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x=?  [#permalink]

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New post 12 Sep 2018, 02:40
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[Math Revolution GMAT math practice question]

\(x=?\)

\(1) x^3+x^2+x=0\)
\(2) x=-2x\)

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Re: x=?  [#permalink]

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New post 12 Sep 2018, 18:53
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.

Thanks in advance .


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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Re: x=?  [#permalink]

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New post 12 Sep 2018, 03:47
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.

Answer: D
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Re: x=?  [#permalink]

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New post 12 Sep 2018, 06:53
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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Re: x=?  [#permalink]

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New post 12 Sep 2018, 09:10
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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Re: x=?  [#permalink]

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New post 12 Sep 2018, 17:58
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.

Thanks in advance .
Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
Posts: 8228
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: x=?  [#permalink]

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New post 14 Sep 2018, 00:37
=>

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.

Therefore, D is the answer.
Answer: D

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=?  [#permalink]

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New post 26 Mar 2019, 03:58
This Question should mention that x is real, then only Statement 1 is sufficient.

Otherwise x = 0 or complex number. It indicates that Statement 1 is not sufficient.
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Re: x=?   [#permalink] 26 Mar 2019, 03:58
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