$4$} \end{align*}$ Let WLOG $r_1r_2=-32$ . Using $(2)$ we have $r_3r_4=-1984/-32=62$ . Using $r_1r_2$ , $r_3r_4$ and equation $(1)$ in equation $(3)$ we have $62(r_1+r_2)-32(r_3+r_4)=-200\tag{$5$}$ Rewriting the equation $(2)$ we have $(r_1+r_2)(r_3+r_4)+30=k\tag{$6$}$ Solving the equation $(1)$ and $(5)$ in terms of variable $r_1+r_2$ and $r_3+r_4$ we get $4, 14$ . Putting these value in the equation $(6)$ we get out answer $k=86$ .

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HamstPan38825

8857 posts

HamstPan38825

#41 h Dec 19, 2022, 2:49 AM • 2 Y

Y by centslordm, cubres

Let $x^4-18x^3+kx^2+200x-1984 = (x^2+ax-32)(x^2+bx+32)$ for some $a, b$ . Equating the $x^3$ and $x$ coefficients yields
$\begin{align*} a+b &= -18 \\ 62a-32b &= 200 \end{align*}$ so $a = -4, b = -14$ , and $k = 56-32+62=\boxed{86}$ .

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cinnamon_e

703 posts

cinnamon_e

#42 h Apr 20, 2023, 4:52 AM • 1 Y

Y by cubres

solution

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mathmax12

6003 posts

mathmax12

#43 h Apr 21, 2023, 1:08 PM • 1 Y

Y by cubres

By Veita the product of the other 2 roots are 62, let the 4 roots be $a,b,c,d$ by vieta's we have $a+b+c+d=18,ab+ac+Ad+bc+bd+cd$ , and $abc+acd+abd+bcd=-200$

And then we can do some calculator from here to get that $k=\boxed{86}$

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Rounak_iitr

455 posts

Rounak_iitr

#44 h May 31, 2023, 11:49 AM • 1 Y

Y by cubres

Assume the roots of the equation $x^4-18x^3+kx^2+200x-1984=0$ are $\alpha ,\beta ,\gamma ,\delta$ Given product of any two roots of this equation is -32. By Vietas Relation we get $\alpha+\beta+\gamma+\delta=18$ $\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\delta+\beta\delta+\gamma\delta=k$ $\alpha\beta\gamma+\beta\gamma\delta+\alpha\beta\delta+\alpha\gamma\delta=-200$ $\alpha\beta\gamma\delta=-1984$ Lets assume $\alpha\beta=-32$ from product of roots we get $\alpha\beta\gamma\delta=-1984\implies\gamma\delta=62$ $\alpha\beta\gamma+\beta\gamma\delta+\alpha\gamma\delta+\alpha\beta\delta=-200\implies62(\alpha+\beta)-32(\gamma+\delta)\implies62(\alpha+\beta)-32(18-(\alpha+\beta))=-200\implies94(\alpha+\beta)=376\implies\boxed{\alpha+\beta=4}$ From these we also get $\boxed{\gamma+\delta=14}$ Therefore, $\alpha\beta+\beta\gamma+\gamma\alpha+\delta\alpha+\beta\delta+\gamma\delta=k\implies(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta=14.4+62-32=56+30\implies86.$ Is the correct answer...
Vietas Relation :love:

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pqr.

174 posts

pqr.

#45 h Aug 1, 2023, 2:59 AM • 2 Y

Y by Rounak_iitr, cubres

Let the roots be $r_1, r_2, r_3, r_4$ . By Vieta, if $r_1r_2=-32$ , then $r_3r_4=62$ . Additionally, Vieta gives $r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4=-200 \implies -32(r_3+r_4)+62(r_1+r_2).$ But notice that $r_3+r_4=18-(r_1+r_2)$ again by Vieta. Then, solving we get that $r_1+r_2=4$ and $r_3+r_4=14$ . Finally, by Vieta, we see have $k=r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=(r_1r_2+r_3r_4)+(r_1r_3+r_1r_4+r_2r_3+r_2r_4)=-32+62+(r_1+r_2)(r_3+r_4)=30+56=86,$ and we're done.

