Weird solutions

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Levido

Junior Member
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Dec 22, 2019
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I have an equation which is relatively basic. It’s just I’ve never really had positive and negative solutions fully explained to me, and an answer to this would properly explain it. See the question below.762602EE-0249-425E-ACA0-F7D09E146D84.jpeg
The working circles in red is my initial attempt, which is all mathematically correct as far as I can tell, yet the solutions are 0,2 as well as -2 whereas in the right solution on the right the solutions are correct and are just 0 and 2. Please show me where I’ve gone wrong in my working.
B1EC09E0-5393-4FF7-9098-0F164E1DB958.jpeg
 
Levido said:
I have an equation which is relatively basic. It’s just I’ve never really had positive and negative solutions fully explained to me, and an answer to this would properly explain it. See the question below.View attachment 15648
The working circles in red is my initial attempt, which is all mathematically correct as far as I can tell, yet the solutions are 0,2 as well as -2 whereas in the right solution on the right the solutions are correct and are just 0 and 2. Please show me where I’ve gone wrong in my working.
View attachment 15649
Click to expand...
The domain of x1/2 and x3/2 is x > = 0
 
Levido said:
I understand this, could you link this to my working to show where I went wrong, an assumption I’ve made which violates the domain of x
Click to expand...
On your 3rd line you write:

x3 - 4x = 0

Domain of [x3 - 4x] is NOT restricted to positive 'x' only! Right there you changed the problem-domain.
 
When you square both sides you can have trouble.

Consider a rather simple equation, x=-2. So x equals -2 !!!. Now square both sides and you get x^2=4. Now the solution to x^2 = 4 is x=2 or x=-2. But only x=-2 is the solution.

The problem lies in the fact that whether you have x=2 or x=-2 when you square both sides you get x=4.
Similarly, if you x^2=x+2 or x^2 = -(x+2), when you square both sides you get x^4=(x+2)^2

As Subhotosh said, the 2nd (3rd) line does not have the same restriction as the original problem.
 
By the way, I hope that you saw that 0 and 2 are clearly solutions to this equation without doing any work--just by visual inspection. You need to carry out the work you did just to see if there are any other solutions.
 
Levido said:
I understand this, could you link this to my working to show where I went wrong, an assumption I’ve made which violates the domain of x
Click to expand...
You give, as your solution, x= 0, x= 2, and x= -2. The point is that -2 is not in the domain of the square root function. Only x= 0 and x= 2 are valid solutions.

The "assumption" you made was that the equations "\(\displaystyle f(x)= g(x)\)" and "\(\displaystyle f^2(x)= g^2(x)\)" have the same solutions. Any solution of "\(\displaystyle f(x)= g(x)\)" is a solution of "\(\displaystyle f^2(x)= g^2(x)\)" so squaring both sides of an equation is a valid method for solving square roots but not the other way around so you have to check your solutions in the original equation.

Generally raising to a power or multiplying both sides by something involving x may introduce "extraneous roots". For example, x= 2 has the obvious solution 2. But \(\displaystyle x^2= 4\) has both 2 and -2 as roots while \(\displaystyle x^2= 2x\) has both 2 and 0 as solutions and \(\displaystyle x(x-3)= 2(x- 3)\), so \(\displaystyle x^2- 3x= 2x- 6\) or \(\displaystyle x^2- 5x+ 6= 0\) has both 2 and 3 as roots.
 
Thanks so much everyone you’ve all really helped me understand this. I wish my teachers cared as much as the people on this board do. Such fast and high quality responses. You’re all amazing.
 
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