- Thread starter Indranil
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If it were x^2 = 4, what would you do?If x^2 = root (y), then how to calculate the value of 'x'?

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|x| = sqrt(√y)

x = ±sqrt(√y)

I'll let you think about what happens, when we take the square root of a square root. If you get stuck, try switching to exponential form. :cool:

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Be careful, using that notation for square root; some people might not understand.If x^2 = roo͏t (y), then how to calculate the value of 'x'?

sqrt(y) is standard notation.

Also, "calculate the value of x" is not what you're thinking. We cannot find a value for x, unless we're first given a value for y.

What you're asking is, "How to solve this equation for x?" :cool:

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I would do it in two waysIf it were x^2 = 4, what would you do?

1. x^2 = 4 = 2^2 = 2.

2. x^2 = 4 = √4 = 2

x^2 = √y = y^(1/2)

|x| = sqrt(√y)

x = ±sqrt(√y)

I'll let you think about what happens, when we take the square root of a square root. If you get stuck, try switching to exponential form. :cool:

so, x^2 = y^(1/2) Now What can I do?

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You are misusing the "=" sign so badly that what you write is meaningless. (This is, unfortunately, common among students who never learned what the equal sign really means.) It is not true that 2^2 = 2, or that 4 = √4. Please be careful to write what you mean.I would do it in two ways

1. x^2 = 4 = 2^2 = 2.

2. x^2 = 4 = √4 = 2

I think what you mean is something like this:

1. x^2 = 4 = 2^2, so x = 2.

2. x^2 = 4, so x = √4 = 2

But it has already been pointed out that this does not give you the complete solution set, because you are ignoring the possibility that x could be negative. You can't blindly "take

The first way, you are imagining that, since x^2 = 2^2, the x must be 2 -- that is, assuming that the square function is one-to-one, so that any result can be attained in only one way. That is not true. There are two numbers whose square is 4, namely 2 and -2.

The second way, the square root of the left side, √(x^2), is not x, but |x|, because for example if x=-2, then √((-2)^2) = √4 = 2 = |-2|, not -2 itself.

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No, this is not true and I bet you know it! Equal signs MUST be valid.I would do it in two ways

1. x^2 = 4 = 2^2 = 2.

2. x^2 = 4 = √4 = 2

1. x^2 = 4 = 2^2 = 2. NO, 2^2 does NOT equal 2!!!

2. x^2 = 4 = √4 = 2. NO, 4 does not equal √4 !!!

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This I why we ask. Thank you.I would do it in two ways

1. x^2 = 4 = 2^2 = 2.

2. x^2 = 4 = √4 = 2

First, don't do that. Please factor and solve completely so that you don't have to remember anything important. Let the notation help you.

x^2 = 4

x^2 - 4 = 0

(x-2)(x+2) = 0

x= 2 or x = -2

See how we managed BOTH solutions without any pain?

Okay, now why is your original problem ANY different? (Hint: It isn't.)

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x^2 = √y, x^2 = y^(1/2)

|x| = sqrt(√y)

x = ±sqrt(√y)

I'll let you think about what happens, when we take the square root of a square root. If you get stuck, try switching to exponential form. :cool:

so, x^2 = y^(1/2) Now What can I do? Still, I am stuck in this problem. In this case, can how to calculate the value of 'x'?

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If NOTHING ELSE is given to you - you have nothing to do of any consequence.x^2 = √y, x^2 = y^(1/2)

so, x^2 = y^(1/2) Now What can I do? Still, I am stuck in this problem. In this case, can how to calculate the value of 'x'?

What else is given to you - like value of "y"?

Why aren't you posting the COMPLETE problem - as it was given to you?

If x^2 = √16, then how to calculate the value of 'x' in exponential form? I have done below in the root formIf NOTHING ELSE is given to you - you have nothing to do of any consequence.

What else is given to you - like value of "y"?

Why aren't you posting the COMPLETE problem - as it was given to you?

x^2 = √16, x= √(√16), x= √4 =2 Is the method is correct? if correct, please solve this problem in exponential form. I have tried below

x^2 = 16^1/2, x^2 = 4, x^2 = 2^2, x = 2 Am I correct? If correct, can you please solve it in any other methods?