This post has been edited 2 times. Last edited by pqr., Aug 1, 2023, 2:59 AM

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exp-ipi-1

1074 posts

exp-ipi-1

#46 h Aug 1, 2023, 9:22 PM • 2 Y

Y by Rounak_iitr, cubres

easiest usamo ever

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joshualiu315

2513 posts

joshualiu315

#47 h Oct 13, 2023, 4:59 PM • 1 Y

Y by cubres

solution

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peppapig_

280 posts

peppapig_

#48 h Dec 24, 2023, 6:55 PM • 1 Y

Y by cubres

Let the roots of the quartic be $a$ , $b$ , $c$ and $d$ . WLOG, let $ab=-32$ , and by Vieta's, we have that $abcd=-1984$ , meaning that $cd=-62$ . Let $x=a+b$ and $y=c+d$ . By Vieta's, we have that
$18=a+b+c+d=(a+b)+(c+d)=x+y,$ and
$-200=abc+abd+acd+bcd=cd(a+b)+ab(c+d)=62x-32y.$ Solving the system of equations gives us that $x=4$ and $y=14$ . Again using Vieta's, we have that
$k=ac+ad+bc+bd+ab+cd=(a+b)(c+d)+ab+cd=xy-32+62,$ or $xy+30$ , giving us that $k$ is $86$ , finishing the problem.

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Markas

105 posts

Markas

#49 h May 5, 2024, 1:21 PM • 1 Y

Y by cubres

By Vieta's we have that 1) $x_1 + x_2 + x_3 + x_4 = 18$ , 2) $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = k$ , 3) $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -200$ , 4) $x_1x_2x_3x_4 = -1984$ . Now we can assume $x_1x_2$ = -32 $\Rightarrow$ $x_3x_4$ = 62. We now plug this into 3) and we get $-32x_3 - 32x_4 + 62x_1 + 62x_2 = 62(x_1 + x_2) - 32(x_3 + x_4)$ and after using 1) we get $62(x_1 + x_2) - 32(18 - x_1 - x_2) = -200$ . Let $x_1 + x_2 = a$ , we get $62a - 32(18 - a) + 200 = 94a - 576 + 200 = 94a - 376 = 0$ $\Rightarrow$ a = 4 $\Rightarrow$ $x_1 + x_2 = 4$ and $x_3 + x_4 = 14$ . Now we plug in everything we know in 2) so $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = - 32 + x_1(x_3 + x_4) + x_2(x_3 + x_4) + 62 = 30 + (x_1 + x_2)(x_3 + x_4) = 30 + 4.14 = 30 + 56 = 86 = k$ $\Rightarrow$ k = 86 $\Rightarrow$ the answer is 86.

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combowomborhombo

11 posts

combowomborhombo

#50 h Aug 18, 2024, 8:46 PM • 1 Y

Y by cubres

The product of all the roots in this polynomial is $-1984$ . Since we know the product of one pair of roots, we can find the product of the other pair of roots as $62$ . We can split the original polynomial into two quadratic polynomials:

$x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + 62)$
By matching coefficients, we find the following relationships:

$\begin{align*} a + b &= -18, \\ ab + 30 &= k, \\ 62a - 32b &= 200. \end{align*}$
Using the first and last relationships, we solve for $a$ and $b$ :

$(a, b) = (-4, -14).$
Substituting into the equation $ab + 30 = k$ , we find:

$k = \boxed{86}.$

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P162008

106 posts

P162008

#51 h Jan 1, 2025, 9:39 PM • 1 Y

Y by cubres

Let's assume that the biquadratic equation x⁴ - 18x³ + kx² + 200x - 1984 = 0 has roots a,b,c and d respectively such that ab = -32.

By Vieta's Relation we've

abcd = -1984 —————(1)

abc + bcd + cda + dab = -200

ab(c + d) + cd(a + b) = -200 —————(2)

ab + ac + ad + bc + bd + cd = k

ab + cd + (a + b)c + (a + b)d = k

ab + cd + (a + b)(c + d) = k —————(3)

and, (a + b) + (c + d) = 18

or, (a + b) = 18 - (c + d) —————(4)

Substituting ab = -32 in equation (1), we get

(-32)cd = -1984

Therefore, cd = 62

On substituting ab = -32, cd = 62 and (a + b) = 18 - (c + d) in equation (2), we get

-32(c + d) + 62[18 - (c + d)] = -200

1116 - 62(c + d) - 32(c + d) = -200

94(c + d) = 1316

Therefore, (c + d) = 14

Similarly, (a + b) = 18 - (c + d) = 4

Therefore, k = -32 + 62 + 4(14) = 86

This post has been edited 5 times. Last edited by P162008, Jan 28, 2025, 11:57 PM
Reason: Typo

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