Since the roots of the given polynomial are four consecutive odd integers , then let the roots be- $2k+1$ , $2k+3$ , $2k+5$ and $2k+7$

By Viete's relation, we have -

$(2k+1)+(2k+3)+(2k+5)+(2k+7) = 0 \Rightarrow 8k+16 = 0 \Rightarrow k = -2$

Thus we get the roots of the polynomial as - $-3,-1,1,3$

Now, substituting $1$ (or $-1$ ) in the given equation gives us -

$1-a+b=0$

Substituting $3$ (or $-3$ ) in the given equation gives us -

$81-9a+b=0$

Solving the above two equations, gives us -

$a = 10$ and $b = 9$

Thus $a+b=\boxed{19}$

If $r$ is a solution, so is $-r$ . Thus, the four consecutive integers are $-3,-1,1,3$ . Plugging in $x=3,1$ , we obtain $b=9a-81$ $b=a-1$ Thus, $a-1=9a-81\implies a=10\implies b=9$ . Indeed, the equation $x^4-10x^2+9$ has roots at $x=\pm 1,3$ , and our answer is $10+9=\boxed{019}$ .

By Vieta's formula's, the sum of the roots is 0. If the roots are consecutive odd numbers that add to 0, then they are -3,-1,1,3. Finally, $(x-3)(x-1)(x+1)(x+3)=x^4-10x^2+9$ so $a+b=\boxed{19}$ .

Let the roots of the equation be $w-2$ , $w$ , $w+2$ and $w+4$ with $w$ being odd. Using Vieta's formula with respect to the coefficient on the $x^3$ term, which is zero, we get:

$0 = w-2+w+w+2+w+4$ , meaning that $w=-1$ . Thus our four roots are -3, -1, 1 and 3. Thus our polynomial is $(x+3)(x+1)(x-1)(x-3)$ , which when expanded and simplified, leaves us with $x^4 -10x^2 +9$ . Therefore our answer is $10+9=\boxed{19}$

Let the 4 roots be $\alpha-2, \alpha, \alpha+2, \alpha+4$ .

By Vieta's Formula,

$\alpha-2+\alpha+\alpha+2+\alpha+4=0$

$\alpha=-1$

Hence, the roots are $-3,-1,1,3$ .

Substitute the root $1$ and $3$ into the equation,

$1-a+b=0$

$81-9a+b=0$

Solve the system of equation,

$a=10$ and $b=9$

Hence, $a+b=19$ .

Let the four roots be (n-3), (n-1), (n+1) and (n+3). So we have (x-n-3)(x-n-1)(x-n+1)(x-n+3) = 0 Upon expansion, we get (x^2 - 2nx - (n^2-1))(x^2 - 2nx - (n^2-9))=0 Since the coefficients of x and x^3 are zero, 2n = 0 hence n=0. Now, a= -(n^2-1)-(n^2-9) = 10 - 2n^2 = 10. b= (n^2-1)(n^2-9) = (-1)(-9) = 9. Hence a+b = 19.

there is a rule( I forgot the name sorry ) check the number of changes in sign i.e. + to - then again from - to + ...(sign behind each terms) .. we can see two such changes so number of positive real roots will be two so the other two must me negative integers. since they are consecutive integers so the numbers are -3 , -1 , 1 , 3 put any two diff values and solve the two equations... u will get a + b = 19

Let the roots be 2n-3, 2n-1, 2n+1, 2n+3 where n is an integer since they are odd. Hence b = multiplication of all four, which is 16n^4- 40n^2 + 9. But n is an integer.
If we try n=0 we get b=9 and roots are -3,-1,3,1
putting x = -3, -1, 1, 3 and solving we always get b = 9 and a = 10. Thus n = 0 was correct try.
Note - I have taken the roots as I did is since it is the most convenient way to take odd numbers.

As co-efficient of $x^{3}$ is 0, so it's clear that the four roots are -3, -1, 1, 3. Then just apply Vieta's formula to get the Value of a, b.

here a = 10, b = 9, a + b = 19

Just take an example of the roots being -3,-1,1,3 as it is applicable for any 4 consecutive odd integers with equation of the form x^4-a.x^2+b=0 .

The equation with the above roots are x^4-10.x^2+9=0 ====> a+b=19

by vietas theoram, sum of roots=0; also roots are 4 consecutive odd integers therefore the roots are -3,-1,1,3 implies the polynomial is (x-3)(x-1)(x+1)(x+3) therefore a=10&b=9

So simple...

Taking the roots as 2N-3 ; 2N-1 ; 2N+1 ; 2N+3

coefficient of x^3=0 =summation of roots = -8N

Since coefficient of x^3=0 thus we get n=0; the roots are -3,-1,+1,+3

By calculating; we get b=9 and a=10

If we think that $S = (1, 3, 5, 7)$

We have:

$x^{4} - ax^{2} + b = 0 => 1^{4} - a\times1^{2} + b = 0 =>$

$1 - a + b = 0 => a = b + 1$

Nice... we have an equation:

$a = b + 1$

The next value of $x$

$x^{4} - ax^{2} + b = 0$ replece the $x$ by $3$

$3^{4} - a\times3^{2} + b = 0$

$81 - 9a + b = 0 => 9a = 81 + b$

replace $a$

$9 (b + 1) = 81 + b => 9b + 9 = 81 + b$

$9b - b = 81 - 9 => 8b = 72 => b = \frac{72}{8} => b = 9$

Replace $b$ in first equation:

$a = b + 1 => a = 9 + 1 => a = 11$

So:

$a + b = 10 + 9 => \boxed{a + b = 19}$