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I'm new to quadratic inequalities. I was trying to solve this following problem -

$$x^2 - 13x + 40 \ge 0 $$ $$(x-5)(x-8) \ge 0 $$

When we consider both of these expressions positive -

$$(x-5) \ge 0$$ and $$(x-8) \ge 0 $$ we get $x \ge 5$ and $x \ge 8$

And I was taught to simplify this as $x \ge 8$. I know this expression also indicates that $x$ is greater than $5$, but it doesn't show that $x$ can be equal to $5$. Or when simplifying expressions with greater than or equal to sign, does equal to doesn't have much significance here.

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    $\begingroup$ If $x\ge 5$ and $x\ge 8$, then $x$ cannot be equal to $5$, since it is required to be at least $8$. The point is that any real number that satisfies the second inequality automatically satisfies the first, so the first is superfluous. $\endgroup$ – Brian M. Scott Sep 12 '20 at 5:56
  • $\begingroup$ I think an important point you might be missing is that "or equal to" doesn't mean that actually has to be a possibility! For example, we can say that for all real numbers $x$, $x^2 \geq -1$ is a true statement, even though it actually can't be less than $0$ so it will never equal $-1$. There's actually a special distinction for "or equal to" inequalities that actually can be equal, they're called sharp inequalities. $\endgroup$ – Alex Jones Sep 12 '20 at 14:42
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enter image description here

The blue region is is x ≥ 5

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The red region is x ≥ 8

As you are solving the inequality with AND , it refers to the intersection of both the areas

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This is the reasoning behind why you were taught to simplify x≥5 and x≥8 as x≥8

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  • $\begingroup$ I have been specific to your doubt. I know that you missed the solution x ≤ 5 or maybe you didn't mention it, I just answered your "why". $\endgroup$ – Soumyadwip Chanda Sep 12 '20 at 6:08
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    $\begingroup$ I didn’t mention x ≤ 5 intentionally. Thanks for your answer. $\endgroup$ – Russell Sep 12 '20 at 6:42
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Believe it or not,

$$x\ge 5\land x\ge 8\iff x\ge 8$$ is a true expression.

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You are missing the case when $x-5\le 0$ AND $x-8\le 0$. This yield a solution $x\le 5$. So the final answer is $x\ge 8$ OR $x\le 5$. Equivalent $$x\in(-\infty,5]\cup[8,\infty)$$

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Your question is the wrong way round.

$x\ge 5$ includes every instance of $x\ge 8$, so if you want both to be true you only have to check $x\ge 8$. Then $x\ge 5$ is automatically satisfied as well.

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