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First, you're not quite right, because x could be either +2 or -2, as has been empathized. PLEASE don't forget that; I think some of us may be thinking you are refusing to pay attention, and getting frustrated.If x^2 = √16, then how to calculate the value of 'x' in exponential form? I have done below in the root form

x^2 = √16, x= √(√16), x= √4 =2 Is the method is correct? if correct, please solve this problem in exponential form. I have tried below

x^2 = 16^1/2, x^2 = 4, x^2 = 2^2, x = 2 Am I correct? If correct, can you please solve it in any other methods?

But for the moment, let's suppose you were also told that x>0, so you don't need to worry about that.

As written, x^2 = √16, I would first simplify the right side, so x^2 = 4. Then there are several ways to get to the answer; your way is valid only if you know that x>0! If you don't, then you need to either remember to include the negative case, or use the factoring method you have been shown:

x^2 = 4

x^2 - 4 = 0

(x + 2)(x - 2) = 0

x + 2 = 0 or x - 2 = 0

x = -2 or x = 2

But suppose now that you were given x^2 = √y, where you don't know y, and you are told to solve for x in terms of y, "in exponential form". Then you might do this, using exponents all the way:

x^2 = y^(1/2)

(x^2)^(1/2) = (y^(1/2))^(1/2)

x^(2*1/2) = y^(1/2 * 1/2)

x^1 = y^(1/4)

x = y^(1/4)

Now, in doing this, I assumed that x>0. If I couldn't do that, I would have to be aware that, just as there are two square roots, there are two 1/2 powers (since that means the same thing), so I would do this:

x^2 = y^(1/2)

(x^2)^(1/2) = ±(y^(1/2))^(1/2)

x^(2*1/2) = ±y^(1/2 * 1/2)

x^1 = ±y^(1/4)

x = ±y^(1/4)

(Note that the right side was given as √y, that means only the positive root, so I didn't use ± at the start.)

It would also be possible to express this in terms of factoring, though that is a little more awkward than with numbers.

If you are asking this because of a specific problem you have been given, it will be very helpful if you can quote the problem and tell us the context, so we can be sure what issues matter.

Oh for goodness' sake, if you had given the problem completely and exactly originally, a great deal of time could have been saved.If x^2 = √16, then how to calculate the value of 'x' in exponential form? I have done below in the root form

x^2 = √16, x= √(√16), x= √4 =2 Is the method is correct? if correct, please solve this problem in exponential form. I have tried below

x^2 = 16^1/2, x^2 = 4, x^2 = 2^2, x = 2 Am I correct? If correct, can you please solve it in any other methods?

\(\displaystyle x^2 = \sqrt{16} \implies x^2 = 16^{(1/2)} \implies (x^2)^{(1/2)}= \pm (16^{(1/2)})^{(1/2)} \implies\)

\(\displaystyle x^1 = \pm 16^{(1/4)} \implies x = \pm 2.\)

Moreover your solution is numerically incomplete: minus 2 is also a solution.

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Are you sure that you are quoting the given problem EXACTLY - verbatim?If x^2 = √16, then how to calculate the value of 'x' in exponential form? I have done below in the root form

x^2 = √16, x= √(√16), x= √4 =2 Is the method is correct? if correct, please solve this problem in exponential form. I have tried below

x^2 = 16^1/2, x^2 = 4, x^2 = 2^2, x = 2 Am I correct? If correct, can you please solve it in any other methods?

In the case it is - can you please explain what does "in exponential form" mean in the given context!

Better yet - give us the whole context - i.e. - where and how this problem was given to you?

Was the problem given to you in English - or are you translating?

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When I suggested switching to exponential form, I was referring only to the simplification of sqrt(√y).x^2 = √y = y^(1/2)

so, x^2 = y^(1/2) Now What can I do?

x^2 = √y

|x| = sqrt(√y)

x = ±[y^(1/2)]^(1/2)

x = ±y^(1/4)

In the last step, I used a property of exponents.

|x| = sqrt(√y)

x = ±[y^(1/2)]^(1/2)

In the last step, I used a property of exponents.

The given equation has been "solved for x".

The result shows the relationship between quantities x and y. It is also a formula for finding x, IF you already know a value for y.

EG:

Given that y = 2401, find all values of x.

x = ±2401^(1/4)

x = ±7

x = ±7

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How's that? Did you intend to say, "so that you don't forget anything important"?… so that you don't have to remember anything important …