Prove there is a value for x

[Jorreke]

Hello! I'm trying to solve exercise 19b (see attached image). Previously, it was shown that there is no value for x if the square root of x is negative, so this is how I approached the exercise. I tried to solve it in 2 different ways, and both times the square root is obviously positive. However, the answer in the book says there is no x. Is the book wrong? I attached my solutions also.
Thanks,

Joris

 

[stapel]

Hello! I'm trying to solve exercise 19b (see attached image).

I'll type it out, so people can see what you're talking about:

19. Solve the following, if possible:

a) [imath]\sqrt{x-2\,} = 3 + 2\sqrt{x\,}[/imath]

b) [imath]\sqrt{x-2\,} = 3 - 2\sqrt{x\,}[/imath]

c) [imath]\sqrt{x+3\,} = 1 + \sqrt{x\,}[/imath]

d) [imath]\sqrt{x+3\,} = 1 -\sqrt{x\,}[/imath]

e) [imath]\sqrt{x-4\,} = 3 + \sqrt{x\,}[/imath]

f) [imath]\sqrt{x-4\,} = 3 - \sqrt{x\,}[/imath]

Previously, it was shown that there is no value for x if the square root of x is negative, so this is how I approached the exercise. I tried to solve it in 2 different ways, and both times the square root is obviously positive. However, the answer in the book says there is no x. Is the book wrong? I attached my solutions also.

Unfortunately, I can't read sideways well, and I'm having trouble making out what you posted.

Eliz.

 

[blamocur]

I'll type it out, so people can see what you're talking about:

19. Solve the following, if possible:

a) [imath]\sqrt{x-2\,} = 3 + 2\sqrt{x\,}[/imath]

b) [imath]\sqrt{x-2\,} = 3 - 2\sqrt{x\,}[/imath]

c) [imath]\sqrt{x+3\,} = 1 + \sqrt{x\,}[/imath]

d) [imath]\sqrt{x+3\,} = 1 -\sqrt{x\,}[/imath]

e) [imath]\sqrt{x-4\,} = 3 + \sqrt{x\,}[/imath]

f) [imath]\sqrt{x-4\,} = 3 - \sqrt{x\,}[/imath]


Unfortunately, I can't read sideways well, and I'm having trouble making out what you posted.

Eliz.

In the same spirit I am contributing a rotated version of the OP's image

 

[khansaheb]

Hello! I'm trying to solve exercise 19b (see attached image). Previously, it was shown that there is no value for x if the square root of x is negative, so this is how I approached the exercise. I tried to solve it in 2 different ways, and both times the square root is obviously positive. However, the answer in the book says there is no x. Is the book wrong? I attached my solutions also.
Thanks,

Joris

a) √(x - 2) = 3 + 2√x.............................................Square both sides

(x - 2) = (3 + 2√x )^2 ........................................ expand and simplify and substitute for √x and get a quadratic equation and solve it..... and continue

 

[blamocur]

You did a good job but forgot that a quadratic equation has 2 roots, and only one of them satisfies the original equation.

 

[Dr.Peterson]

Hello! I'm trying to solve exercise 19b (see attached image). Previously, it was shown that there is no value for x if the square root of x is negative, so this is how I approached the exercise. I tried to solve it in 2 different ways, and both times the square root is obviously positive. However, the answer in the book says there is no x. Is the book wrong? I attached my solutions also.
Thanks,

Joris

b) [imath]\sqrt{x-2\,} = 3 - 2\sqrt{x\,}[/imath]

If we graph the two sides of the equation, we can see that there is a solution. So the book is wrong if it says there is no solution. (Does it really?)



Checking it, [imath]\sqrt{2.024-2}=0.155[/imath], and [imath]3-2\sqrt{2.024}=0.155[/imath].

On the other hand, your first solution is about 6.643, which is not correct. You omitted a plus-or-minus, and the negative sign gives the correct solution:

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You didn't finish solving the second way. But in both, you are ignoring the fact that squaring (in the first step and later) loses information about signs, producing likely extraneous solutions. There is no guarantee that whatever you find will actually work.

I'm not sure what you are saying the book is teaching about "if the square root of x is negative"; I think you are misinterpreting it. Can you show us what they said?