Solve the following inequality. $$ x^{2}-16 x+550 $$

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Solve the following inequality. $$ x^{2}-16 x+55>0 $$

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1. Write the inequality: $$ x^2 - 16x + 55 > 0 $$ 2. Factor the quadratic expression. We look for two numbers whose product is $$55$$ and sum is $$-16$$. These numbers are $$-5$$ and $$-11$$ because: $$ (-5)(-11) = 55 \quad \text{and} \quad (-5) + (-11) = -16. $$ Thus, the inequality becomes: $$ (x-5)(x-11) > 0 $$ 3. Identify the critical points where the expression is equal to zero: $$ x = 5 \quad \text{and} \quad x = 11. $$ 4. Determine the sign of the product in the intervals determined by the critical points: - For $$x < 5$$, both $$x-5$$ and $$x-11$$ are negative, so their product is positive. - For $$5 < x < 11$$, $$x-5$$ is positive while $$x-11$$ is negative, so their product is negative. - For $$x > 11$$, both $$x-5$$ and $$x-11$$ are positive, so their product is positive. 5. Since we require $$(x-5)(x-11) > 0$$, the inequality holds in the intervals: $$ x < 5 \quad \text{or} \quad x > 11. $$ 6. Therefore, the solution is: $$ (-\infty, 5) \cup (11, \infty) $$ The solution to the inequality $$ x^{2} - 16x + 55 > 0 $$ is all real numbers less than 5 or greater than 11.

Solve the following inequality.

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