If a, b, c are three positive real numbers such that the ratio a : b : c is (q + r – p) : (r + p – q) : (p + q – r) where p, q and r are any constants, then find the value of (q – r)a + (r – p)b + (p – q)c.
@shattereddream said:If a, b, c are three positive real numbers such that the ratio a : b : c is (q + r – p) : (r + p – q) : (p + q – r) where p, q and r are any constants, then find the value of (q – r)a + (r – p)b + (p – q)c.
0?
a=b=c=1
and p=q=r=1
The equations x^ 2 + ax + b = 0 and x^ 2 + bx + a = 0, where a ‰ b, have a common root. Then the equation whose roots are the other two roots of the given equations is
@shattereddream said:If a, b, c are three positive real numbers such that the ratio a : b : c is (q + r – p) : (r + p – q) : (p + q – r) where p, q and r are any constants, then find the value of (q – r)a + (r – p)b + (p – q)c.
zero.
@shattereddream said:The equations x^ 2 + ax + b = 0 and x^ 2 + bx + a = 0, where a ‰ b, have a common root. Then the equation whose roots are the other two roots of the given equations is
-a = k + l
b = kl
-b = k + m
a = km
b = kl
-b = k + m
a = km
=> k + l = -km
and k+m = -kl
=> l - m = k(l - m)
=> k =1
Roots of new equation = m and l
x^2 - (-1)x + ab = 0 ?
x^2 + x + ab = 0..
@INDRESH90 said:What is the highest possible value of 'n' for which 3^1024 – 1 is divisible by 2n?
(2+1)^1024-1
1024c0 2^1024*1+....................+1024C1023 2^1*1+1024C1024 2^0*1-1
this expression shall be divisible by 2048 i.e 2^11
1024c0 2^1024*1+....................+1024C1023 2^1*1+1024C1024 2^0*1-1
this expression shall be divisible by 2048 i.e 2^11
@shattereddream said:If a, b, c are three positive real numbers such that the ratio a : b : c is (q + r – p) : (r + p – q) : (p + q – r) where p, q and r are any constants, then find the value of (q – r)a + (r – p)b + (p – q)c.
0 ?
a = b = c = 1
p = q = r = 1
@hexagon said:@naga25french sorry for that Find remainder when 33333333......86 times is divided by 13?i didnot understand how to apply ferm thm to the above question as explained by you
http://www.pagalguy.com/news/remainders-reloaded-euler-fermat-wilsons-theorems-cat-2011-a-19067
Hope this helps 
@jaspunit said:how to find number of points of intersection of the diagonals of a regular figure,we can find small ones easily like - square - 1hexagon - 15,any short cut?if we have to find - of regular octagon?
If it regular then is the centre of the circumcircle
@mailtoankit said:-1 kaise aaya ??.....i m getting x^2 - (a + b)x + ab = 0...
k + l = -km
And we got k = 1
Put value, 1 + l = -m
=> m + l = -1 = Sum of roots
@jaspunit said:If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?148157
15?
1/R + 1/K = 1/12
So
12/R + 12/ K = 1
So possible cases are
24, 24
28, 21
21, 28
30, 30
20, 30
36, 18
18, 36
48, 16
16, 48
60, 15
15, 60
84, 14
14, 84
156, 13
13, 156
regards
scrabbler
1/R + 1/K = 1/12
So
12/R + 12/ K = 1
So possible cases are
24, 24
28, 21
21, 28
30, 30
20, 30
36, 18
18, 36
48, 16
16, 48
60, 15
15, 60
84, 14
14, 84
156, 13
13, 156
regards
scrabbler
@ScareCrow28 said:k + l = -kmAnd we got k = 1Put value, 1 + l = -m=> m + l = -1 = Sum of roots
got it bhai....pehele dhang se pada nahi..
.....thanks..
.....thanks..