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Solving the following equation for x: ##(x-5)^{-2} = 9##

  • Thread starter malknar
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  • #1
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Homework Statement


An exercise in Lang's Basic Mathematics which asked to solve for x in the equation ##(x - 5)^{-2} = 9##

I tried to solve it and got ##x = \frac {16} 3##

which is correct according to the book, but I've found that the book states that x also equals \frac {14}3, and I can't figure out how.

Homework Equations




The Attempt at a Solution


##(x-5)^{-2} = 9##

This yields: ##x-5 = 9^{-1/2} =\frac 1 {9^{1/2}} = \frac13##

Which yields: ##x=\frac 13 +5 = \frac {1+15} 3 =\frac {16}3##

But I can't figure out how can x equals ##\frac {14}3## too.
 

Answers and Replies

  • #2
33,262
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Homework Statement


An exercise in Lang's Basic Mathematics which asked to solve for x in the equation ##(x - 5)^{-2} = 9##

I tried to solve it and got ##x = \frac {16} 3##

which is correct according to the book, but I've found that the book states that x also equals \frac {14}3, and I can't figure out how.

Homework Equations




The Attempt at a Solution


##(x-5)^{-2} = 9##

This yields: ##x-5 = 9^{-1/2} =\frac 1 {9^{1/2}} = \frac13##
It's better to write this as ##(x - 5)^2 = \frac 1 9##, which leads to ##x - 5 = \pm \sqrt{\frac 1 9}##
Quadratic equations can have two solutions.
malknar said:
Which yields: ##x=\frac 13 +5 = \frac {1+15} 3 =\frac {16}3##

But I can't figure out how can x equals ##\frac {14}3## too.
 
  • #3
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What is ##(-\frac{1}{3})^2##?
 
  • #4
301
15
I am thinking about that old math movie with young actor Edward James Olmos. He was playing the role of high school teacher.

Stand and deliver

"Negative times negative is positive!!!" The result would be 1/9


For the original poster's benefit. The following is compiled from my own high school era old math factbook.

second degree equations.
  • you can evaluate whether the quadratic has two roots, or one root, or none of the real-numbered roots. (roots belonging to ℝ)
  • calculate the quadratic into the form of: $$Ax^2 + Bx+C= 0$$
$$x=\frac{-B±\sqrt{B^2-4AC}}{2A}$$

this is called the discriminant part of the equation ##D=\sqrt{B^2-4AC}##

if D>0, then the equation has two non-equal roots
if D=0, then the equation has a twin-root
if D<0, then the equation doesn't have real-numbered roots (roots in the real numbers ℝ)
This last fact is because negative inside the squareroot sign is not defined.
E.g. ##\sqrt{(-5)}= ~~not ~~defined~~ in~~ real~~ numbers##

negative inside the squareroot is defined in the complex numbers, I think, but I have no knowledge of the complex numbers myself. (complex numbers symbol is this one ℂ)
 
  • #5
epenguin
Homework Helper
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In short if A2 = B, then A = ± √B
 
  • #6
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I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering ##-3 = 9^{1/2}##
 
  • #7
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I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering ##-3 = 9^{1/2}##
The symbol ##\sqrt{x}## denotes the positive square root of x, by established convention. In this case ##\sqrt{9} = +3##. However, the equation ##x^2 = 9## has two solutions: 3 and -3.
 
  • #8
Ray Vickson
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I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering ##-3 = 9^{1/2}##
No: ##9^{1/2}## is ##+3##, NOT ##-3##. Essentially by definition, the functions ##x^{1/2}## and ##\sqrt{x}## mean the positive root.

There is no really good language for it, but saying "a" square root instead of "the" square root comes close to capturing it: there are two roots to the equation ##x^2 = a \:(a > 0)##; these are ##x = \sqrt{a}## and ##x = -\sqrt{a}##
 
  • #9
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I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering ##-3 = 9^{1/2}##

  • watch this khan academy video it will be helpful for this problem and also for the purpose of increased understanding.
  • review what is the binomial square formula such as:
  • ##(a+b)^2= a^2+2ab+b^2##
  • ##(a-b)^2 = a^2-2ab+b^2##
 
  • #10
James R
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An exercise in Lang's Basic Mathematics which asked to solve for x in the equation ##(x - 5)^{-2} = 9##
Alternatively you could do this:
##\frac{1}{(x-5)^2}=9##
##9(x-5)^2 = 1##
##9(x^2 - 10 x + 25) = 1##
##9x^2 - 90 x + 224 = 0##
##(3x - 14)(3x - 16)=0##
##x=\frac{14}{3}## or ##x=\frac{16}{3}##.
 

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