Exihibit a counterexample for the following. Every prime number is odd.
1 answer:
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Answer:
1) x ≤ 2 or x ≥ 5
2) -6 < x < 2
Step-by-step explanation:
1) We have x^2 - 7x + 10, so let's factor this as if this were a regular equation:
x^2 - 7x + 10 = (x - 2)(x - 5)
So, we now have (x - 2)(x - 5) ≥ 0
Let's imagine this as a graph (see attachment). Notice that the only place that is above the number line is considered greater than 0, and that's when x ≤ 2 or x ≥ 5 (the shaded region).
2) Again, we have x^2 + 4x - 12, so factor this as if this were a regular equation:
x^2 + 4x - 12 = (x + 6)(x - 2)
So now we have (x + 6)(x - 2) < 0
Now imagine this as a graph again (see second attachment). Notice that the only place that is below 0 (< 0) is when -6 < x < 2 (the shaded region).
Hope this helps!
