The set of all solutions of the inequation x^ | vedclass

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Question

(C) Given inequation is $x^2 - 2x + 5 \leq 0$. We can rewrite the expression by completing the square: $x^2 - 2x + 1 + 4 \leq 0$ $(x - 1)^2 + 4 \leq 0

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$\phi$

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(C) Given inequation is $x^2 - 2x + 5 \leq 0$. We can rewrite the expression by completing the square: $x^2 - 2x + 1 + 4 \leq 0$ $(x - 1)^2 + 4 \leq 0$ Since $(x - 1)^2 \geq 0$ for all real $x$, it follows that $(x - 1)^2 + 4 \geq 4$. Therefore, the expression $(x - 1)^2 + 4$ is always positive and can never be less than or equal to $0$. Thus, there are no real values of $x$ that satisfy the given inequation. The set of all solutions is the empty set, denoted by $\phi$.

The set of all solutions of the inequation $x^2 - 2x + 5 \leq 0$ in $R$ is

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