Vieta's Formula

This week, we learn about Vieta's Formula, which allows us to relate the roots of a polynomial to its coefficients.

How would you use Vieta's Formula to solve the following?

  1. If α,β \alpha, \beta are the roots to the equation x2+2x+3=0 x^2 +2x+3 = 0 , what is the quadratic equation whose roots are (α1α)2 (\alpha - \frac{1}{\alpha} )^2 and (β1β)2 ( \beta - \frac{1}{\beta} ) ^ 2 ?

  2. Share a problem which requires understanding of Vieta’s Formula.

Note by Calvin Lin
5 years, 9 months ago

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37 votes

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Show that j=1ncot2(jπ2n+1)  =  13n(2n1) \sum_{j=1}^n \cot^2\Big(\tfrac{j \pi}{2n+1}\Big) \; = \; \tfrac13n(2n-1) for any integer n1n \ge 1. What is j=1ncot4(jπ2n+1)? \sum_{j=1}^n \cot^4\Big(\tfrac{j \pi}{2n+1}\Big) \qquad \mbox{?}

Mark Hennings - 5 years, 9 months ago

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I like this one. The idea is to show that j=1n(x2cot2(jπ2n+1))=12n+1Im(x+i)2n+1 \prod_{j=1}^n \left( x^2 - \cot^2( \frac{j\pi}{2n+1} ) \right) = \frac1{2n+1} {\rm Im} (x+i)^{2n+1} by comparing roots; then compare coefficients of x2n2 x^{2n-2} .

Patrick Corn - 5 years, 8 months ago

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Another way to look at it is to (courtesy of de Moivre's Theorem) find the degree nn polynomial fn(X)f_n(X) such that sin(2n+1)x  =  sin2n+1xfn(cot2x) \sin(2n+1)x \; = \; \sin^{2n+1}x f_n(\cot^2x) and identify the roots of fn(X)f_n(X).

What is cool is that you can use these identities to obtain low-level proofs of the series n=11n2  =  16π2n=11n4  =  190π4 \sum_{n=1}^\infty \frac{1}{n^2} \; = \; \tfrac16\pi^2 \qquad \sum_{n=1}^\infty \frac{1}{n^4} \; = \; \tfrac{1}{90}\pi^4 since you can use them to obtain error bounds on the partial sums.

Mark Hennings - 5 years, 8 months ago

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For the given problem ,we use Vieta's Formula and get :

α+β=ba=2 \alpha + \beta = - \dfrac{b}{a} = -2 and αβ=ca=3 \alpha \cdot \beta = \dfrac{c}{a} = 3

[Now - considering the case that Calvin sir wanted it to be (α1α)2 (\alpha - \dfrac{1}{\alpha} )^2 ]

Let the new equation be -

Ax2+Bx+C=0 Ax^2 + Bx + C = 0

Thus by again applying * Vieta's Formula * ,

We get the following relations between the assumed coefficients and the* required *roots :

BA=(α1α)2+(β1β)2 -\dfrac{B}{A} = \left(\alpha - \dfrac{1}{\alpha} \right)^2 + \left(\beta - \dfrac{1}{\beta} \right)^2

BA=α22+(1α)2+β22+(1β)2 -\dfrac{B}{A} = \alpha ^2 - 2 + \left(\dfrac{1}{\alpha} \right)^2 + \beta^2 - 2 + \left(\dfrac{1}{\beta} \right)^2

To obtain the value of LHS , we convert RHS in the form of α+β\alpha + \beta and αβ \alpha \cdot \beta .

