Cubic Arithmetic Progressions

For the sake of simplicity, let the roots be $a-d$ , $a$ , and $a+d$ , where $d$ is the common difference in the arithmetic progression. By Vieta's Formulas, we have $(a-d) + a + (a+d) = 12 \Rightarrow a = 4$ . Furthermore, Vieta's gives us

$a(a-d) + a(a+d) + (a-d)(a+d) = 34$

$a^2 - ad + a^2 + ad + a^2 - d^2 = 34$

$3a^2 - d^2 = 34$

$3(4)^2 - d^2 = 34$

$d = \pm \sqrt{14}$

Because of the symmetry in the progression, both of our values for $d$ give the same roots: $4-\sqrt{14}$ , $4$ , and $4+\sqrt{14}$ . Applying Vieta's again, the product of these roots is the opposite of the constant term. Thus, we have

$-N = -4(4-\sqrt{14})(4+\sqrt{14})$

$N = 4(16-14)$

$N = \boxed{8}$

By applying Vieta's Formula to the cubic equation, $-12 = -(a+b+c)$ , where $a, b$ , and $c$ are the roots of the equation. In order to be in an arithmetic progression, we must have values of $(4-x), 4$ , and $(4+x)$ for some value $x$ . Using Vieta's Formula again, we find that $34 = (ab+bc+ac) = (4-x)(4) + (4+x)4 + (4-x)(4+x) = 4(8) + 16 - x^2$ . $x^2$ must be the square root of $14$ , so by applying Vieta's once again, we know that $-N = -(abc) = -(4)(4-x)(4+x) = -4(16-14) = -8$ . Thus, $N=8$ .

Let $a,a+d,a+2d$ be the roots of the cubic equation by using Vieta's Theorem, $a+(a+d)+(a+2d)=12$ , $3a+3d=12$ , $a+d=4$ . Since $a+d$ is one our roots,thus 4 is one of the roots of our cubic equation. This means that $f(4)=0$ . By substitution, $f(4)=(4)^3 -12(4)^2 + 34(4) - N =0$ , so $8-N=0$ or $N=8$ .

[Edits for clarity, Latex edits - Calvin]

Since the function f(x) is cubic then there is atmost 3 roots.

Since the roots are in arithmetic progression, then the roots are

(a-d) , a , and (a+d) , where d is the common difference.

With the relationship of the roots and the coeffecients of the fuction, we know that,

(a-d) + a + (a+d) = -(-12)

                  3a = 12

                    a = 4

Hence, 4 is one of the roots. Thus (x-4) is one of the factors of f(x).

To solve for N we can use synthetic division or remainder theorem with the use of (x-4) as a divisor or x=4, respectively.

Hence,

f(4) = 4^3 -12(4^2) + 34(4) - N = 0

    = 64 - 192 + 136 - N = 0

N = 8.

Given cubic equation f(x)=x3−12x2+34x−N Consider a - d, a, a + d are the three roots of the given cubic polynomial which are in A.P. Therefore by Vieta's rule, sum of the roots = - b/a i.e. a - d + a + a + d = 12 => 3a = 12 => a = 4 Now sum of product of roots (taken two roots at a time) = c/a i.e. (a - d) a + a(a + d) + (a - d)(a + d) = 34
=> (4 - d) 4 + 4(4 + d) + (4 - d)(4 + d) = 34 (using a = 4) => 16 - 4d + 16 + 4d + 16 - d2 = 34 => 48 - d2 = 34 => d2 = 14 Now product of roots (taken all at a time) = - d/a i.e. (a - d)a(a + d) = N => (4 - d) 4(4 + d) = N (using a = 4) => 4(16 - d2) = N => 4(16 - 14) = N (using d2 = 14) => N = 4(2) => N = 8 * Note that d2 represnts square of d

As we know (x-a) (x-b) (x-c)=0 where a,b,c are roots of the equation,then the simplified form is x^3 - (a + b + c) x^2 + (ab + bc + ca) x -abc = 0.

As we can observe abc = N,b*(a + c) + ca = 34 and a + b + c = 12

As a,b,c are in A.P. 2b = a + c.

Hence b = 4 and 2b^2 + ca = 32 i.e. ac = 2. Hence ac*b = abc = 8

Let the roots be $r, r+d$ and $r+2d$ . Then by Vieta's formulas, $r+r+d+r+2d=3r+3d=12$ , and $r+d=4$ . Also, by Vieta's, $r(r+d)+r(r+2d)+(r+d)(r+2d)$ $=4r+r(r+2d)+4(r+2d)=34$ . But $r+2d=r+d+d=d+4$ , so $4r+r(r+2d)+4(r+2d)=4r+r(d+4)+4(d+4)$ $=4r+rd+4r+4d+16=rd+8r+4d+16=34$ . Noting that $rd+8r+4d+32=(r+4)(d+8)=(r+4)(r+2d+4)$ , we can add 16 to both sides of the Vieta equation to get $r+4)(r+2d+4)=50$ Multiplying this equation by $r+d+4=8$ , we get the nice symmetrical-looking equation $r(r+d)(r+2d)+4(r(r+d)+r(r+2d)+(r+d)(r+2d))$ $+16(r+r+d+r+2d)+64=400$ . Reusing Vieta's formulas yields $N+4*34+16*12+64=400$ . Subtracting it all out yields $N=8$ .

