Teorema Vieta

Misalkan$P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{1}x+a_{0}$ adalah sukubanyak (polinom) dengan akar-akar $x_{1},\ x_{2},\ x_{3},\ ...,\ x_{n-1},\ x_{n}$
maka berlaku:
$x_{1}+x_{2}+x_{3}+...+x_{n-1}+x_{n}=-\frac{a_{n-1}}{a_{n}}$
$x_{1}x_{2}+x_{1}x_{3}+...+x_{2}x_{3}+x_{2}x_{4}+...+x_{n-1}x_{n}=+\frac{a_{n-2}}{a_{n}}$
$x_{1}x_{2}x_{3}+x_{1}x_{3}x_{4}+...+x_{2}x_{3}x_{4}+x_{2}x_{4}x_{5}+...+x_{n-2}x_{n-1}x_{n}=-\frac{a_{n-3}}{a_{n}}$

$...\textrm{dan seterusnya}$

$x_{1}x_{2}x_{3}\ .\ ...\ .\ x_{n-1}x_{n}=(-1)^n\ .\ \frac{a_{0}}{a_{n}}$

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Contoh 1:

Persamaan kuadrat $x^{2}+5x-7=0$ memiliki akar-akar $x_{1}$ dan $x_{2}$. 

Tentukan nilai dari ${x_{1}}^3+{x_{2}}^3$.
Jawaban:

$n=2,\ a_{n}=a_{2}=1,\ a_{n-1}=a_{1}=5,\ a_{0}=-7$

$x_{1}+x_{2}=-\frac{a_{1}}{a_{2}}=-\frac{5}{1}=-5$

$x_{1}x_{2}=+\frac{a_{0}}{a_{2}}=+\frac{-7}{1}=-7$

sehingga diperoleh

$\begin{array}{rcl}{x_{1}}^3+{x_{2}}^3&=&(x_{1}+x_{2})^3-3{x_{1}}^2x_{2}-3x_{1}{x_{2}}^2\\&=&(x_{1}+x_{2})^3-3x_{1}x_{2}(x_{1}+x_{2})\\&=&(-5)^3-3(-7)(-5)\\&=&-125-105\\&=&-230\end{array}$

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Contoh 2:

Persamaan sukubanyak $x^{4}-5x^3-16x^2+41x-15=0$ memiliki akar-akar $a,\ b,\ c,$ dan $d$. 

Tentukan nilai dari: 

a. $a^2+b^2+c^2+d^2$

b. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$

Jawaban:

$n=4,\ a_{n}=a_{4}=1,\ a_{n-1}=a_{3}=-5,\ a_{n-2}=a_{2}=-16,\ a_{n-3}=a_{1}=41,\ a_{0}=-15$

$a+b+c+d=-\frac{a_{3}}{a_{4}}=-\frac{-5}{1}=5$

$ab+ac+ad+bc+bd+cd=+\frac{a_{2}}{a_{4}}=+\frac{-16}{1}=-16$

$abc+abd+acd+bcd=-\frac{a_{1}}{a_{4}}=-\frac{41}{1}=-41$

$abcd=+\frac{a_{0}}{a_{4}}=+\frac{-15}{1}=-15$

sehingga diperoleh 

jawaban a: 

$\begin{array}{rcl}a^2+b^2+c^2+d^2&=&(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)\\&=&(5)^2-2(-16)\\&=&25+32\\&=&57\end{array}$

jawaban b: 

$\begin{array}{rcl}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}&=&\frac{abc+abd+acd+bcd}{abcd}\\&=&\frac{-41}{-15}\\&=&\frac{41}{15}\end{array}$

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Contoh 3:

Jika $\alpha ,\ \beta ,$ dan $\gamma$ adalah akar-akar dari $x^{3}-x-1=0$, tentukan nilai dari $\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }$. 

Jawaban:

$n=3,\ a_{n}=a_{3}=1,\ a_{n-1}=a_{2}=0,\ a_{n-2}=a_{1}=-1,\ a_{0}=-1$

maka

$\alpha +\beta +\gamma =-\frac{a_{2}}{a_{3}}=-\frac{0}{1}=0$
$\alpha \beta +\alpha \gamma +\beta \gamma =+\frac{a_{1}}{a_{3}}=+\frac{-1}{1}=-1$
$\alpha \beta \gamma =-\frac{a_{0}}{a_{3}}=-\frac{-1}{1}=1$

sehingga diperoleh

$\begin{array}{rcl}\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }&=&\frac{(1+\alpha )(1-\beta )(1-\gamma )+(1+\beta )(1-\alpha )(1-\gamma )+(1+\gamma )(1-\alpha )(1-\beta )}{(1-\alpha )(1-\beta )(1-\gamma )}\\&=&\frac{3-(\alpha +\beta +\gamma )-(\alpha \beta +\alpha \gamma +\beta \gamma )+3\alpha \beta \gamma }{1-(\alpha +\beta +\gamma )+(\alpha \beta +\alpha \gamma +\beta \gamma )-\alpha \beta \gamma }\\&=&\frac{3-(0)-(-1)+3.1}{1-(0)+(-1)-1}\\&=&\frac{7}{-1}\\&=&-7\end{array}$

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