BA=(α+β)22αβ+[1α+1β]22αβ4 -\dfrac{B}{A} = ( \alpha + \beta )^2 - 2 \alpha \beta + \left[ \dfrac{1}{\alpha} + \dfrac{1}{\beta} \right]^2 - \dfrac{2}{\alpha\beta} - 4

BA=(α+β)22αβ+[α+βαβ]22αβ4 -\dfrac{B}{A} = ( \alpha + \beta )^2 - 2 \alpha \beta + \left[ \dfrac{\alpha + \beta}{ \alpha \beta} \right]^2 - \dfrac{2}{\alpha\beta} - 4

Thus by putting the values we obtained initially , we have -

BA=(2)223+[23]2234 -\dfrac{B}{A} = ( -2)^2 - 2 \cdot 3 + \left[ \dfrac{-2}{ 3} \right]^2 - \dfrac{2}{3} - 4

This gives us : [CAUTION : I cancelled the negative signs of both sides]

BA=569 \dfrac{B}{A} =\boxed{\dfrac{56}{9}}

Similarly we get :

CA=(α1α)2×(β1β)2 \dfrac{C}{A} = \left(\alpha - \dfrac{1}{\alpha} \right)^2 \times \left(\beta - \dfrac{1}{\beta} \right)^2

CA=(αβ)2+[1αβ]22(α2+β2)+(α2+β2)22α2β2α2β22(α2+β2α2β2)+4 \dfrac{C}{A} = (\alpha \beta)^2 + \left[ \dfrac{1}{\alpha \beta} \right]^2 - 2(\alpha^2 + \beta^2) + \dfrac{(\alpha^2 + \beta^2 )^2-2 \alpha^2 \beta^2 }{\alpha^2 \beta^2} - 2\left( \dfrac{\alpha^2 + \beta^2 }{\alpha^2 \beta^2 } \right) + 4

Thus , we put the values of α+β \alpha + \beta , αβ \alpha \beta and α2+β2 \alpha^2 + \beta^2 [Refer above]

We get - CA=161=1449 \dfrac{C}{A} = \dfrac{16}{1} = \boxed{\dfrac{144}{9}}

Thus by comparing the values of A A , B B and C C -

We get A=9 A = 9 B=56 B = 56 C=144 C = 144

And hence , the required quadratic equation is -

9x2+56x+144=0 \boxed{ 9x^2 + 56x + 144 = 0 }

Priyansh Sangule - 5 years, 8 months ago

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An easier way is to first find the quadratic equation which has roots α1α \alpha - \frac {1}{\alpha} and β1β \beta - \frac {1}{\beta}

By Vieta's formula, we have α+β=2 \alpha + \beta = -2 and αβ=3 \alpha \beta = 3 , then α2+β2=(α+β)22αβ=(2)22(3)=2 \alpha^2 + \beta^2 = ( \alpha + \beta)^2 - 2 \alpha \beta = (-2)^2 - 2(3) = -2

Now consider a quadratic equation f(x)=ax2+bx+cf(x)=ax^2+bx+c with integers a>0,b,ca>0,b,c which has roots α1α \alpha - \frac {1}{\alpha} and β1β \beta - \frac {1}{\beta}

The sum of roots is α1α+β1β=α+β(1α+1β)=(α+β)(α+βαβ)=(α+β)(11αβ)=(2)(113)=43 \alpha - \frac {1}{\alpha} + \beta - \frac {1}{\beta} = \alpha + \beta - ( \frac {1}{\alpha} + \frac {1}{\beta} ) = ( \alpha + \beta ) - ( \frac { \alpha + \beta }{\alpha \beta } ) = ( \alpha + \beta ) (1 - \frac {1}{\alpha \beta} ) = (-2)(1 - \frac {1}{3} ) = - \frac {4}{3}

And the product of roots is

(α1α)(β1β)=αβ(αβ+βα)+1αβ=αβα2+β21αβ=3213=4 (\alpha - \frac {1}{\alpha})(\beta - \frac {1}{\beta}) = \alpha \beta - ( \frac {\alpha}{\beta} + \frac {\beta}{\alpha} ) + \frac {1}{\alpha \beta} = \alpha \beta - \frac {\alpha^2 + \beta^2 - 1}{\alpha \beta} = 3 - \frac {-2-1}{3} = 4

Then f(x)=3(x2(43)x+4)=3x2+4x+12 f(x) = 3( x^2 - (- \frac {4}{3} ) x + 4 ) = 3x^2 +4x + 12 has roots α1α \alpha - \frac {1}{\alpha} and β1β \beta - \frac {1}{\beta}

