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Simple Quadratic

  • Thread starter Vagrant
  • Start date
  • #1
195
1

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?
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
  • #3
195
1
It's x^2-6x+9 riased to the power x^2-4
 
  • #4
1,341
3
What's [tex](x^2-6x+9)^0[/tex] ?
 
  • #5
195
1
1. How will that help?
 
  • #6
655
0
anything to power 0 is 1
 
  • #7
195
1
so? Ho is it relevant to the ques?
 
  • #8
1,341
3
1. How will that help?
Well anything raised to the 0 power is going to equal 1. So, how would you get [tex](x^2-6x+9)^{x^2- 4}[/tex] to be [tex](x^2-6x+9)^0[/tex]?
 
  • #9
195
1
OK I get your point. x=2,-2. But what is wrong with my solution?
 
  • #10
1,341
3
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.
 
  • #11
195
1
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?
 
  • #12
1,341
3
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.
 
  • #13
195
1
Why not or?
I couldn't graph the func. still haven't learnt enough.
 
  • #14
1,341
3
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:

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

So, 4 is also a solution
 
  • #15
182
0
Its simple. You get 1 on RHS iff the exponent is 0 OR/AND the base is itself 1.
So, [itex]\ x^2-6x-9=1 \ \mbox{OR/AND} \ x^2-4=0 [/itex]
 
  • #16
195
1
Thanks for the help.
 

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