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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Nov 2007 21:51:15 IST
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Real roots of the equation x2 + 5|x| + 4=0 are (a) -1,-4 (b) 1,4 (c) -4,4 (d) None of these Ans- d Please provide solution. Rates assured
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28 Nov 2007 21:57:05 IST
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this eqn is |x|^2+5|x|+4=0
hence |x|1 + |x|2=-5 which is not possible for real nos
so no real root
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28 Nov 2007 21:57:34 IST
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....(mod is used to make a number +ve and square of a number is always +ve)
so
this equation can also be written as
mod(x^2)+5mod(x)+4=0
solve and u get:
[mod(x)+4][mod(x)+1]
so
no real roots
nudge if hav a doubt
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28 Nov 2007 22:02:09 IST
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@ netkid
u got (|x|+4)(|x|+1)=0
this gives |x|=-4 or |x|=-1
now, how can the modulus of a real no. ever be negative
there r no real solutions
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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28 Nov 2007 22:08:39 IST
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x^2 + 5|x| + 4 =0
x>=0
|x|=x
x^2 + 5x +4 =0
(x+4)(x+1) =0
x=-4,-1 not possible because we took x>=0
if x<0
|x|=-x
x^2 -5x + 4 =0
(x-4)(x-1)=0
x=1,4 , not possible because we took x<0
so no real roots
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