## Algebra 1

In order to solve the equation $x^{2}$-25=0 we will factor the left side of the equation by applying the rule that states $a^{2}$-$b^{2}$=(a+b)(a-b) If we set $a^{2}$=$x^{2}$ and $b^{2}$=25, we can solve for a and b. $a^{2}$=$x^{2}$ After we square root both sides of the equation, we will get a=x $b^{2}$=25 After we square root both sides of equation, we will go b=5 Therefore, $x^{2}$-25=(x+5)(x-5) And if $x^{2}$-25=0, then we can set x+5=0 and x-5=0 For x+5=0, we can solve for x by subtracting 5 from both sides of the equation, and getting x=-5 For x-5=0, we can solve for x by adding 5 to both sides of the equation, and getting x=5
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