# How $x \ge 8$ represents every value of $x \ge 5$?

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.

• 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. – Brian M. Scott Sep 12 '20 at 5:56
• 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. – Alex Jones Sep 12 '20 at 14:42

The blue region is is x ≥ 5

The red region is x ≥ 8

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

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

• 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". – Soumyadwip Chanda Sep 12 '20 at 6:08
• I didn’t mention x ≤ 5 intentionally. Thanks for your answer. – Russell Sep 12 '20 at 6:42

Believe it or not,

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

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 $$x\in(-\infty,5]\cup[8,\infty)$$

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.