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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # x=?  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6504 GMAT 1: 760 Q51 V42 GPA: 3.82 x=? [#permalink] ### Show Tags 12 Sep 2018, 01:40 00:00 Difficulty: 65% (hard) Question Stats: 44% (01:15) correct 56% (01:25) wrong based on 54 sessions ### HideShow timer Statistics [Math Revolution GMAT math practice question] $$x=?$$ $$1) x^3+x^2+x=0$$ $$2) x=-2x$$ _________________ 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$99 for 3 month Online Course"
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12 Sep 2018, 02:47
MathRevolution wrote:
[Math Revolution GMAT math practice question]

$$x=?$$

$$1) x^3+x^2+x=0$$
$$2) x=-2x$$

Statement 1: $$x^3+x^2+x=0$$
$$x(x^2+x+1)=0$$
Now the lease value of$$x^2+x+1$$ is $$+\frac{3}{4}$$
therefore x must be equal to 0
Sufficient.

Statement 2: x=-2x
3x=0
x=0
Sufficient.

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12 Sep 2018, 05:53
MathRevolution wrote:
[Math Revolution GMAT math practice question]

$$x=?$$

$$1) x^3+x^2+x=0$$
$$2) x=-2x$$

$$? = x$$

$$\left( 1 \right)\,\,\,x\left( {{x^2} + x + 1} \right) = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered} x = 0 \hfill \\ \,\,\,{\text{OR}} \hfill \\ {x^2} + x + 1 = 0 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\, \Rightarrow }\limits^{\left( * \right)} \,\,\,\,\,x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

$$\left( * \right)\,\,\,\,{x^2} + x + 1 = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\Delta = {\left( 1 \right)^2} - 4 \cdot 1 \cdot 1 < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^2} + x + 1 > 0\,\,\,\,{\text{for}}\,\,{\text{all}}\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,$$

$$\left( 2 \right)\,\,\,x = - 2x\,\,\,\,\, \Rightarrow \,\,\,\,3x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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12 Sep 2018, 08:10
IMO D.

1) This gives us x(x^2 + x + 1) = 0, so either x=0 or x^2 + x + 1. But there is no value of x that fits the equation as x^2 can either be +ve or 0, therefore the only value possible for x=0. Sufficient.

2) The only value that fits the equation is x=0. Sufficient.
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12 Sep 2018, 16:58
fskilnik wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

$$x=?$$

$$1) x^3+x^2+x=0$$
$$2) x=-2x$$

$$? = x$$

$$\left( 1 \right)\,\,\,x\left( {{x^2} + x + 1} \right) = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered} x = 0 \hfill \\ \,\,\,{\text{OR}} \hfill \\ {x^2} + x + 1 = 0 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\, \Rightarrow }\limits^{\left( * \right)} \,\,\,\,\,x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

$$\left( * \right)\,\,\,\,{x^2} + x + 1 = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\Delta = {\left( 1 \right)^2} - 4 \cdot 1 \cdot 1 < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^2} + x + 1 > 0\,\,\,\,{\text{for}}\,\,{\text{all}}\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,$$

$$\left( 2 \right)\,\,\,x = - 2x\,\,\,\,\, \Rightarrow \,\,\,\,3x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Sir how did you reach the second statement?

If the discriminant is negative then the equation involves complex numbers but how to arrive at the equation
x^3+x^2+x>1.
Can you please explain this concept with diagram.

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12 Sep 2018, 17:53
1
arvind910619 wrote:
fskilnik wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

$$x=?$$

$$1) x^3+x^2+x=0$$
$$2) x=-2x$$

$$? = x$$

$$\left( 1 \right)\,\,\,x\left( {{x^2} + x + 1} \right) = 0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \begin{gathered} x = 0 \hfill \\ \,\,\,{\text{OR}} \hfill \\ {x^2} + x + 1 = 0 \hfill \\ \end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\, \Rightarrow }\limits^{\left( * \right)} \,\,\,\,\,x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

$$\left( * \right)\,\,\,\,{x^2} + x + 1 = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\Delta = {\left( 1 \right)^2} - 4 \cdot 1 \cdot 1 < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^2} + x + 1 > 0\,\,\,\,{\text{for}}\,\,{\text{all}}\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,$$

$$\left( 2 \right)\,\,\,x = - 2x\,\,\,\,\, \Rightarrow \,\,\,\,3x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Sir how did you reach the second statement?

If the discriminant is negative then the equation involves complex numbers but how to arrive at the equation
x^3+x^2+x>1.
Can you please explain this concept with diagram.

Hi arvind910619!

Thank you for your interest in my solution.

01. The expression $${x^3} + {x^2} + x$$ may be written in the form $$x\left( {{x^2} + x + 1} \right)$$ and this last "factorized version" is better for our purposes, because a product is zero if and only if (at least) one of the factors is zero, therefore: $$x\left( {{x^2} + x + 1} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x = 0\,\,\,\,{\text{or}}\,\,\,{x^2} + x + 1 = 0$$

02. The equation $${x^2} + x + 1 = 0$$ has no real solutions, because its discriminant ("delta") is negative. Therefore the curve (parabola) does not intercept the horizontal axis. From the fact that the parabola is concave upward (see image attached), we are sure $${x^2} + x + 1$$ is always positive, by the way.

03. Finally, from the fact that $$x\left( {{x^2} + x + 1} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x = 0\,\,\,\,{\text{or}}\,\,\,{x^2} + x + 1 = 0$$ and also $${x^2} + x + 1 \ne 0$$ for all real values of $$x$$, we are sure we must have $$x = 0$$ , as mentioned.

I hope you got all details!

Regards and success in your studies,
Fabio.

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13 Sep 2018, 23:37
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
$$x^3+x^2+x=0$$
$$=> x(x^2+x+1)=0$$
$$=> x = 0$$ since $$x^2+x+1 ≠ 0$$
Condition 1) is sufficient.

Condition 2)
$$x = -2x$$
$$=> 3x = 0$$
$$=> x = 0$$
Condition 2) is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: x=? &nbs [#permalink] 13 Sep 2018, 23:37
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