Thus f(x)=0f(\sqrt{x}) = 0 has roots (α1α)2 (\alpha - \frac {1}{\alpha})^2 and (β1β)2 (\beta - \frac {1}{\beta})^2

3(x)2+4x+12=0 \Rightarrow 3 ( \sqrt{x} )^2 + 4 \sqrt{x} + 12 = 0

3x+12=4x \Rightarrow 3x + 12 = -4 \sqrt{x}

(3x+12)2=16x \Rightarrow (3x+12)^2 = 16x

9x2+56x+144=0 \Rightarrow 9x^2 + 56x + 144 = 0 has roots (α1α)2 (\alpha - \frac {1}{\alpha})^2 and (β1β)2 (\beta - \frac {1}{\beta})^2

Pi Han Goh - 5 years, 8 months ago

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Good approach!

I like how you use f(x) f ( \sqrt{x}) to square the roots.

Calvin Lin Staff - 5 years, 8 months ago

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Do you really mean for ββ1\beta-\beta^{-1} to be squared and αα1\alpha-\alpha^{-1} not?

Mark Hennings - 5 years, 8 months ago

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Edited, thanks!

Calvin Lin Staff - 5 years, 8 months ago

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Let m,n>1m, n > 1 and m,nZm, n \in \mathbb{Z}. Suppose that the product of the solutions for x of the equation 8(lognx)(logmx)7lognx6logmx2013=08(\log_n x)(\log_m x) - 7\log_n x - 6 \log_m x - 2013 = 0 is the smallest possible integer. Compute m+nm + n

Source: AMC

Michael Tong - 5 years, 9 months ago

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Hint: Write logab\log_a b as logbloga\frac{log_b}{log_a}.

..

Solution:

Using the hint, we have (8(logn)(logm))(logx)2(7logn+6logm)logx2013=0(\frac{8}{(\log n)(\log m)}) (\log x)^2 - (\frac {7}{\log n} + \frac {6}{\log m})\log x - 2013 = 0. This is quadratic in logx\log x. Using vieta's formulas, we have that the sum of the roots a1=logx1,a2=logx2a_1 = \log x_1, a_2 = \log x_2 is equal to 7logn+6logm8lognlogm=7logm+6logn8 \frac{\frac {7}{\log n} + \frac {6}{\log m}}{\frac{8}{\log n \log m}} = \frac {7 \log m + 6 \log n}{8}.

Using log properties, this expression is equal to logm7n68=logm7n68\frac {\log m^7 n^6}{8} = \log \sqrt[8] {m^7 n^6} Because a1+a2=logx1+logx2=logx1x2 a_1 + a_2 = \log x_1 + \log x_2 = \log x_1 x_2, the product of the solutions for xx is equal to m7n68 \sqrt[8]{m^7 n^6}. To minimize this as an integer and m,n>1m , n > 1, we have m=2a,n=2bm = 2^a, n = 2^b and find the smallest multiple of 88 which can be written as 7a+6b7a + 6b for a,b1a, b \geq 1. Since 62(mod8)6 \equiv -2 \pmod 8 and 71(mod8)7 \equiv -1 \pmod 8 it is easy to find a solution in a=2,b=3a = 2, b = 3. Thus, x1x2x_1x_2 is minimized at m=8,n=4m = 8, n = 4, and m+n=12m + n = 12.

Michael Tong - 5 years, 8 months ago

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Hard problem, nice solution!

Guilherme Dela Corte - 5 years, 8 months ago

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The zeros of f(x)=x6+2x5+3x4+5x3+8x2+13x+21f(x) = x^6 + 2x^5 + 3x^4 + 5x^3 + 8x^2 + 13x + 21 are distinct complex numbers. Compute the average value of A+BC+DEFA + BC + DEF over all possible permutations (A,B,C,D,E,F)(A, B, C, D, E, F) of these six numbers.

