## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 44

#### Answer

$n=\dfrac{-1\pm\sqrt{1+8S}}{2}$

#### Work Step by Step

$S=\dfrac{n(n+1)}{2}$; for $n$ Multiply the whole equation by $2$: $2S=n(n+1)$ Evaluate the product on the right: $2S=n^{2}+n$ Rewrite the equation like this: $n^{2}+n-2S=0$ Solve for $n$ using the quadratic formula, which is: $n=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ For this particular equation, $a=1$, $b=1$ and $c=-2S$ $n=\dfrac{-1\pm\sqrt{(1)^{2}-4(1)(-2S)}}{2(1)}=\dfrac{-1\pm\sqrt{1+8S}}{2}$

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