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

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

Similar question to practice: new-algebra-set-149349-60.html#p1200948
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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.

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

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
Answer is B

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 ?

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:
https://magoosh.com/gmat/2013/gmat-math- ... -radicals/

Mike
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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:
https://magoosh.com/gmat/2013/gmat-math- ... -radicals/

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.

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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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.

We can solve this question in a matter of seconds if we recognize that we aren't required to actually solve the given equation.
Instead, we can test each answer choice.

I. -1
Is x = -1 a solution to the equation?
Let's plug it in and find out: \(2(-1) - 2 = \sqrt{3(-1)^2+13}\)
STOP!
The left side of the equation evaluates to be -4
Since the square root of a number cannot be negative, we know that x = -1 is NOT a solution.

This means we can eliminate answer choices A, C, D and E.

By the process of elimination, the correct answer is B

Mathematically , both (-4)^2 and (4)^2 can give 16 . So (16)^1/2 could be -4, and -1 could also be a root. Could someone please help me understand that what am I missing here ?

Rather than squaring both sides to get the value of 'x, it is much easier to substitute the options.

A simple simplification: \((2x -2)^2 = 3x^2\) + 13

I: -1: 2x - 2 = 2(-1) - 2 = -4 = \((-4)^2\) = 16 => \(3x^2 + 13 = 3(-2)^2 + 13\) - more than 16 - NOT POSSIBLE.

II: 4: 2x - 2 = 2(4) - 2 = 6 = \((6)^2 = 36 => 3x^2 + 13 = 3(6)^2\) + 13 - more than 36 - NOT POSSIBLE.

III: 9: 2x - 2 = 2(9) - 2 = 16 = \((16)^2 = 256 => 3x^2 + 13 = 3(9)^2\) + 13 = 243 + 13 = 256 - POSSIBLE.

Answer B

[quote="mikemcgarry"]\(2x - 2 = \sqrt{3x^2+13}\)

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

Hi mikemcgarry,

Awesome question! Great solution as well. I have one silly question to ask. Since it's given that \(2x - 2\) is equal to \(\sqrt{3x^2+13}\) (which can't be negative), can we assume that \(-1\) cannot satisfy \(2x - 2\) because it makes the expression \(2x - 2\) negative?

And since all answers include I, I went for B. Appreciate your thoughts on this.

Thank you!

im9572

Hi

We know that sqrt(x^2)= |x|

by this logic. sqrt(4^2)=|4| which can equal -4.

Found this dichotomy as we are asked to consider |x| =sqrt(x^2) but also asked not to consider sqrt of any number as a negative value

Why can't under root 16 = -4
Afterall, (-4) (-4) = 16
So I and III should be the right answer. Can someone help me here please?

Ask yourself this:
What is the range of the function:
\(\sqrt{x}\)
It is only positive - on the real number line. Therefore, whatever value is outputted from a square root function must be positive.
It is true that -4*-4=16, but that does not mean that square root of 16 is -4.

Given: \(2x - 2 = \sqrt{3x^2+13}\)

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

I. -1
2x-2 = -4
\(\sqrt{3x^2+13}= \sqrt{16}=4\)
II. 4
2x-2 = 6
\(\sqrt{3x^2+13}= \sqrt{61}\)
III. 9
2x-2 = 16
\(\sqrt{3x^2+13}= \sqrt{243+13} = \sqrt{256} = 16\)

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

IMO B