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Math question: Quite easy (maybe at least for you ;))

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Intern
Joined: 06 Jan 2013
Posts: 8

Kudos [?]: 1 [1], given: 7

Math question: Quite easy (maybe at least for you ;)) [#permalink]

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06 Jan 2013, 04:42
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Hey Guys,

as I have found this forum today I would like to share my thoughts on the GMAT with you. I´ve began preparing for the test about a week ago and right now I´m still fighting easy problems.

I got a qestion that needs two equations two be created. I got the first equation, which is:

\sqrt{3-2x} = \sqrt{2x}+1

When I square both sides, I do get

3-2x = ?

My problem is that I don´t know any rule that leads to the correct term on the right side, which Is (as indicated in the explanations)

3-2x = 2x+2*\sqrt{2x}+1

Anyone any idea how this works?

Maybe some of you can help me with this.

Max

Kudos [?]: 1 [1], given: 7

Current Student
Joined: 02 Jan 2013
Posts: 57

Kudos [?]: 63 [0], given: 2

GMAT 1: 750 Q51 V40
GPA: 3.2
WE: Consulting (Consulting)
Re: Math question: Quite easy (maybe at least for you ;)) [#permalink]

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06 Jan 2013, 06:02
Hi there, Try squaring the equation one more time:

$$\sqrt{3-2x} = \sqrt{2x} -1$$

Squaring both sides:
$$3-2x = 2x - 2.\sqrt{2x} +1$$
=> $$\sqrt{2x} =2x-1$$
Squaring both sides once again:

$$2x = 4 x^{2} -4x + 1$$
=> $$4 x^{2} -6x +1 = 0$$
=> $$x =\frac{3+- \sqrt{5}}{4}$$

Now for the tricky part. Since you squared a couple of times the original equation you certainly changed the nature of the equation (meaning, you may have inserted extra solutions that aren't in the original equation). So, you have to test your solutions in the original equation to see which ones still aply.

For the solution $$k = \frac{3- \sqrt{5}}{4}$$ you get that $$\sqrt{2k}-1$$is negative. Which is no good, since the left side of the original equation has to be always positive. The only solution that fits is

$$x = \frac{3+\sqrt{5}}{4}$$
Cheers!

Kudos [?]: 63 [0], given: 2

Manager
Joined: 13 Dec 2012
Posts: 194

Kudos [?]: 110 [0], given: 29

Re: Math question: Quite easy (maybe at least for you ;)) [#permalink]

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06 Jan 2013, 06:11
maxkay47 wrote:
Hey Guys,

as I have found this forum today I would like to share my thoughts on the GMAT with you. I´ve began preparing for the test about a week ago and right now I´m still fighting easy problems.

I got a qestion that needs two equations two be created. I got the first equation, which is:

\sqrt{3-2x} = \sqrt{2x}+1

When I square both sides, I do get

3-2x = ?

My problem is that I don´t know any rule that leads to the correct term on the right side, which Is (as indicated in the explanations)

3-2x = 2x+2*\sqrt{2x}+1

Anyone any idea how this works?

Maybe some of you can help me with this.

Max

3 - 2x =( \sqrt{2x} + 1) * ( \sqrt{2x} + 1)

= 2x + \sqrt{2x} + \sqrt{2x} + 1

= 2x + 2 \sqrt{2x} + 1
_________________

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Kudos [?]: 110 [0], given: 29

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Joined: 16 Oct 2010
Posts: 7736

Kudos [?]: 17766 [0], given: 235

Location: Pune, India
Re: Math question: Quite easy (maybe at least for you ;)) [#permalink]

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08 Jan 2013, 21:27
maxkay47 wrote:
Hey Guys,

as I have found this forum today I would like to share my thoughts on the GMAT with you. I´ve began preparing for the test about a week ago and right now I´m still fighting easy problems.

I got a qestion that needs two equations two be created. I got the first equation, which is:

\sqrt{3-2x} = \sqrt{2x}+1

When I square both sides, I do get

3-2x = ?

My problem is that I don´t know any rule that leads to the correct term on the right side, which Is (as indicated in the explanations)

3-2x = 2x+2*\sqrt{2x}+1

Anyone any idea how this works?

Maybe some of you can help me with this.

Max

How do you square a sum of two terms?

$$(a + b)^2 = a^2 + 2*a*b + b^2$$ Use this. If you don't know this, check out an algebra book first. This is a basic algebraic identity and you need to know this and similar identities for GMAT.

$$(\sqrt{2x}+1)^2 = (\sqrt{2x})^2+2*\sqrt{2x}*1+1^2$$
_________________

Karishma
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Kudos [?]: 17766 [0], given: 235

Re: Math question: Quite easy (maybe at least for you ;))   [#permalink] 08 Jan 2013, 21:27
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