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Algebra
20 Feb 2008 23:57:29 IST
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aNdRoMeDa
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21 Feb 2008 00:04:37 IST
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is d answer dat d roots r real??
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21 Feb 2008 01:13:12 IST
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hope its multiple ans quesn....else d soln doesnt turn out
expand d eqn nd u get
3X^2 -2(a+b+c)X +(ab+bc+ca)=0
discrim. D=4(b+c+a)^2 - 12(ab+bc+ca)
now, if i subs b=c=a=1, i get D=0 nd real roots
if i substitute b=0,a=1, c=1, i get D>0 => roots r real nd +ve
if i subs a=1,b=0,c=-1, i again get D>0 =>real nd +ve roots
if i subs a=c=1, b=-1, i get D<0 => complex roots
expand d eqn nd u get
3X^2 -2(a+b+c)X +(ab+bc+ca)=0
discrim. D=4(b+c+a)^2 - 12(ab+bc+ca)
now, if i subs b=c=a=1, i get D=0 nd real roots
if i substitute b=0,a=1, c=1, i get D>0 => roots r real nd +ve
if i subs a=1,b=0,c=-1, i again get D>0 =>real nd +ve roots
if i subs a=c=1, b=-1, i get D<0 => complex roots
21 Feb 2008 01:29:06 IST
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I will do it the fundamental way.
Put X=a
=> f(a) = (a-b)(a-c)
Put X=b
=>f(b) = (b-c)(b-a)
Put X=c
=>f(c) = (c-a)(c-b)
Now there are 6 cases:
1) a>c , b>c , a>b
2) a>c , b>c , b>a
3) a>c , b<c
4) a<c , b>c
5) a<c , b<c , a>b
6) a<c , b<c , b>a
Using the above knowledge and taking cases we find that always f(x).f(y)<0, which implies there is a root between x and y, hence roots are real.
Put X=a
=> f(a) = (a-b)(a-c)
Put X=b
=>f(b) = (b-c)(b-a)
Put X=c
=>f(c) = (c-a)(c-b)
Now there are 6 cases:
1) a>c , b>c , a>b
2) a>c , b>c , b>a
3) a>c , b<c
4) a<c , b>c
5) a<c , b<c , a>b
6) a<c , b<c , b>a
Using the above knowledge and taking cases we find that always f(x).f(y)<0, which implies there is a root between x and y, hence roots are real.
21 Feb 2008 11:05:47 IST
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if (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
3x2-2x(a+b+c)+(ab+bc+ca)=0
this is quardadic equation so we will find D
4(a+b+c)2-12(ab+bc+ca)=4[(a+b+c)2-3(ab+bc+ca)]=4{a2+b2+c2-ab+bc+ca}
which is equal to
4/2{ (a-b)2+(b-c)2+(c-a)2} which is greater than 0
so the roots of this equation are real.
21 Feb 2008 13:24:35 IST
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here
just multiply all first then it becomes
3x2 -2(a+b+c)x+ab+ac+bc =0
discriminant=4(a+b+c)2-4.3(ab+ac+bc )
=4(a2+b2+c2-ab-ac-bc)
=2[(a-b)2+(b-c)2+(c-a)2 which is either equal to 0 or greater than 0
so its nature of roots are real .
so ans is A
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