For n 2 and n in N,the product of the roots | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $y = x^2 - 2nx$. The equation becomes $(x-n)(y^2 + (2n^2-5)y + (n^4-5n^2+4)) = 0$. Factoring the quadratic in $y$: $y^2 + (2n^2-5)y + (n^2-1)(

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Let $y = x^2 - 2nx$. The equation becomes $(x-n)(y^2 + (2n^2-5)y + (n^4-5n^2+4)) = 0$. Factoring the quadratic in $y$: $y^2 + (2n^2-5)y + (n^2-1)(n^2-4) = (y + n^2-1)(y + n^2-4) = 0$. Substituting $y$ back: $(x-n)(x^2 - 2nx + n^2 - 1)(x^2 - 2nx + n^2 - 4) = 0$. This simplifies to $(x-n)((x-n)^2 - 1)((x-n)^2 - 4) = 0$. $(x-n)(x-n-1)(x-n+1)(x-n-2)(x-n+2) = 0$. The roots are $x = n, n+1, n-1, n+2, n-2$. The product of the roots $P = n(n+1)(n-1)(n+2)(n-2) = (n-2)(n-1)n(n+1)(n+2)$. This is the product of $5$ consecutive integers,which is always divisible by $5! = 120$.

For $n > 2$ and $n \in N$,the product of the roots of $(x-n)((x^2-2nx)^2 + (2n^2-5)(x^2-2nx) + (n^4-5n^2+4)) = 0$ is divisible by

Similar Questions

The solution to the equation $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$ is:

Let $S = \{ x : x \in R \text{ and } (\sqrt{3} + \sqrt{2})^{x^2 - 4} + (\sqrt{3} - \sqrt{2})^{x^2 - 4} = 10 \}$. Then $n(S)$ is equal to

The positive root of $x^{2}-78.8=0$ after the first approximation by the Newton-Raphson method,assuming the initial approximation to the root is $14$,is:

If $72^x \cdot 48^y = 6^{xy}$,where $x$ and $y$ are non-zero rational numbers,then $x+y$ equals

The roots of the equation $x^{2/3} + x^{1/3} - 2 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module