Simple Quadratic

[Vagrant]

Homework Statement

Solve for all real x: (x^2-6x+9)^(x^2-4)=1

Homework Equations

The Attempt at a Solution

I raised both sides by 1/x^2-4, so got x^2-6x+9=1, solved it to get x=2,4.
It's a question from a college enterance exam sample paper, so I wondered if it could be that simple. Is my Solution correct?

 

[Dick]

I'm assuming you didn't mean to put '^' between the two factors. In which case just remember a*b=0 if and only if a=0 or b=0.

 

[Vagrant]

It's x^2-6x+9 riased to the power x^2-4

 

[Feldoh]

What's $$(x^2-6x+9)^0$$ ?

 

[Vagrant]

1. How will that help?

 

[thomas49th]

anything to power 0 is 1

 

[Vagrant]

so? Ho is it relevant to the ques?

 

[Feldoh]

1. How will that help?

Well anything raised to the 0 power is going to equal 1. So, how would you get $$(x^2-6x+9)^{x^2- 4}$$ to be $$(x^2-6x+9)^0$$?

 

[Vagrant]

OK I get your point. x=2,-2. But what is wrong with my solution?

 

[Feldoh]

OK I get your point. x=2,-2. But what is wrong with my solution?

Well then if you can turn the exponent to 0, then wouldn't the equation also be satisfied when x=-2? (You already have 2 as a solution) Do you see what I'm getting at?

If you don't, graph the equation (x^2-6x+9)^(x^2-4)-1=0 and look at the x-intercepts.

 

[Vagrant]

Ok if I take log on both sides then I get (x^2-4)log(x^2-6x+9)=log1=0.
then i get x^2-4=0, for which i get x=2,-2 or x^2-6x+9=1 for which i get x=2,4. so is the answer x=2,-2 or x=2,4?

 

[Feldoh]

Ok if I take log on both sides then I get (x^2-4)log(x^2-6x+9)=log1=0.
then i get x^2-4=0, for which i get x=2,-2 or x^2-6x+9=1 for which i get x=2,4. so is the answer x=2,-2 or x=2,4?

Not quite, the answer is going to be x=-2,2,4.

 

[Vagrant]

Why not or?
I couldn't graph the func. still haven't learned enough.

 

[Feldoh]

It's going to be x=-2,2,4 because if x=-2 or 2 then the exponent becomes 0, and anything raised to the 0 power will equal 1. x=4 is a solution because:

$$(x^2-6x+9)^{x^2- 4}=1$$
$$((x^2-6x+9)^{x^2- 4})^{1/x^2- 4}=1^{1/x^2- 4}$$
$$x^2-6x+9=1$$
$$(x-2)(x-4)=0$$
$$x=2,4$$

So, 4 is also a solution

 

[f(x)]

Its simple. You get 1 on RHS iff the exponent is 0 OR/AND the base is itself 1.
So, $\ x^2-6x-9=1 \ \mbox{OR/AND} \ x^2-4=0$

 

[Vagrant]

Thanks for the help.