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 $k$.
Problem
Source: 1984 USAMO #1
Tags: algebra, polynomial, AMC, USAMO
11.08.2009 20:04
i think you shuold put the solution and problem in different posts...
11.08.2009 20:10
11.08.2009 20:24
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.
11.08.2009 23:28
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)
11.09.2009 06:12
16.02.2012 00:50
18.10.2016 01:09
Nvm this doesn't work
26.11.2019 19:24
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 $n=14$. $k = mn + 30 = \boxed{86}$.
11.05.2020 16:37
I wish USAMO was this easier nowadays
11.05.2020 16:38
Here is a good polynomial's handout: Did you get this from YCMA? https://youthconway.com/handouts/polynomials.pdf
11.05.2020 16:40
srisainandan6 wrote: Here is a good polynomial's handout: Did you get this from YCMA? https://youthconway.com/handouts/polynomials.pdf yup
20.07.2020 17:16
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
20.07.2020 18:21
There is a much easier way (in my opinion)
Shoot this is the same as #6
20.07.2020 18:24
@above i think you meant 86 but nice solution
20.07.2020 19:04
OlympusHero wrote: There is a much easier way (in my opinion) We can factor the quartic as $(x^2+ax-32)(x^2+bx+62)$. Expanding this gives $x^4+(a+b)x^3+(ab+30)x^2+(62a-32b)x-1984$. We have $a+b=-18$, $62a-32b=200$. We know that $32a+32b=-576$, so $94a=-376$, which means that $a=-4$. So, $b=-14$, and we need to compute $ab+30=\boxed{84}$, the answer. Shoot this is the same as #6 nice solution but imo system of equations is kinda easier.
20.07.2020 19:25
mathstats wrote: @above i think you meant 86 but nice solution Yeah, you're right, fixed. ATGY wrote: OlympusHero wrote: There is a much easier way (in my opinion) We can factor the quartic as $(x^2+ax-32)(x^2+bx+62)$. Expanding this gives $x^4+(a+b)x^3+(ab+30)x^2+(62a-32b)x-1984$. We have $a+b=-18$, $62a-32b=200$. We know that $32a+32b=-576$, so $94a=-376$, which means that $a=-4$. So, $b=-14$, and we need to compute $ab+30=\boxed{84}$, the answer. Shoot this is the same as #6 nice solution but imo system of equations is kinda easier. Maybe for you, but this method is easier than you think. It is called the Method of Undetermined Coefficients. In fact, you can use it to prove Vieta's Formulas.
20.07.2020 19:31
Hello, OlympusHero's Method is faster, but ATGY's is easier to think of.
20.07.2020 19:38
sshi wrote: Hello, OlympusHero's Method is faster, but ATGY's is easier to think of. Really? What first came to my mind was my method, I never even thought about ATGY's
20.07.2020 19:43
OlympusHero wrote: sshi wrote: Hello, OlympusHero's Method is faster, but ATGY's is easier to think of. Really? What first came to my mind was my method, I never even thought about ATGY's oof lol ATGY's method is what I immediately thought of.
15.08.2021 02:40
As seen in my solution, both methods ultimately boil down to the same system.
15.08.2021 03:18
Let the roots be $a,b,c,d$. Then $ab=-32,cd=62$. Next, note \[(a+b)+(c+d) = 18\]\[-200 = abc+abd+acd+bcd = -32(c+d)+ 62(a+b)\]Thus, \[94(a+b) = 18\cdot 32 - 200 = 376\]Thus, $a+b=4,c+d=14$. Now, we may answer extract \[k = ab+cd+(a+b)(c+d) = 30+56 = 86\]
24.08.2021 03:42
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 $\boxed{k=86}$.
18.10.2022 22:50
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 $k=86$.
19.12.2022 05:49
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 $k = 56-32+62=\boxed{86}$.
20.04.2023 07:52
21.04.2023 16:08
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 $k=\boxed{86}$
31.05.2023 14:49
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
01.08.2023 05:59
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.
02.08.2023 00:22
easiest usamo ever
13.10.2023 19:59
24.12.2023 21:55
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.
05.05.2024 16:21
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.
18.08.2024 23:46
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: \[ k = \boxed{86}. \]
02.01.2025 00:39
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 Now, ab + cd + (a + b)(c + d) = k On substituting the values of ab,cd,(a + b) and (c + d), we get k = -32 + 62 + 4(14) k = 30 + 56 = 86 Therefore, the required value of k is 86.