# what is the polynomial that has the roots xk+yh, if k=1,2 and h=1,2?the equation x^2+ax+b=0 has the roots xk and the equation y^2+cy+d=0 has the roots yh

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An equation of the form x^2 + ax + b = 0 cannot have roots equal to xk, the roots are numeric terms that do not contain the variable.

So, the roots of x^2 + ax + b = 0 are x = 1 and x = 2. Similarly the roots of the equation y^2 + cy + d = 0 are numeric terms and for the problem given are equal to y = 1 and y = 2.

The roots of the polynomial that have to be determined are also numeric and cannot be of the form xk + yh.

Assuming the polynomial to be determined is bivariate and the roots are 2 and 3 for x and 3 and 4 for y, we get:

(x - 2)(x - 3)(y - 3)(y - 4) = 0

=> (x^2 - 5x + 6)(y^2 - 7y + 12) = 0

=> x^2*y^2 - 7x^2*y + 12x^2 - 5x*y^2 + 35xy - 60x + 6y^2 - 42y + 72 = 0

**The required polynomial is x^2*y^2 - 7x^2*y + 12x^2 - 5x*y^2 + 35xy - 60x + 6y^2 - 42y + 72 = 0**

THE SOLUTION IS NOT GOOD.

Actually xk is x1 and x2, k=1,2

The solution in my book is not numerical.

solution: z^4 +2(a+c)z^3+(a^2+c^2+2b+2d+3ac)z^2+(2ab+2ad+a^2c+2bc+2dc+ac^2)z+b^2+abc+bc^2+acd+d^2