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If x is an integer, what is the value of x?
(1) x^2 - 4x + 3 < 0
(2) x^2 + 4x +3 > 0
(1) x^2 - 4x + 3 < 0
(2) x^2 + 4x +3 > 0
30 May 2010, 09:31
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Hussain15
Bunuel
dimitri92
If x is an integer, what is the value of x?
(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0
(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0
Given: \(x=integer\).
(1) \(x^2-4x+3<0\) --> \(1<x<3\)--> as \(x\) is an integer then \(x=2\). Sufficient.
(2) \(x^2+4x+3>0\) --> \(x<-3\) or \(x>-1\) --> multiple values are possible for integer \(x\). Not sufficient.
Answer: A.
With reference to your statement \(1<x<3\) above, I calculated the range as \(x<1\) or \(x<3\)
What I did wrong?
How to solve quadratic inequalities - Graphic approach.
\(x^2-4x+3<0\) is the graph of parabola and it look likes this:
Attachment:
en.plot (1).png [ 3.79 KiB | Viewed 166319 times ]
If the sign were ">": \(x^2-4x+3>0\). First find the roots (\(x_1=1\) and \(x_2=3\)). ">" sign means in which range of \(x\) the graph is above x-axis. Answer is \(x<1\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
This approach works for any quadratic inequality. For example: \(-x^2-x+12>0\), first rewrite this as \(x^2+x-12<0\) (so that the coefficient of x^2 to be positive. It's possible to solve without rewriting, but easier to master one specific pattern).
\(x^2+x-12<0\). Roots are \(x_1=-4\) and \(x_1=3\) --> below ("<") the x-axis is the range for \(-4<x<3\) (between the roots).
Again if it were \(x^2+x-12>0\), then the answer would be \(x<-4\) and \(x>3\) (to the left of the smaller root and to the right of the bigger root).
Hope it helps.
23 May 2010, 01:46
Kudos
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dimitri92
If x is an integer, what is the value of x?
(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0
(1) x2 - 4x + 3 < 0
(2) x2 + 4x +3 > 0
Given: \(x=integer\).
(1) \(x^2-4x+3<0\) --> \(1<x<3\) --> as \(x\) is an integer then \(x=2\). Sufficient.
(2) \(x^2+4x+3>0\) --> \(x<-3\) or \(x>-1\) --> multiple values are possible for integer \(x\). Not sufficient.
Answer: A.
