Product of Values on Roots

Note that $f(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ . Now we look at the product we must find. We may factor each $g(x_i)$ as $(-\sqrt{2}i-x_i)(\sqrt{2}i-x_i)$ . If we take the product of all the $g(x_i)$ , we have two products, the first being the equivalent of $f(-2i)$ and the second equivalent to $f(2i)$ . Evaluating both we get $(29+8\sqrt{2}i)(29-8\sqrt{2}i)$ , which equals 969, our desired answer.

Translation of the above:

$\begin{aligned} \prod_{i=1}^4 g(x_i) & = \prod_{i=1}^4 (-\sqrt{2}i-x_i)(\sqrt{2}i-x_i) = \left[ \prod_{i=1}^4 (-\sqrt{2}i-x_i) \right] \left[\prod_{i=1}^4 (\sqrt{2}i-x_i) \right] \\ & = f(\sqrt{2}i)f(-\sqrt{2}i) = 969 \end{aligned}$

Please note: ∏ represents multiplication of indicated terms for varying index; analogous to ∑. j is used as an index for eg. in xj, j is the index. g(x)=(x+√2 i)(x-√2 i) To find : ∏g(xj) ; j=1,2,3,4 . We can write that f(x)=∏(x-xj) ; j=1,2,3,4 f(√2)=∏(√2-xj) ; j=1,2,3,4 =∏(xj-√2) ; j=1,2,3,4 f(-√2)=∏(-√2-xj) ; j=1,2,3,4 =∏(√2+xj) ; j=1,2,3,4 f(√2)f(-√2)=∏(xj+√2)(xj-√2) ; j=1,2,3,4 =∏g(xj) ; j=1,2,3,4 Therefore ∏g(xj)= (29 + 8√2i)(29-8√2i) =969

By Vieta's formula we know that $\sum x_i=4, \sum_{i\neq j}x_ix_j=-12, \sum \frac{x_1x_2x_3x_4}{x_i}=0, x_1x_2x_3x_4=1$ . $\sum_{i\neq j}(x_ix_j)^2=(\sum_{i\neq j}x_ix_j)^2-6x_1x_2x_3x_4-2 \sum \frac{x_1x_2x_3x_4}{x_i}(x_1+x_2+x_3+x_4-x_i) =144-6-8\sum frac{x_1x_2x_3x_4}{x_i}+2a_1a_2a_3a_4=146$ . $\sum (\frac{x_1x_2x_3x_4}{x_i})^2=(\sum \frac{x_1x_2x_3x_4}{x_i})^2-2x_1x_2x_3x_4\sum_{i\neq j}x_ix_j=0-24=-24$ . $\sum x_i^2=(\sum x_i)^2-2\sum_{i\neq j}x_ix_j=16+24=40$ . Expanding the term we want we get $16+(x_1x_2x_3x_4)^2+4\sum_{i\neq j}(x_ix_j)^2+2\sum (\frac{x_1x_2x_3x_4}{x_i})^2+8\sum x_i^2=1+16+584+320+48=969$ .

Because $x_1, x_2, x_3,$ and $x_4$ are roots of $f(x)$ , we know that $f(x) = (x-x_1)(x-x_2)(x-x_3)(x-x_4) \\ = x^4 - (x_1 + x_2 + x_3 + x_4)x^3 \\ + (x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4)x^2 \\- (x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4)x + (x_1 + x_2 + x_3 + x_4).$

Also, $f(x) = x^4 - 4x^3 -12 x^2 +1$ , so $x_1 + x_2 + x_3 + x_4 = 4$ , $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = -12$ , $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = 0,$ and $x_1 + x_2 + x_3 + x_4 = 1$ .

Expanding the product $g(x_1)g(x_2)g(x_3)g(x_4)$ , we get $(x_1^2+2)(x_2^2+2)(x_3^2+2)(x_4^2+2)\\ = (x_1x_2x_3x_4)^2 + 2((x_1x_2x_3)^2 + (x_1x_2x_4)^2 + (x_1x_3x_4)^2 + (x_2x_3x_4)^2) \\ + 4((x_1x_2)^2 + (x_1x_3)^2 + (x_1x_4)^2 + (x_2x_3)^2 + (x_2x_4)^2 + (x_3x_4)^2) \\+ 8(x_1^2 + x_2^2 + x_3^2 +x_4^2) +16$ .

The value of the first term, $(x_1x_2x_3x_4)^2$ , can easily be found using Vieta's Formula, so $(x_1x_2x_3x_4)^2 = 1^2 = 1$ .

The value of the second term, $2((x_1x_2x_3)^2 + (x_1x_2x_4)^2 + (x_1x_3x_4)^2 + (x_2x_3x_4)^2)$ , is equal to $2((x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4)^2 - 2(x_1x_2x_3x_4)(x_1+x_2+x_3+x_4))$ By Vieta's, we have that this term is equal to $2( 0^2 - 2\cdot 1\cdot (-12)) = 48$ .

The value of the third term, $4((x_1x_2)^2 + (x_1x_3)^2 + (x_1x_4)^2 + (x_2x_3)^2 + (x_2x_4)^2 + (x_3x_4)^2)$ , is equal to $(4((x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4)^2 \\ - 2(x_1 + x_2 + x_3 + x_4)(x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4) \\ + 2(x_1x_2x_3x_4))$ . By Vieta’s, we have that this term is equal to $4( (-12)^2 - 2\cdot 4\cdot 0 +2\cdot 1) = 584$ .

The value of the fourth term, $8(x_1^2 + x_2^2 + x_3^2 +x_4^2)$ , is $8((x_1 + x_2 + x_3 + x_4)^2-2(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3+ x_2x_4+x_3x_4)) \\=8(4^2 - 2\cdot (-12)) = 320.$

Adding the terms gives us $1 + 48 + 584 + 320 +16 = \boxed {969}$ .