ARML 2012, Team Problems, #6

Cody Johnson - 5 years, 8 months ago

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First, there are (63)(32)=60{6 \choose 3} {3 \choose 2} = 60 permutations. So, this is equal to 10(A+B+C+D+E+F)+4(AB+BC+CD+DE+EF+...)+3(ABC+ABD+ABE+...)10(A + B + C + D + E + F) + 4(AB + BC + CD + DE + EF + ...) + 3(ABC + ABD + ABE + ...). Using Vieta's formula, this is equal to 10(2)+4(3)+3(5)=2310(-2) +4(3) + 3(-5) = -23. Taking the average value, this is 2360\frac{-23}{60}.

Michael Tong - 5 years, 8 months ago

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(GCDC)
If a=05a×xa=b=15(xxb)\displaystyle \sum_{a=0}^5 a \times x^a = \displaystyle \prod_{b=1}^{5} (x-x_b), evaluate b=15xb5\displaystyle \sum_{b=1}^5 -x_b \: ^5.

EDIT: The answer is 113113.

Guilherme Dela Corte - 5 years, 8 months ago

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The desired polynomial is 5x5+4x4+3x3+2x2+x5x^5 + 4x^4 + 3x^3 + 2x^2 + x. We wish to find the sum of the fifth powers of the roots a,b,c,d,ea, b, c, d, e. I am confident that a solution can be found using the expansion of (a+b+c+d+e)5(a + b + c + d + e)^5 and further and further factorization, and chasing the solution further in this way is trivial. I'll edit the post if I think of more clever way to find the solution.

Michael Tong - 5 years, 8 months ago

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You can first find the values of a2+b2+c2+d2+e2a^2 + b^2 + c^2 + d^2 + e^2 and a3+b3+c3+d3+e3 a^3 + b^3 + c^3 + d^3 + e^3 . Then find their product. It's much less work compared to actually expanding (a+b+c+d+e)5 (a+b+c+d+e)^5 . But still look ridiculously long compared to Newton's Sums.

Pi Han Goh - 5 years, 8 months ago

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@Pi Han Goh For 2n32 \leq n \leq 3, the absolute value of b=1nxbn \displaystyle \sum_{b=1}^n x_b \: ^n still looks very easy and nice to do by Vieta's Formulae.

For n4n \geq 4, things begin to get "olympic", thus, as Pi Han Goh stated, making Newton's Sums use much preferable.

Guilherme Dela Corte - 5 years, 8 months ago

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Hey I just wanted to ask, in the 4th and final question in Vieta's formula, how did you find out the roots by just seeing the equation ? Please do explain.

Aejeth Lord - 5 years, 8 months ago

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It becomes a reciprocal polynomial, which is very easy to solve. Those polynomials have roots which multiplied together give 1. Generally, x1x_1 and 1x1\frac{1}{x_1}.

Guilherme Dela Corte - 5 years, 8 months ago

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Can you explain ? Please

Aejeth Lord - 5 years, 8 months ago

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@Aejeth Lord Calvin explained it a little on his last worked example.

Note that for this polynomial (which Calvin didn't write, but is implicit) 16x4170x3+357x2170x+16x=0   16x^4-170x^3+357x^2-170x+16x=0 \; the coefficients are symmetrical (the first coefficient is the same as the fifth one, the second is the same as the fourth and the third is the same as the third). It guarantees us that if, let's say, α\alpha is a root, its reciprocal (which is α1=1α\alpha ^{-1} = \frac{1}{\alpha}) will also be a root.

Because the independent coefficient (x0x^0) is not zero, we can assure that no root equals zero. Therefore we can divide the whole equation by a power of xx, which here would be x2x^2. (Can you figure out why by yourself?)

Rearranging the equation, we get 16(x2+1x2)170(x+1x)+357=016 (x^2 + \frac{1}{x^2}) - 170(x + \frac{1}{x}) + 357 = 0. To simplify it, we can call y=x+1xy = x + \frac{1}{x} and therefore y2=x2+2+1x2y^2 = x^2 + 2 + \frac{1}{x^2}, thus getting a new form 16(y22)170y+357=016y2170y+325=016 (y^2 - 2) - 170y + 357 = 0 \Rightarrow 16y^2 - 170y +325 = 0, a quadratic equation easily solvable by any known formula.

