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

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In summary, the Lang's Basic Mathematics exercise asks to solve for x in the equation ##(x - 5)^{-2} = 9##. The attempted solution yielded x = \frac 13 + 5 = \frac 16 3. However, x also equals \frac {14}3, which the book states. The book also states that there are two solutions to the equation x^2 = a \:(a > 0). The first solution is x = \sqrt {a} and the second solution is x = -\sqrt {a}.
  • #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.
 
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  • #2
malknar said:

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
What is ##(-\frac{1}{3})^2##?
 
  • #4
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 1/9For 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
In short if A2 = B, then A = ± √B
 
  • #6
I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering ##-3 = 9^{1/2}##
 
  • #7
malknar said:
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
malknar said:
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
malknar said:
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
malknar said:
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}##.
 

1. What is the first step in solving this equation?

The first step in solving this equation is to take the reciprocal of both sides, which gives us (x-5)^2 = 1/9.

2. How can I simplify the left side of the equation?

To simplify the left side, we can expand the exponent by using the FOIL method, which gives us x^2 - 10x + 25 = 1/9.

3. What is the next step in solving the equation?

The next step is to move the constant term to the right side by subtracting 25 from both sides, which gives us x^2 - 10x + 16 = 0.

4. How do I factor the left side of the equation?

To factor the left side, we can use the AC method by finding two numbers that multiply to give us 16 and add to give us -10. In this case, those numbers are -8 and -2. So the factored form is (x-8)(x-2) = 0.

5. What are the solutions to this equation?

The solutions are x = 8 and x = 2, as those are the values that make the equation true when substituted back in. So the final answer is x = 8 or x = 2.

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

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