2 Answers | Add Yours
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
We’ve answered 319,814 questions. We can answer yours, too.Ask a question