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18 May 2011 20:04:36 IST
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no. of distinct real solutions of the equation x(x^2-1)(x+2)+1=0
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no. of distinct real solutions of the equation x(x^2-1)(x+2)+1=0


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NugoRama's Avatar

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Joined: 11 Mar 2009
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18 May 2011 22:45:12 IST
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2.

Scorching goIITian

Joined: 2 Mar 2011
Posts: 216
19 May 2011 14:19:45 IST
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is it 2


New kid on the Block

Joined: 18 May 2011
Posts: 18
19 May 2011 17:35:46 IST
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How you guys got the answer?

 

Inferno's Avatar

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Joined: 19 Apr 2010
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19 May 2011 18:23:45 IST
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x(x^2-1)(x+2)+1=0

x(x+1)(x-1)(x+2)+1=0

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

Put (x^2+x)=u

u(u-2)+1=0

u^2-2u+1=0

u=1

x^2+x=1

x^2+x-1=0

x=[(-1)+-(5^(1/2))]/2

Hence 2 real values.

Are you from Brilliant Tutorials.


New kid on the Block

Joined: 18 May 2011
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19 May 2011 18:30:24 IST
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Yes inferno i am from brilliant tutorials i also thought that answer is 2 i just wnated to check thank you :)

 

and can you please tell me your name

Inferno's Avatar

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Joined: 19 Apr 2010
Posts: 359
20 May 2011 06:25:05 IST
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I am also from Brilliant Tutorials. If you have more doubts then after posting them do nudge me once.
And one more big thing that you did not know. The solutions,SOLUTIONS,SSOOLLUUTTIIOONNSS to the WNICE are online. Also I did not copy my answer from there. It's my own solution. There also the same procedure is followed. So, don't doubt me.
What will you do with my name. We don't know each other.
hemang's Avatar

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Joined: 27 Dec 2010
Posts: 1474
20 May 2011 09:43:10 IST
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two is the answer.



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