Question 250958
This one is a bit tricky. We have to think out how we can take a number and reise it to a power and get a result of 1. There are three possibilities I can think of:<ul><li>1 to <i>any</i> power is 1</li><li>Any number (excpet zero) to the zero power is 1</li><li>-1 to any even power is 1</li></ul>
So we will look for solutions from all three possibilities.
1 to any power is 1:
Solve {{{x^2 -5x + 5 = 1}}}
Subtract 1 from each side:
{{{x^2 -5x + 4 = 0}}}
Factor:
{{{x -4)(x-1)= 0}}}
Use the Zero Product Property:
{{{x -4 = 0}}} or {{{x - 1 = 0}}}
Solve:
{{{x = 4}}} or {{{x = 1}}}<br>
Any number but zero to the zero power is 1
For this we will find the x's that make {{{x^2 -9x +20 = 0}}} and then check to make sure they do not make {{{x^2-5x+5}}} zero, too. Solving {{{x^2 -9x +20 = 0}}}...
Factor:
{{{(x-4)(x-5) = 0}}}
Use the Zero Product Property:
{{{x-4 = 0}}} or {{{x-5 = 0}}}
Solve:
{{{x = 4}}} or {{{x = 5}}}
Now we check to see if either of these make {{{x^2 - 5x +5}}} zero:
for x = 4:
{{{(4)^2 - 5(4) + 5}}}
{{{16 - 5(4) + 5}}}
{{{16 - 20 + 5}}}
{{{1}}} This is not 0 so x = 4 is OK.
for x = 5:
{{{(5)^2 - 5(5) + 5}}}
{{{25 - 5(5) + 5}}}
{{{25 - 25 + 5}}}
{{{5}}} This is not 0 so x = 5 is OK.<br>
-1 to an even power is 1
Solve {{{x^2 -5x + 5 = -1}}} and then see if {{{x^2 -9x + 20}}} is even for those x's.
Solve {{{x^2 -5x + 5 = -1}}}:
Add 1 to each side:
{{{x^2 -5x + 6 = 0}}}
Factor:
{{{(x-3)(x-2) = 0}}}
Use the Zero Product Property:
{{{x=3 = 0}}} or {{{x-2 = 0}}}
Solve:
x = 3 or x = 2
Now we will see if either of these make the exponent, {{{x^2 -9x + 20}}}, even:
for x = 3:
{{{(3)^2 -9(3) + 20}}}
{{{9 -9(3) + 20}}}
{{{9 -27 + 20}}}
{{{2}}} This is even so x = 3 is a solution.
Now we will see if either of these make the exponent, {{{x^2 -9x + 20}}}, even:
for x = 2:
{{{(2)^2 -9(2) + 20}}}
{{{4 -9(2) + 20}}}
{{{4 -18 + 20}}}
{{{2}}} This is also even so x = 2 is also a solution.<br>
So the solutions are: x = 4, x = 1, x = 5, x = 3 and x = 2.<br>