The number of distinct solutions of the equat | vedclass

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(C) Given equation is $x^{11}-x^7+x^4-1=0$. Factorizing the expression: $x^7(x^4-1) + 1(x^4-1) = 0$ $(x^4-1)(x^7+1) = 0$. Case $(i)$: $x^4-1=0 \implie

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$10$

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(C) Given equation is $x^{11}-x^7+x^4-1=0$. Factorizing the expression: $x^7(x^4-1) + 1(x^4-1) = 0$ $(x^4-1)(x^7+1) = 0$. Case $(i)$: $x^4-1=0 \implies x^4=1$. The roots are $x = e^{i(2k\pi/4)}$ for $k=0, 1, 2, 3$. These are $1, i, -1, -i$. There are $4$ distinct roots. Case $(ii)$: $x^7+1=0 \implies x^7=-1$. The roots are $x = e^{i((2k+1)\pi/7)}$ for $k=0, 1, 2, 3, 4, 5, 6$. There are $7$ distinct roots. To find the total number of distinct solutions,we check for common roots: If $x^4=1$ and $x^7=-1$,then $x^4=1 \implies |x|=1$. Also $x^7=-1 \implies |x|=1$. If $x$ is a common root,then $x^4=1$ and $x^7=-1$. Then $x^8 = (x^4)^2 = 1^2 = 1$. Since $x^8 = x^7 \cdot x = 1$,we have $(-1) \cdot x = 1$,so $x = -1$. Check if $x = -1$ satisfies both: $(-1)^4 = 1$ (True) and $(-1)^7 = -1$ (True). So,$x = -1$ is the only common root. Total distinct solutions = (Number of roots of $x^4-1=0$) + (Number of roots of $x^7+1=0$) - (Number of common roots) Total = $4 + 7 - 1 = 10$.

The number of distinct solutions of the equation $x^{11}-x^7+x^4-1=0$ is

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