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What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?

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What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post Updated on: 25 Oct 2018, 23:27
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[Math Revolution GMAT math practice question]

What is the number of roots of the equation \(\frac{(x^3-1)}{(x^2+x+1)}=x\)?

\(A. 1\)
\(B. 2\)
\(C. 3\)
\(D. 4\)
\(E. no solution\)

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Originally posted by MathRevolution on 22 Oct 2018, 01:29.
Last edited by MathRevolution on 25 Oct 2018, 23:27, edited 1 time in total.
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post 22 Oct 2018, 02:24
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the number of roots of the equation \(\frac{(x^3-1)}{(x^2+x+1)}=x\)?

\(A. 0\)
\(B. 1\)
\(C. 2\)
\(D. 3\)
\(E. no solution\)


You know, \(a^3-b^3=(a-b)(a^2+ab+b^2)\)
Factorizing numerator of the given expression(LHS), we have
\(\frac{(x^3-1)}{(x^2+x+1)}=x\)
Or, \(\frac{(x-1)(x^2+x+1)}{(x^2+x+1)}=x\)
Or, x-1=x
Substracting x from both LHS & RHS,
-1=0
Therefore, No Solution.
Ans. (E)

Another approach:-
\(\frac{(x^3-1)}{(x^2+x+1)}=x\)
Or, \((x^3-1)=x(x^2+x+1)\)
Or, \(x^3-1=x^3+x^2+x\)
Or, \(x^3+x^2+x=x^3-1\)
Or, \(x^2+x+1=0\) is a quadratic equation with a=b=c=1
\(D=b^2-4ac=1^2-4*1*1=-3<0\); so no real roots.

Hence, No Solution exists.

Ans. (E)
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post 22 Oct 2018, 04:31
Ans is E as after solving we will get x2+x+1=0 and D is negative for this equation. Hence No real solution is possible.
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post 22 Oct 2018, 08:09
MathRevolution wrote:
[Math Revolution GMAT math practice question]

What is the number of roots of the equation \(\frac{(x^3-1)}{(x^2+x+1)}=x\)?

\(A. 0\)
\(B. 1\)
\(C. 2\)
\(D. 3\)
\(E. no solution\)


\(\frac{(x^3-1)}{(x^2+x+1)}=x\)

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

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

\(-1 = x^6-x^3+x\)

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

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

\(B^2 - 4 *C = 1^2 - 4 *1 *1 = -3\)

since the root is less than 0, there are no solutions

pushpitkc, interesting if I couldn't use a factoring method, like product of C equals the sum of B :?
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post 24 Oct 2018, 01:18
=>

The original condition \(\frac{(x^3-1)}{(x^2+x+1)}=x\) is equivalent to \(x – 1 = x\) as shown below:

\(\frac{(x^3-1)}{(x^2+x+1)}=x\)
\(=> \frac{(x-1) (x^2+x+1)}{(x^2+x+1)} = x\) by factoring
\(=> x – 1 = x\)

But \(x – 1 = x\) has no solution.

Therefore, the answer is E.
Answer: E
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post 24 Oct 2018, 01:24
MathRevolution wrote:
=>

The original condition \(\frac{(x^3-1)}{(x^2+x+1)}=x\) is equivalent to \(x – 1 = x\) as shown below:

\(\frac{(x^3-1)}{(x^2+x+1)}=x\)
\(=> \frac{(x-1) (x^2+x+1)}{(x^2+x+1)} = x\) by factoring
\(=> x – 1 = x\)

But \(x – 1 = x\) has no solution.

Therefore, the answer is E.
Answer: E



math revoultion
I completely agree that there are no solutions for the equation. Bbt Answer option A tells us the number of solutions is 0, which is the same as No solution for any equation. Only if an equation has 0 solutions, will it have no equations.
Where am I wrong here?
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?  [#permalink]

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New post 25 Oct 2018, 23:27
nitesh50 wrote:
MathRevolution wrote:
=>

The original condition \(\frac{(x^3-1)}{(x^2+x+1)}=x\) is equivalent to \(x – 1 = x\) as shown below:

\(\frac{(x^3-1)}{(x^2+x+1)}=x\)
\(=> \frac{(x-1) (x^2+x+1)}{(x^2+x+1)} = x\) by factoring
\(=> x – 1 = x\)

But \(x – 1 = x\) has no solution.

Therefore, the answer is E.
Answer: E



math revoultion
I completely agree that there are no solutions for the equation. Bbt Answer option A tells us the number of solutions is 0, which is the same as No solution for any equation. Only if an equation has 0 solutions, will it have no equations.
Where am I wrong here?


You are right.
0 should be eliminated from the choices.
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Re: What is the number of roots of the equation (x^3-1)/(x^2+x+1)=x?   [#permalink] 25 Oct 2018, 23:27
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