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If x-5=x+5 , x=?

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If x-5=x+5 , x=? [#permalink]

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New post 12 Sep 2017, 02:11
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A
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If \(x-5=\sqrt{x}+\sqrt{5}\) , x=?

A. \(6+2\sqrt{5}\)
B. \(6-2\sqrt{5}\)
C. \(6+\sqrt{5}\)
D. \(6-\sqrt{5}\)
E. \(5+2\sqrt{6}\)
[Reveal] Spoiler: OA

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Re: If x-5=x+5 , x=? [#permalink]

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New post 12 Sep 2017, 03:44
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Given: \(x-5 = \sqrt{x} + \sqrt{5}\)

=> \((\sqrt{x})^2 - (\sqrt{5})^2 = \sqrt{x} + \sqrt{5}\)
=> \((\sqrt{x} + \sqrt{5})(\sqrt{x} - \sqrt{5}) = \sqrt{x} + \sqrt{5}\)
=> \((\sqrt{x} - \sqrt{5}) = (\sqrt{x} + \sqrt{5}) / (\sqrt{x} + \sqrt{5})\)
=>\((\sqrt{x} - \sqrt{5}) = 1\)
=> \(\sqrt{x} = \sqrt{5} + 1\)

Squaring both sides
=>\(x= 1+5 +2*1* \sqrt{5}\)
=>\(x= 6+2\sqrt{5}\)

Answer: A

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Re: If x-5=x+5 , x=? [#permalink]

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New post 14 Sep 2017, 00:55
=>
x-5=√x+√5
(√x+√5)(√x-√5)=√x+√5
√x-√5 = 1
√x = 1+√5
x = 1 + 2√5 + 5 = 6 + 2√5

Ans: A
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Re: If x-5=x+5 , x=? [#permalink]

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New post 16 Sep 2017, 23:52
MathRevolution wrote:
=>
x-5=√x+√5
(√x+√5)(√x-√5)=√x+√5
√x-√5 = 1
√x = 1+√5
x = 1 + 2√5 + 5 = 6 + 2√5

Ans: A


I tried to square both sides but it didn't work, I don't understand why, is it wrong to square both sides

x-5=√x+√5 ...by squaring both sides
(x-5)^2= (√x+√5)^2
(x-5)(x+5)=(x+5)+2√x√5
(x-5)= 1+2√x√5
x= 1+2√x√5+5
x=6+2√x√5

Can someone help me to find what I did wrong, thx
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Re: If x-5=x+5 , x=? [#permalink]

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New post 17 Sep 2017, 00:21
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NNDD wrote:
MathRevolution wrote:
=>
x-5=√x+√5
(√x+√5)(√x-√5)=√x+√5
√x-√5 = 1
√x = 1+√5
x = 1 + 2√5 + 5 = 6 + 2√5

Ans: A


I tried to square both sides but it didn't work, I don't understand why, is it wrong to square both sides

x-5=√x+√5 ...by squaring both sides
(x-5)^2= (√x+√5)^2
(x-5)(x+5)=(x+5)+2√x√5
(x-5)= 1+2√x√5
x= 1+2√x√5+5
x=6+2√x√5

Can someone help me to find what I did wrong, thx



x-5=√x+√5 ...by squaring both sides
(x-5)^2= (√x+√5)^2
(x-5)(x+5)=(x+5)+2√x√5 => Here you used wrong formula, =>\((a-b)^2 = a^2+b^2-2ab\) and \(a^2 -b^ 2 = (a-b)(a+b)\)
Here after squaring correct expansion is :
(x-5)^2= (√x+√5)^2
\(x^2+25-2*25*x\)= \((x+5)+2\sqrt{x}\sqrt{5}\)
\(x^2+25-50x = x+5+2\sqrt{x}\sqrt{5}\)

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Re: If x-5=x+5 , x=? [#permalink]

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New post 17 Sep 2017, 02:15
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Nikkb wrote:
NNDD wrote:
MathRevolution wrote:
=>
x-5=√x+√5
(√x+√5)(√x-√5)=√x+√5
√x-√5 = 1
√x = 1+√5
x = 1 + 2√5 + 5 = 6 + 2√5

Ans: A


I tried to square both sides but it didn't work, I don't understand why, is it wrong to square both sides

x-5=√x+√5 ...by squaring both sides
(x-5)^2= (√x+√5)^2
(x-5)(x+5)=(x+5)+2√x√5
(x-5)= 1+2√x√5
x= 1+2√x√5+5
x=6+2√x√5

Can someone help me to find what I did wrong, thx



x-5=√x+√5 ...by squaring both sides
(x-5)^2= (√x+√5)^2
(x-5)(x+5)=(x+5)+2√x√5 => Here you used wrong formula, =>\((a-b)^2 = a^2+b^2-2ab\) and \(a^2 -b^ 2 = (a-b)(a+b)\)
Here after squaring correct expansion is :
(x-5)^2= (√x+√5)^2
\(x^2+25-2*25*x\)= \((x+5)+2\sqrt{x}\sqrt{5}\)
\(x^2+25-50x = x+5+2\sqrt{x}\sqrt{5}\)


Thanks a lot for explaining this, this a serious misconception from me, appreciate it

I now realized that this can't be solved by squaring both sides, as it gave complicated calculations and more square roots...How can we decide which path we go ...squaring both sides versus simplifying one side as it was shown in the provided answer

Thanks,
NNDD
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Re: If x-5=x+5 , x=? [#permalink]

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New post 17 Sep 2017, 02:37
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NNDD wrote:

Thanks a lot for explaining this, this a serious misconception from me, appreciate it

I now realized that this can't be solved by squaring both sides, as it gave complicated calculations and more square roots...How can we decide which path we go ...squaring both sides versus simplifying one side as it was shown in the provided answer

Thanks,
NNDD



You will get used to such type of questions with practice.
Generally everyone initially go with squaring option. We are used to work with integers so we start with expanding in integer format. But after solving few such questions you can used to square roots and formulas related to them :)

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Re: If x-5=x+5 , x=? [#permalink]

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New post 18 Sep 2017, 05:47
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MathRevolution wrote:
If \(x-5=\sqrt{x}+\sqrt{5}\) , x=?

A. \(6+2\sqrt{5}\)
B. \(6-2\sqrt{5}\)
C. \(6+\sqrt{5}\)
D. \(6-\sqrt{5}\)
E. \(5+2\sqrt{6}\)


The key to solving this problem is recognizing that (x - 5) can be expressed as a difference of squares. Recall that an expression like (x^2 - 36), a difference of squares, can be factored as (x - 6)(x + 6). In a similar way, we can factor x - 5 as (√x + √5)(√x - √5). Thus, we have:

(√x + √5)(√x - √5) = √x + √5

Dividing both sides by (√x + √5), since it can’t be 0, we have:

√x - √5 = 1

√x = 1 + √5

x = (1 + √5)^2

x = 1 + 2√5 + 5

x = 6 + 2√5

Answer: A
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Re: If x-5=x+5 , x=?   [#permalink] 18 Sep 2017, 05:47
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