Attention now: solving 16y2170y+325=016y^2 - 170y +325 = 0 gives us 52\frac{5}{2} and 658\frac{65}{8}, the values of yy, not xx. We have to plug the two back into y=x+1xy = x + \frac{1}{x}, leading us to two more quadratics, getting finally 18,12,2,8\boxed{\dfrac{1}{8}, \dfrac{1}{2}, 2, 8}.

EDIT: Sorry for being too long. Hope you understood.

Guilherme Dela Corte - 5 years, 8 months ago

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@Guilherme Dela Corte Thanks! Let me add in your explanations.

Calvin Lin Staff - 5 years, 8 months ago

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@Calvin Lin Woohoo, a personal accomplishment achieved! \o/

I just felt it'd be much better to explain the full story right from the start to one who isn't familiar to reciprocal polynomials. We have three more other cases, which I really want to work with, but now is not the adequate time (since we are only analising Vieta's Formulae).

Guilherme Dela Corte - 5 years, 8 months ago

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@Guilherme Dela Corte Thank you for explaining, now may I ask, can we proof for a general symmetric equation without solving it, that it has symmetric roots ?

Aejeth Lord - 5 years, 8 months ago

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@Aejeth Lord I'd highly recommend you work through the proof yourself using the example Guilherme provided. The general case is exactly the same. Just apply the method he used to the polynomial p(x)=anxn+...+an2+1xn2+1+an2xn2+an2+1xn21+...+anx0p(x) = a_nx^n + ... + a_{\frac{n}{2} + 1}x^{\frac{n}{2} + 1} + a_{\frac{n}{2}}x^{\frac{n}{2}} + a_{\frac{n}{2} + 1}x^{\frac{n}{2} - 1} + ... + a_nx^0 with aiC,an0,na_i \in \mathbb{C}, a_n \neq 0, n an even positive integer (this is the general symmetric polynomial). It may seem intimidating, but just work through it in the exact same way Guilherme did in his example. Remember, also, that you have it easier: he went through and calculated the roots. All you need to do is show that if rr is a root, so is 1r\frac{1}{r}. You don't care at all what those roots actually are, so don't do any of the substitutions with yy. It may be useful to note that r+1r=1r+11rr + \frac{1}{r} = \frac{1}{r} + \frac{1}{\frac{1}{r}}. Good luck :)

Sotiri Komissopoulos - 5 years, 8 months ago

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@Aejeth Lord Let's consider a polynomial like the ones we worked with, P(x)=αx2n+βx2n1++ω++βx+αP(x) = \alpha x^{2n} + \beta x^{2n-1} + \dots + \omega + \dots + \beta x + \alpha . The condition for it to be an even degree polynomial is that α0\alpha \neq 0. Therefore the product of its roots (which is αα=1\frac{\alpha}{\alpha} = 1, but let's pretend we don't know that) is not zero, so none of the roots equal zero, allowing us to do P(x)xn=0\dfrac{P(x)}{x^n} = 0.

Its form now is P(x)=α(xn+xn)+β(xn1+x1n)++ωP'(x) = \alpha (x^n + x^{-n}) + \beta (x^{n-1} + x^{1-n}) + \dots + \omega. Making more changes because x0x \neq 0, we can express it as a function of y=x+x1y = x + x^{-1}, becoming Q(y)=ω+ψy+χ(y22)+  Q(y) = \omega + \psi y + \chi (y^2 - 2) + \dots \;.

This new polynomial has roots y=A,  B,  ,  Zy = A, \; B, \; \dots, \; Z, let's say. After finding them, we plug them back into the equation y=x+x1y = x + x^{-1}. Let's pick AA for this example. Once again because x0x \neq 0 we can multiply the yy equation by xx, finding a quadratic x2Ax+1=0x^2 - Ax + 1 = 0. Note that, for any root of Q(x)Q(x), the product of this quadratic's roots will always be 1, guaranteeing us that if P(x0)=0P(x_0) = 0, then P(1x0)=0P(\frac{1}{x_0}) = 0.

