How do we prove If n≥6, then n^2−6n≥0. I know it is true, just do not know how to prove it.
Prove or disprove: If n≥6, then n^2−6n≥0.
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algebra-precalculus
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3$n^2-6n=n(n-6)$. The product of two nonnegative numbers is nonnegative. – 2017-02-18
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1It would be good to include in your question what leads you to "know it is true"? – 2017-02-18
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0No need to factor as egreg did. If $n\geq6$ then $n$ is positive, so multiplying through by $n$ gives $n^2\geq 6n$. – 2017-02-18
1 Answers
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When $n \geq 6$, you have $n \geq 0$ and $n-6 \geq 0$. Now $n^2 - 6n = n(n-6)$ and a product of nonnegative numbers is nonnegative.