## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 21

#### Answer

$5 +\sqrt{5}\approx 7.236$ $5 -\sqrt{5}\approx 2.764$

#### Work Step by Step

Set the $RHS$ to 0 by adding 20 to both sides $k^{2}-10k=-20\qquad.../+20$ $k^{2}-10k+20=0$ To factor the trinomial we search for integer factors of $20$ whose sum is $-10$ ... ... none We can always use the quadratic formula: (For $ax^{2}+bx+c=0,\ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ ) $k=\displaystyle \frac{-(-10)\pm\sqrt{(-10)^{2}-4(1)(20)}}{2(1)}$ $=\displaystyle \frac{10\pm\sqrt{100-80}}{2}=\frac{10\pm\sqrt{20}}{2}$ $=\displaystyle \frac{10\pm\sqrt{4\times 5}}{2}=\frac{10\pm 2\sqrt{5}}{2}$ $=\displaystyle \frac{2(5\pm\sqrt{5})}{2}$ $k=5\pm\sqrt{5}$ Solutions: $5 +\sqrt{5}\approx 7.236$ and $5 -\sqrt{5}\approx 2.764$.

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