Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 41

Answer

The four real solutions are: $x = -2,~~$ $x=2,~~$ $x=-3,~~$ or $~~x=3$

Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$ Then the next step is to rewrite the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ We can consider the given equation with $x^2$ in place of $x$: $x^4-13x^2+36 = 0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -13~$ and $~r\times s = 36$. We can see that $(-4)+(-9) = -13~$ and $(-4)\times (-9) = 36$ We can factor the equation as follows: $x^4-13x^2+36 = 0$ $(x^2-4)~(x^2-9) = 0$ The general form of factoring the difference of two squares is as follows: $x^2-y^2 = (x+y)~(x-y)$ We can use difference of squares to continue factoring the equation as follows: $x^4-13x^2+36 = 0$ $(x^2-4)~(x^2-9) = 0$ $(x+2)~(x-2)~(x+3)~(x-3) = 0$ $x+2 = 0,~~$ $x-2=0,~~$ $x+3=0,~~$ or $~~x-3=0$ $x = -2,~~$ $x=2,~~$ $x=-3,~~$ or $~~x=3$
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