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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5562
GMAT 1: 800 Q59 V59
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If x-5=x+5 , x=? [#permalink]

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12 Sep 2017, 02:11
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Difficulty:

45% (medium)

Question Stats:

67% (01:40) correct 33% (02:29) wrong based on 83 sessions

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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. $$5+2\sqrt{6}$$

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 02 Jul 2017 Posts: 294 GMAT 1: 730 Q50 V38 Re: If x-5=x+5 , x=? [#permalink] ### Show Tags 12 Sep 2017, 03:44 2 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5562 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: If x-5=x+5 , x=? [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 3 month Online Course"
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Re: If x-5=x+5 , x=? [#permalink]

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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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Using Forum Tags: https://gmatclub.com/forum/using-forum-tags-158411.html#p1909410

Senior Manager
Joined: 02 Jul 2017
Posts: 294
GMAT 1: 730 Q50 V38
Re: If x-5=x+5 , x=? [#permalink]

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17 Sep 2017, 00:21
1
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
$$x^2+25-2*25*x$$= $$(x+5)+2\sqrt{x}\sqrt{5}$$
$$x^2+25-50x = x+5+2\sqrt{x}\sqrt{5}$$
Intern
Status: Enjoying the Journey
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Re: If x-5=x+5 , x=? [#permalink]

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17 Sep 2017, 02:15
1
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
_________________

High achievement always takes place in the framework of high expectation Charles Kettering
If we chase perfection we can catch excellence Vince Lombardi

Using Forum Tags: https://gmatclub.com/forum/using-forum-tags-158411.html#p1909410

Senior Manager
Joined: 02 Jul 2017
Posts: 294
GMAT 1: 730 Q50 V38
Re: If x-5=x+5 , x=? [#permalink]

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17 Sep 2017, 02:37
1
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
Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2544
Re: If x-5=x+5 , x=? [#permalink]

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18 Sep 2017, 05:47
2
1
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

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

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