Guilherme Dela Corte - 5 years, 8 months ago

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In addition, if a polynomial is of the form P(x)=Axn+Bxn1+ +Yx+ZP(x) = Ax^n + Bx^{n - 1} + \cdots\ + Yx + Z and another polynomial is of the form Q(x)=Zxn+Yxn1+Bx+A Q(x) = Zx^n + Yx^{n-1} \cdots + Bx + A, then the roots of Q are the reciprocals of the roots of P.

Michael Tong - 5 years, 8 months ago

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The product of two of the four zeros of the quadrtic equation

x^4 - 18x^3 + kx^2 + 200x - 1984 = 0 is -32 . then find k

tushar patil - 5 years, 8 months ago

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Answer is 86... By some algebraic manipulation..

John Ashley Capellan - 5 years, 8 months ago

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I don't really get vieta's theorem. I know that it us used to solve quadratics and that I am able to use it. The wiki shows some very complicated definition with complex numbers involved. Can someone please help clarify my doubts? Thanks very much.

Tan Yong Boon - 4 years, 10 months ago

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*it is

Tan Yong Boon - 4 years, 10 months ago

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Perhaps a slightly simpler solution for part 1.

Let the required quadratic equation be: x2Ax+B=0x^2 - Ax + B = 0, then we have:

A=(α1α)2+(β1β)2=α22+1α2+β22+1β2=α2+β24+1α2+1β2=(α+β)22αβ4+(1α+1β)22αβ=(2)22(3)4+(α+βαβ)223=464+4923=569B=(α1α)2(β1β)2=(αβαββα+1αβ)2=(3α2+β2βα+13)2=(3(2)22(3)3+13)2=(3+23+13)2=42=16 \begin{aligned} A & = \left(\alpha - \frac{1}{\alpha} \right)^2 + \left(\beta - \frac{1}{\beta} \right)^2 = \alpha^2 - 2 + \frac{1}{\alpha^2} + \beta^2 - 2 + \frac{1}{\beta^2} = \alpha^2 + \beta^2 - 4 + \frac{1}{\alpha^2} + \frac{1}{\beta^2} \\ & = (\alpha + \beta)^2 - 2 \alpha \beta - 4 + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^2 - \frac{2}{ \alpha \beta} = (-2)^2 - 2(3) - 4 + \left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 - \frac{2}{3} \\ & = 4 - 6 - 4 + \frac{4}{9} - \frac{2}{3} = -\frac{56}{9} \\ B & = \left(\alpha - \frac{1}{\alpha} \right)^2 \left(\beta - \frac{1}{\beta} \right)^2 = \left(\alpha\beta - \frac{\alpha}{\beta} - \frac{\beta}{\alpha} + \frac{1}{\alpha \beta} \right)^2 = \left(3 - \frac{\alpha^2 + \beta^2} {\beta \alpha} + \frac{1}{3} \right)^2 \\ & = \left(3 - \frac{(-2)^2 - 2(3)} {3} + \frac{1}{3} \right)^2 = \left(3 + \frac{2} {3} + \frac{1}{3} \right)^2 = 4^2 = 16 \end{aligned}

x2Ax+B=0x2+569x+16=09x2+56x+16=0\Rightarrow x^2 - Ax + B = 0 \\ x^2 + \dfrac{56}{9}x + 16 = 0 \\ \boxed{9x^2 + 56x + 16 = 0}

Chew-Seong Cheong - 4 years ago

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what is the next two numbers for the series: 0,2,24......?

CHIKU PRASAD Lenka - 5 years, 8 months ago

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This question has infinite answers. I'd answer π100,e200\pi^{100}, e^{200}. 246 and 2468 could be nice guesses, but we cannot state this is the absolute true answer.

EDIT: Also, not related to Vieta's Formula.

Guilherme Dela Corte - 5 years, 8 months ago

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i think so answer to this question should be 246

tushar patil - 5 years, 8 months ago

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what is the next two numbers

Tushar, we'd have to find the next one (24682468).

Guilherme Dela Corte - 5 years, 8 months ago

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