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08 Nov 2009, 14:02
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If the equation x^3 - ax^2 +bx -a = 0 has three real roots then the following id true
A. a=11
B. a not equal to 1
C. b = 1
D. b not equal to 1
How to solve equations with power 3?
Any specific tips to handle questions with such equations?
A. a=11
B. a not equal to 1
C. b = 1
D. b not equal to 1
How to solve equations with power 3?
Any specific tips to handle questions with such equations?
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08 Nov 2009, 16:05
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papillon86
If the equation x^3 - ax^2 +bx -a = 0 has three real roots then the following id true
a) a=11
b) a not equal to 1
c) b = 1
d) b not equal to 1
How to solve equations with power 3?
Any specific tips to handle questions with such equations?
a) a=11
b) a not equal to 1
c) b = 1
d) b not equal to 1
How to solve equations with power 3?
Any specific tips to handle questions with such equations?
I don't think we need to know how to solve cubic equation for GMAT.
For above question:
\(x^3 - ax^2 +bx -a = 0\)
\(x^2*(x-a)+(bx-a)=0\)
We are told that this equation has THREE roots. If look closer we notice that when \(b=1\) we can write the equation as:
\(x^2*(x-a)+(x-a)=0\) --> \((x^2+1)*(x-a)=0\) and this equation has only ONE root \(x=a\).
Hence given equation to have three roots b must not equal to 1.
Answer: D.
11 May 2011, 23:23
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assuming b= 1
x^2(x-a)+ 1(x-a) = (x^2+1) * (x-a)
x^2 != -1 meaning b != 1.
Hence D.
x^2(x-a)+ 1(x-a) = (x^2+1) * (x-a)
x^2 != -1 meaning b != 1.
Hence D.
