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What is the solution of 4√x - 4 = 3 * 2√x?

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What is the solution of 4√x - 4 = 3 * 2√x?  [#permalink]

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New post Updated on: 09 Sep 2018, 21:12
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What is the solution of 4√x - 4 = 3 * 2√x?

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Originally posted by rahul16singh28 on 09 Sep 2018, 21:02.
Last edited by Bunuel on 09 Sep 2018, 21:12, edited 1 time in total.
Renamed the topic and edited the question.
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Re: What is the solution of 4√x - 4 = 3 * 2√x?  [#permalink]

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New post 10 Sep 2018, 00:18
rahul16singh28 wrote:
What is the solution of 4√x - 4 = 3 * 2√x?

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4√x - 4 = 3 * 2√x
\(4\sqrt{x}-6\sqrt{x}=4.................-2\sqrt{x}=4\)
square both sides..
\(4x=16...............x=4\)

4√x - 4 = 3 * 2√x...........- 4 = 2√x....
Only possible if x is perfect square

D
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Re: What is the solution of 4√x - 4 = 3 * 2√x?  [#permalink]

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New post 10 Sep 2018, 06:28
rahul16singh28 wrote:
What is the solution of 4√x - 4 = 3 * 2√x?

1. 1
2. 2
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4. 4
5. 5

\(4√x - 4 = 3 * 2√x\)

Or, \(4√x - 4 = 6√x\)

So, \(2√x = - 4\)

Or, \(4x = 16\) {Squaing both sides}

Or, \(x = 4\), Answer must be (D)
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What is the solution of 4√x - 4 = 3 * 2√x?  [#permalink]

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New post 11 Sep 2018, 16:49
Abhishek009 wrote:
rahul16singh28 wrote:
What is the solution of 4√x - 4 = 3 * 2√x?

1. 1
2. 2
3. 3
4. 4
5. 5

\(4√x - 4 = 3 * 2√x\)

Or, \(4√x - 4 = 6√x\)

So, \(2√x = - 4\)

Or, \(4x = 16\) {Squaing both sides}

Or, \(x = 4\), Answer must be (D)


Hi Abhishek009, thanks for the lucid explanation but I realised on plugging in 4 into the equation. The LHS does not equal the RHS. Why is this

Posted from my mobile device
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Re: What is the solution of 4√x - 4 = 3 * 2√x?  [#permalink]

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New post 11 Sep 2018, 18:27
1
Kem12 wrote:
Abhishek009 wrote:
rahul16singh28 wrote:
What is the solution of 4√x - 4 = 3 * 2√x?

1. 1
2. 2
3. 3
4. 4
5. 5

\(4√x - 4 = 3 * 2√x\)

Or, \(4√x - 4 = 6√x\)

So, \(2√x = - 4\)

Or, \(4x = 16\) {Squaing both sides}

Or, \(x = 4\), Answer must be (D)


Hi Abhishek009, thanks for the lucid explanation but I realised on plugging in 4 into the equation. The LHS does not equal the RHS. Why is this

Posted from my mobile device


Hi Kem12,
This question is flawed.

Given, \(4√x - 4 = 3 * 2√x\)

Or, \(4√x - 4 = 6√x\)

So, \(2√x = - 4\)
Divide both sides by -2, we have
\(\sqrt{x}=-2\)

Notice that \(\sqrt{f(x)}\) can't be negative for all \(x\in \mathbb{R}\).

Therefore, NO SOLUTION.
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Re: What is the solution of 4√x - 4 = 3 * 2√x?  [#permalink]

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New post 11 Sep 2018, 20:02
PKN wrote:
Kem12 wrote:
Abhishek009 wrote:
What is the solution of 4√x - 4 = 3 * 2√x?

1. 1
2. 2
3. 3
4. 4
5. 5

\(4√x - 4 = 3 * 2√x\)

Or, \(4√x - 4 = 6√x\)

So, \(2√x = - 4\)

Or, \(4x = 16\) {Squaing both sides}

Or, \(x = 4\), Answer must be (D)


Hi Abhishek009, thanks for the lucid explanation but I realised on plugging in 4 into the equation. The LHS does not equal the RHS. Why is this

Posted from my mobile device


Hi Kem12,
This question is flawed.

Given, \(4√x - 4 = 3 * 2√x\)

Or, \(4√x - 4 = 6√x\)

So, \(2√x = - 4\)
Divide both sides by -2, we have
\(\sqrt{x}=-2\)

Notice that \(\sqrt{f(x)}\) can't be negative for all \(x\in \mathbb{R}\).

Therefore, NO SOLUTION.


You are right. This is a poor quality question.

Topic is locked.
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Re: What is the solution of 4√x - 4 = 3 * 2√x? &nbs [#permalink] 11 Sep 2018, 20:02
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