Denote the two roots of the polynomial $g$ by $y_1=\sqrt{2}i$ and $y_2=-\sqrt{2}i$ . By the product formula $g(x)=(x-y_1)(x-y_2)$ , the product $g(x_1)g(x_2)g(x_3)g(x_4)$ can be written as $(x_1-y_1)(x_1-y_2)(x_2-y_1)(x_2-y_2)(x_3-y_1)(x_3-y_2)(x_4-y_1)(x_4-y_2)$ Changing the order of terms in all eight factors, and grouping the factors differently, we get $(y_1-x_1)(y_1-x_2)(y_1-x_3)(y_1-x_4)(y_2-x_1)(y_2-x_2)(y_2-x_3)(y_2-x_4)$ By the product formula for $f(x)$ , this equals $f(y_1)f(y_2)= (29+8\sqrt{2}i)(29-8\sqrt{2}i)=969$ .

Note: This product is called the resultant of $f$ and $g$ . Its symmetric definition is related to the Weil reciprocity law

If $a,b,c,d$ are the roots of $f(x)$ , $a^2,b^2,c^2,d^2$ are the roots of $f(\sqrt{x})$ and $a^2+2,b^2+2,c^2+2,d^2+2$ are the roots of $f(\sqrt{x-2})$ i.e. $g(a),g(b),g(c),g(d)$ are the roots. $f(\sqrt{(x-2)})=0 => x^4-48x^3+378x^2-1356x+969=0$ product of the roots is 969.

Sum of roots (x1, x2, x3, x4) taken 1 at a time = 4 --> a

Sum of roots taken 2 at a time = -12 --> b

Sum of roots taken 3 at a time = 0 --> c

Sum of roots taken 4 at a time(products of 4 roots) = 1 --> d

Now g(x1)g(x2)g(x3)g(x4) = (x1^2 + 2) (x2^2 + 2) (x3^2 + 2) (x4^2 + 2)

= d^2 + 2 * (sum of squares of roots taken 3 at a time) + 4 * (sum of squares of roots taken 2 at a time) + 8 * (sum of squares of roots taken 1 at a time) + 16

= d^2 + 2 * r + 4 * s + 8 * t +16

squaring equation 1 will give me t = 40

squaring equation 3 will give me r = 24

squaring equation 2 will give me s = 146

and d = 1

so final answer would be 1 + 2 * 24 + 4 * 146 + 8 * 40 + 16 = 969

First we construct the polynomial with roots $x_1^2, x_2^2, x_3^2, x_4^2$ . These must satisfy $(x)^2-4(x)\sqrt{x}-12(x)+1=0$ for each $x = x_1^2, x_2^2, x_3^2, x_4^2$ . Expanding, we get $(x^2-12x+1)^2=16x^3$ , so $x^4-40x^3+146x^2-24x+1=0$ .

Now we need to construct the polynomial with roots $x_1^2+2, x_2^2+2, x_3^2+2, x_4^2+2$ . Then $(x-2)^4-40(x-2)^3+146(x-2)^2-24(x-2)+1=0$ for each $x = x_1^2+2, x_2^2+2, x_3^2+2, x_4^2+2$ , so $x^4-48x^3+410x^2-1120x+969=0$ . This means that the product of the $g(x_i)$ is the product of the roots of this new polynomial, or $\fbox{969}$ .

let g(x)=y y=x^2+2 x=[y-2]^1/2 ........(1) f(x)=x^4-4x^3-12x^2+1 f(x)=0 for roots and substitute (1) you will get y^4 term as 1 and constant term as 969 therefore products of all y is 969 i.e. g(x1).g(x2).g(x3).g(x4)=969

Finding the roots by graphing in TI 83 and solving.

We first find a polynomial with roots $x^2_1, x^2_2, x^2_3, x^2_4$ Let $y$ be a root of the transformed equation. $y=x^2$ $x^4-4x^3-12x^2+1=0$ $\Rightarrow (x^4-12x^2+1)^2=(4x^3)^2$ Expand to get $x^8-40x^6+146x^4-24x^2+1=0$ Reverting back to $y$ $y^4-40y^3+146y^2-24y+1=0$ Let $h(y)=y^4-40y^3+146y^2-24y+1=0$

Since $x^2_1, x^2_2, x^2_3, x^2_4$ are the roots of $h(y)$ , we have

$h(y)=\displaystyle\prod_{i=1}^4 (y-x^2_i)$

we are required to find $\displaystyle\prod_{i=1}^4 (2+x^2_i)=\displaystyle\prod_{i=1}^4 (-2-x^2_i)=h(-2)$

$\boxed{h(-2)=969}$

given f(x) =(x-x1)(x-x2)(x-x3)(x-x4) then the equation f(sqrt(x-2)) will have roots as x1^2+2, x2^2+2 ,x3^2+2 ,x4^2+2. the product of the roots can be found from this new transformed equation.

Let $y=x^2+2$ , where $x$ is ${any}$ of the four roots of $f(x)$ . The key technique is to get a fourth degree equation in $y$ and determine the required product from it.
We have, $y^2=x^4+4x^2+4$
but $x$ (being a root of $f(x)$ ), satisfies $x^4=4x^3+12x^2-1$ . Substituting this value of $x$ in the above expression, we get $y^2=4x^3+16x^2+3$
$= 4x^2(x+4)+3$
$= 4(y-2)(x+4)+3$
$=4(y-2)x+16y-29$
Transposing the last two terms to LHS and squaring, we get $(y^2-16y+29)^2=16(y-2)^3$ The required product is simply the constant term of the above equation and is given by $29^2+16\times 8 = 969$