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# Which value(s) of x satisfies the equation above?

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Magoosh GMAT Instructor
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Which value(s) of x satisfies the equation above?  [#permalink]

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09 May 2013, 12:20
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55% (hard)

Question Stats:

57% (01:25) correct 43% (01:05) wrong based on 357 sessions

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$$2x - 2 = \sqrt{3x^2+13}$$

Which value(s) of x satisfies the equation above?

I. -1
II. 4
III. 9

(A) I
(B) III
(C) I & II
(D) I & III
(E) I, II, & III

For a discussion of algebraic equations involving radicals, as well as a solution to this question, see this post:

Mike

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Re: Which value(s) of x satisfies the equation above?  [#permalink]

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09 May 2013, 12:27
3
Using numbers:

$$2x - 2 = \sqrt{3x^2+13}$$

I)-1

$$-4= \sqrt{3(-1)^2+13}$$

$$\sqrt{3+13}$$ does not equal $$-4$$ , so $$-1$$ is NOT a possible value.

If we take a look at the possible answer, all contain I except B. So B is the correct answer
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Re: Which value(s) of x satisfies the equation above?  [#permalink]

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09 May 2013, 13:05
1
mikemcgarry wrote:
$$2x - 2 = \sqrt{3x^2+13}$$

Which value(s) of x satisfies the equation above?

I. -1
II. 4
III. 9

(A) I
(B) III
(C) I & II
(D) I & III
(E) I, II, & III

For a discussion of algebraic equations involving radicals, as well as a solution to this question, see this post:

Mike

Similar question to practice: new-algebra-set-149349-60.html#p1200948
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Re: Which value(s) of x satisfies the equation above?  [#permalink]

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10 May 2013, 02:55
1
I solved the equation to get the wrong answer. Squaring both sides made the difference.

Leason Learnt: Such type of questions, better to place the given values & check.
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Which value(s) of x satisfies the equation above?  [#permalink]

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02 Nov 2014, 23:35
1
mikemcgarry wrote:
$$2x - 2 = \sqrt{3x^2+13}$$

Which value(s) of x satisfies the equation above?

I. -1
II. 4
III. 9

(A) I
(B) III
(C) I & II
(D) I & III
(E) I, II, & III

For a discussion of algebraic equations involving radicals, as well as a solution to this question, see this post:

Mike

$$2x-2 = \sqrt{3x^2 + 13}$$

Squaring both sides, we get

$$4x^2 - 4x + 3 = 3x^2 + 13$$

$$x^2 - 4x = 9$$

$$x (x-4) = 9$$

x = 9 or x-4 = 9

x = 9 or x = 13

Since only X = 9 is given and it satisfies the condition
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Re: Which value(s) of x satisfies the equation above?  [#permalink]

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03 Nov 2014, 11:20
2
Ashishmathew01081987 wrote:
mikemcgarry wrote:
$$2x - 2 = \sqrt{3x^2+13}$$

Which value(s) of x satisfies the equation above?

I. -1
II. 4
III. 9

(A) I
(B) III
(C) I & II
(D) I & III
(E) I, II, & III

For a discussion of algebraic equations involving radicals, as well as a solution to this question, see this post:

Mike

$$2x-2 = \sqrt{3x^2 + 13}$$

Squaring both sides, we get

$$4x^2 - 4x + 3 = 3x^2 + 13$$

$$x^2 - 4x = 9$$

$$x (x-4) = 9$$

x = 9 or x-4 = 9

x = 9 or x = 13

Since only X = 9 is given and it satisfies the condition

Hey Ashishmathew,
(2x-2)^2 = 4x^2 +4-8x right ? How come ur LHS of the eqn is [m][b]4x^2 - 4x + 3 ? Or Am I missing out something here ?
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4667
Which value(s) of x satisfies the equation above?  [#permalink]

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03 Nov 2014, 12:00
3
sreelakshmigs wrote:
Ashishmathew01081987 wrote:
mikemcgarry wrote:
$$2x - 2 = \sqrt{3x^2+13}$$

Which value(s) of x satisfies the equation above?

I. -1
II. 4
III. 9

(A) I
(B) III
(C) I & II
(D) I & III
(E) I, II, & III

For a discussion of algebraic equations involving radicals, as well as a solution to this question, see this post:

Mike

$$2x-2 = \sqrt{3x^2 + 13}$$

Squaring both sides, we get

$$4x^2 - 4x + 3 = 3x^2 + 13$$

$$x^2 - 4x = 9$$

$$x (x-4) = 9$$

x = 9 or x-4 = 9

x = 9 or x = 13

Since only X = 9 is given and it satisfies the condition

Hey Ashishmathew,
(2x-2)^2 = 4x^2 +4-8x right ? How come ur LHS of the eqn is $$4x^2 - 4x + 3$$ ? Or Am I missing out something here ?

Dear sreelakshmigs
I'm happy to respond.

First of all, Ashishmathew01081987's solution is not correct at all. You are perfectly correct:
$$(2x-2)^2 = 4x^2 - 8x + 4$$
Also, the factoring thing he does at the end, the steps after x(x - 4) = 9, are 100% incorrect.
If you want to see the correct solution to this problem, see:

Mike
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Re: Which value(s) of x satisfies the equation above?  [#permalink]

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03 Nov 2014, 22:48
Hey Ashishmathew,
(2x-2)^2 = 4x^2 +4-8x right ? How come ur LHS of the eqn is $$4x^2 - 4x + 3$$ ? Or Am I missing out something here ?[/quote]
Dear sreelakshmigs
I'm happy to respond.

First of all, Ashishmathew01081987's solution is not correct at all. You are perfectly correct:
$$(2x-2)^2 = 4x^2 - 8x + 4$$
Also, the factoring thing he does at the end, the steps after x(x - 4) = 9, are 100% incorrect.
If you want to see the correct solution to this problem, see:

Mike [/quote]

Thanks Sree for pointing out my mistake. That was a blunder. Hope that it doesn't happen on the GMAT.

Thanks Mike, I understand why the factoring stuff would have led me into the trap. Plugging numbers is the best option here.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4667
Re: Which value(s) of x satisfies the equation above?  [#permalink]

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03 Nov 2014, 23:34
Ashishmathew01081987 wrote:
Thanks Sree for pointing out my mistake. That was a blunder. Hope that it doesn't happen on the GMAT.

Thanks Mike, I understand why the factoring stuff would have led me into the trap. Plugging numbers is the best option here.

Dear Ashishmathew01081987,
Actually, it's very good to understand the algebra in this problem. Again, you can see a full algebraic solution at the blog to which I linked. Plugging numbers is good sometimes, but it's best not to make that a one-size-fit-all kind of strategy.
Does this make sense?
Mike
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Which value(s) of x satisfies the equation above?  [#permalink]

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12 Nov 2017, 11:24
Zarrolou wrote:
Using numbers:

$$2x - 2 = \sqrt{3x^2+13}$$

I)-1

$$-4= \sqrt{3(-1)^2+13}$$

$$\sqrt{3+13}$$ does not equal $$-4$$ , so $$-1$$ is NOT a possible value.

If we take a look at the possible answer, all contain I except B. So B is the correct answer

I think this might be incorrect.

First, -4 = 16^(1/2) is as correct answer since (-4)(-4) = 16.
Second, the bold part is not correct. x² when x = -1 is the same as -1² = -1 and not (-1)² = 1, so we should have -4 = 10^(1/2). So, option I is incorrect.
Since the only option that says I is incorrect is B, then, B is our answer.
Which value(s) of x satisfies the equation above? &nbs [#permalink] 12 Nov 2017, 11:24
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