Let roots of f(x)=m,n,o. So m,n,o are in Arithmetic progression. By relation between coeff. and roots m+n+o=12....eq.1; mn+no+om=34.....eq.2; mno=N.....eq.3 Let m=a-d;n=a;o=a+d....eq.4.By substituting eqn.4in eqns.1,2 we get a=4 &d^2=14 so N=mno=(a-d)(a)(a+d)=a(a^2-d^2)=4(16-14)=4(2)=8

Assume that, the roots of the equation is :

$x_1, x_2, x_3$ . So, by vieta's formula, we obtain that :

$x_1 + x_2 + x_3 = \frac{-b}{a} = \frac{-(-12)}{1} = 12,$ and $x_1.x_2 + x_1.x_3 + x_2.x_3 = \frac{c}{a} = \frac{34}{1} = 34$

So, since $x_1, x_2, x_3$ form arithmetic progression. Assume that :

$x_1 = a - d$ , and $x_2 = a$ , and $x_3 = a + d$

So, $x_1 + x_2 + x_3 = 12 \implies (a - d) + a + (a+d) = 12$ , so : $3a = 12 \implies a = 4$

So, $x_1 = 4 - d$ , then $x_2 = 4$ , and $x_3 = 4 + d$

From,

$x_1.x_2 + x_1.x_3 + x_2.x_3 = 34$ , so : $(4 - d)4 + (4 - d)(4+d) + (4+d)4 = 34$ , then $48 - d^2 = 34$ , and $d^2 = 14$

And, we know that :

$x_1. x_2. x_3 = -\frac{d}{a} = - \frac{-N}{1} = N$ . So, $(4 - d).4.(4+d) = N$ , implies $N = 4(16 - d^2)$ implies $N = 4(16 - 14)$ implies $N = \boxed{2}$

From Vieta's formula, we know that:

$a+(a+d)+(a+2d)=-(-12)$

$a+d=4$

and

$a(a+d)+a(a+2d)+(a+d)(a+2d)=34$

$3a^2+6ad+2d^2=34$

and

$N=a(a+d)(a+2d)$

After substituting $d=a-4$ into the second equation and simplifying, I found that $d=\pm\sqrt14$ .

When $d$ is positive, you get $(4-\sqrt14)(4)(4+\sqrt14)=8$

When $d$ is negative you also get $(4+\sqrt14(4)(4-\sqrt14)=8$

Let the roots be $(a-d)$ , $a$ & $(a+d)$ . Now, according to Newton's theorem of elementary polynomials (or simply by expanding of form $(x-a)(x-b)(x-c)$ ,we have: $\displaystyle \sum$ A $=-(-12)/1=12$ , $\displaystyle \sum$ AB $=34/1=34$ , ABC $=-(-N)/1=N$ ,(where A,B & C are roots of cubic polynomial). Now $\displaystyle \sum$ A $= a-d + a + a+d = 3a=12 or, a=4$ . Again $\displaystyle \sum$ AB $= a(a-d) + (a-d)(a+d) + a(a+d) =34 or, d^2=14.$ Now ABC $=N= a(a-d)(a+d) = a(a^2-d^2) = 8$ . Ans: 8

for any cubic equation x^3+px^2+qx+r=0 sum of roots: = -p = a-d+a+a+d => 12 = a-d+a+a+d => a=4 sum of roots taking two at a time: = q =(a-d).a +(a+d).a +(a-d).(a+d) =>q= a(2a) + (a^2-d^2) => 34 = 32+ (a^2-d^2) => (a^2-d^2) = 2 multiplication of roots: = r = a.(a-d).(a+d) => N = a.(a^2-d^2) => N = 4.2 =8

suppose the roots of the polynomials are: a-d, a and a+d (as they are in arithmetic progression).

Hence, their summation: 3a = 12 or a = 4 Again, sum of product of 2 roots: a(a-d) + a(a+d) + (a-d)(a+d) = 3a^2 - d^2 = 34. Since a = 4, d = sqrt(14). So N = a(a-d)(a+d) = 8

let roots be a-d,a,a+d thus sum of root=3a=12,a=4 thus roots are 4-d,4,4+d. (4-d)4+4(4+d)+(16-d^2)=34 so d^2= but n=(16-d^2)4 so n=8

let a-d,a,a+d be three roots of eq. then sum of roots=12 a-d+a+a+d=12 a=4 then put x=4 in equation then we get 0=64-192+136-N N=8

By Vieta's Formula, the sum of the roots is equal to $-\frac {-12}{1} = 12$ . Hence, the middle term of the AP is $\frac {12}{3} = 4$ . Since 4 is a root, by the Remainder-Factor Theorem, $f(4) = 0$ . Thus, $4^3 - 12 \times 4^2 + 34 \times 4 -N = 0$ , which gives $N= 8$ .

Note: There is no need to find out all the roots.