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| − | '''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.
| + | ==Problem== |
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| − | == Introduction ==
| + | Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have? |
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| − | Vieta's Formulas were discovered by the French mathematician [[François Viète]].
| + | <math>\textbf{(A)}\ 10\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50</math> |
| − | Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as:
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| − | <center><math>x^2+ax+b=(x-p)(x-q)</math></center>
| + | ==Solution 1== |
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| − | (Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we now have
| + | Denote the number of blueberry and cherry jelly beans as <math>b</math> and <math>c</math> respectively. Then <math>b = 2c</math> and <math>b-10 = 3(c-10)</math>. Substituting, we have <math>2c-10 = 3c-30</math>, so <math>c=20</math>, <math>b=\boxed{\textbf{(D) } 40}</math>. |
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| − | <center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center>
| + | ==Solution 2== |
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| − | Vieta's Formulas are often used when finding the sum and products of the roots of a quadratic in the form <math>ax^2 + bx +c</math> with roots <math>r_1</math> and <math>r_2.</math> They state that:
| + | From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer choice satisfying this is <math>\boxed{\textbf{(D) } 40}</math>. We can check that 40 blueberry and 20 cherry jelly beans indeed does work. |
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| − | <cmath>r_1 + r_2 = -\frac{b}{a}</cmath> and <cmath>r_1 \cdot r_2 = \frac{c}{a}.</cmath>
| + | ==Solution 3 (Fakesolve)== |
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| − | We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.
| + | Note that the number of jellybeans minus <math>2 \cdot 10 = 20</math> is a multiple of <math>4</math> and positive. Therefore, the number of jellybeans is a multiple of <math>4</math> and <math>>20</math>. The only answer choice satisfying this is <math>\boxed{\textbf{(D) } 40}</math>. |
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| − | A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.
| + | Note (not from original author): The problem asks for the number of blueberry jellybeans, not the total number, so we cannot conclude that <math>40</math> is the answer. There are actually <math>60</math> total jellybeans, <math>40</math> of which are blueberry jerrybeans. |
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| − | We can state Vieta's formulas more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,
| + | ==Video Solution== |
| − | where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that
| + | https://youtu.be/QrCSysVHhPU |
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| − | <center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center>
| + | ~savannahsolver |
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| − | Expanding out the right-hand side gives us
| + | ==Video Solution by TheBeautyofMath== |
| | + | With new whiteboard at home: https://youtu.be/zTGuz6EoBWY?t=774 |
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| − | <center><math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math></center>
| + | ~IceMatrix |
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| − | The coefficient of <math> x^k </math> in this expression will be the <math>(n-k)</math>-th [[elementary symmetric sum]] of the <math>r_i</math>.
| + | ==See also== |
| − | | + | {{AMC10 box|year=2017|ab=B|num-b=4|num-a=6}} |
| − | We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that
| + | {{MAA Notice}} |
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| − | <center><math>a_n = a_n</math></center>
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| − | <center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
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| − | <center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center>
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| − | <center><math>\vdots</math></center>
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| − | <center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center>
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| − | More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).
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| − | If we denote <math>\sigma_k</math> as the <math>k</math>-th elementary symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.
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| − | Also, <math>-b/a = p + q, c/a = p \cdot q</math>.
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| − | In short:
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| − | Should <math>f(x)=a_nx^n +a_{n_1}x^{n_1}+a_{n_2}x^{n_2}+\cdots+a_1x+a_0=(x-r_1)(x-r_2)\cdots(x-r_n)</math>, then:
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| − | <cmath>
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| − | \begin{align*}
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| − | s_1&= & r_1+r_2+r_2&+\cdots+r_n & &=-\frac{a_{n-1}}{a_n} \\
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| − | s_2&= & r_1r_2+r_1r_3+r_1r_4&+\cdots+r_{n-2}r_{n-1} & &=\phantom{-}\frac{a_{n-2}}{a_n} \\
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| − | s_3&= & r_1r_2r_3+r_1r_2r_4&+\cdots+r_{n-2}r_{n-1}r_n & &=-\frac{a_{n-3}}{a_n} \\
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| − | & & &\vdots & & \\
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| − | s_n&= & r_1r_2r_3&\cdots r_n & &=(-1)^n\frac{a_0}{a_n}.\\
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| − | \end{align*}
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| − | </cmath>
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| − | ==Proving Vieta's Formula==
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| − | Basic proof:
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| − | This has already been proved earlier, but I will explain it more.
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| − | If we have
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| − | <math>x^2+ax+b=(x-p)(x-q)</math>, the roots are <math>p</math> and <math>q</math>.
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| − | Now expanding the left side, we get: <math>x^2+ax+b=x^2-qx-px+pq</math>.
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| − | Factor out an <math>x</math> on the right hand side and we get: <math>x^2+ax+b=x^2-x(p+q)+pq</math>
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| − | Looking at the two sides, we can quickly see that the coefficient <math>a</math> is equal to <math>-(p+q)</math>. <math>p+q</math> is the actual sum of roots, however. Therefore, it makes sense that <math>p+q= \frac{-b}{a}</math>. The same proof can be given for <math>pq=\frac{c}{a}</math>.
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| − | Note: If you do not understand why we must divide by a, try rewriting the original equation as <math>ax^2+bx+c=(x-p)(x-q)</math>
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| − | SuperJJ
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Problem
Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?
Solution 1
Denote the number of blueberry and cherry jelly beans as 
and 
respectively. Then 
and 
. Substituting, we have 
, so 
, 
.
Solution 2
From the problem, we see that 10 less than one of the answer choices must be a multiple of 3 and positive. The only answer choice satisfying this is 
. We can check that 40 blueberry and 20 cherry jelly beans indeed does work.
Solution 3 (Fakesolve)
Note that the number of jellybeans minus 
is a multiple of 
and positive. Therefore, the number of jellybeans is a multiple of and 
. The only answer choice satisfying this is .
Note (not from original author): The problem asks for the number of blueberry jellybeans, not the total number, so we cannot conclude that 
is the answer. There are actually 
total jellybeans, of which are blueberry jerrybeans.
Video Solution
https://youtu.be/QrCSysVHhPU
~savannahsolver
Video Solution by TheBeautyofMath
With new whiteboard at home: https://youtu.be/zTGuz6EoBWY?t=774
~IceMatrix
See also
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
and 
respectively. Then 
and 
. Substituting, we have 
, so 
, 
.
. We can check that 40 blueberry and 20 cherry jelly beans indeed does work.
is a multiple of 
and positive. Therefore, the number of jellybeans is a multiple of 
. The only answer choice satisfying this is 
is the answer. There are actually 
total jellybeans, 
