Equation solving is confusing

[Loki123]

The problem I have is:

x + √3+√x =3
*keep in mind 3+ √x is inside the square root.

Now logically, I would do this:
√3+√x =3-x /2
3+√x= (3-x)2
√x=(3-x)2 - 3
Square that too and calculate everything. However, It gets very complicated like that and these problems are supposed to be quick solving. I looked at the solution and I do not understand it at all.
It says that the problem has two cases:
1st case x > 1 so it's
x + √3 + √ x > 3

2nd case 0 ≤ x < 1 so it's
x + √3 + √ x < 3

How did we get the inequality now and why?
Thank you in advance

 

[Otis]

x + √3+√x =3
*keep in mind 3+ √x is inside the square root.

Hi Loki. This is how I interpret your footnote:

[imath]x + \sqrt{3 + \sqrt{x}} = 3[/imath]

If that's correct, then we text grouping symbols to show that the sum 3+√x is a radicand (expression inside a radical sign). Like this:

x + √(3 + √x) = 3

Also, we use the caret symbol ^ to show exponents. Like this:

(3 - x)^2

these problems are supposed to be quick solving … I do not understand [the given solution] … It says [there's] two cases [to consider: x>1 and 0≤x<1]

Something is wrong with this exercise. If the equation above is correct, then their approach is flawed because those two cases taken together exclude x=1. But x=1 is the solution.

Please check to be sure that you've copied all the given information correctly. If you have, then your materials contain a mistake.

PS: The equation above is not quick/easy to solve algebraically by hand.

?

 

[Loki123]

Hi Loki. This is how I interpret your footnote:

[imath]x + \sqrt{3 + \sqrt{x}} = 3[/imath]

If that's correct, then we text grouping symbols to show that the sum 3+√x is a radicand (expression inside a radical sign). Like this:

x + √(3 + √x) = 3

Also, we use the caret symbol ^ to show exponents. Like this:

(3 - x)^2


Something is wrong with this exercise. If the equation above is correct, then their approach is flawed because those two cases taken together exclude x=1. But x=1 is the solution.

Please check to be sure that you've copied all the given information correctly. If you have, then your materials contain a mistake.

PS: The equation above is not quick/easy to solve algebraically by hand.

?

Thank you, I checked again. The solution says that x > 1 is for x + √(3 + √x) > 3, 0 ≤ x < 1 is for x + √(3 + √x) < 3 and that the only possible solution is x=1. How did they come up with this?

 

[Dr.Peterson]

The problem I have is:

x + √3+√x =3
*keep in mind 3+ √x is inside the square root.

Now logically, I would do this:
√3+√x =3-x /2
3+√x= (3-x)2
√x=(3-x)2 - 3
Square that too and calculate everything. However, It gets very complicated like that and these problems are supposed to be quick solving. I looked at the solution and I do not understand it at all.
It says that the problem has two cases:
1st case x > 1 so it's
x + √3 + √ x > 3

2nd case 0 ≤ x < 1 so it's
x + √3 + √ x < 3

How did we get the inequality now and why?
Thank you in advance

The "/2" doesn't belong there, and you ignored it. Is it just a typo?

I assume you meant, as Otis suggests, [imath]x + \sqrt{3+\sqrt{x}} = 3[/imath] and [imath]\sqrt{x}=(3-x)^2-3[/imath].

What is the source, that mentions these cases?

The only explanation I can see for them is that they may have seen by inspection that x=1 is a solution, and proved those inequalities to show that the solution is unique. To be sure what they have in mind, we need access to what they said, so can you give us an image or link to it?

 

[Otis]

The solution says that … the only possible solution is x=1

Ah, you hadn't mentioned that bit.

I agree with Dr. Peterson. Their solution approach assumes that you first recognize by inspection x=1 is a solution and then consider (analyze) what happens when x moves away from 1. It's a process of deduction, leading to the two given inequalities. For example:

If x < 1, then √x < 1
But if √x < 1, then 3+√x < 4
But if 3+√x < 4, then √(3+√x) < 2
But if √(3+√x) < 2, then x+√(3+√x) < 3

That bold part says we're adding a number less than 1 to a number less than 2. So their sum must be less than 3. Hence, there can be no solutions when x<1.

The same reasoning (analysis) can be applied in the other case.

Maybe they want you to think beyond general/usual methods, since you're working problems in the category of "quick solving" exercises. What sorts of things has your class been doing?

 

[Otis]

For fun (to see how long it would take me), I solved the equation by hand. I took the same approach as the OP, isolating radicals and squaring (twice). Result: Fourth-degree polynomial; I used the Rational Root Theorem, then I used synthetic division, then I used the Quadratic Formula, and I checked the roots.

x=1, in about 17 min

For me, that was quick. I often mess up and clean up. (They'd picked one with nice numbers.) Their suggested solution method is nice, too.

 

[Loki123]

For fun (to see how long it would take me), I solved the equation by hand. I took the same approach as the OP, isolating radicals and squaring (twice). Result: Fourth-degree polynomial; I used the Rational Root Theorem, then I used synthetic division, then I used the Quadratic Formula, and I checked the roots.

x=1, in about 17 min

For me, that was quick. I often mess up and clean up. (They'd picked one with nice numbers.) Their suggested solution method is nice, too.

I managed to solve it that way. Thank you.