@Subhashdec2 How ........ Please show me ur approach !!
@sa1m007 said:@Subhashdec2 How ........ Please show me ur approach !!
since roots are distinct and |x1|=|x2|
hence x1=-x2
or x1+x2=0
hence b/a =0
b=0
|b|
@sa1m007 said:Given a quadratic equation ax^2+bx+c=0, and X1, X2 are two distinct roots and |X1|=|X2|, then which of the following is always correct
a) |b||c| d) |a|
d ???
@sa1m007 said:Given a quadratic equation ax^2+bx+c=0, and X1, X2 are two distinct roots and |X1|=|X2|, then which of the following is always correcta) |b||c| d) |a|
x^2 - 1 = 0
x= +/-1
=> a=1, b=0 and c =1
so |c| > |b|
hence option a...
@Subhashdec2 said:since roots are distinct and |x1|=|x2|hence x1=-x2or x1+x2=0hence b/a =0b=0|b|d bhi hoga na....
Another Question came like this ........ 1024*2065 + 2564*2653 + 5216*5645 is in octal form ........ find number of digit 6 that comes after multiplication ?? .......... Can u plzz help !!
@sa1m007 said:Another Question came like this ........ 1024*2065 + 2564*2653 + 5216*5645 is in octal form ........ find number of digit 6 that comes after multiplication ?? .......... Can u plzz help !!
what does the bold part mean?
@Dexian said:d bhi hoga na....
c/a = -x1^2
now x1 can be greater than 1
which will give |c|>|a|
x1
|c|
x1=1
|c|=|a|
@Subhashdec2 It means find ........ the number or ..... how many times digit 6 comes in the multiplication !! ....... It came in Elitmus Exam . pLease help !!
log p (log a p)/log a (log p a)=
a)log a p b)-log p a c)log a p d)1
a)log a p b)-log p a c)log a p d)1
Q.2 - In how many ways can a selection be made of 5 letters out of 5As, 4Bs, 3Cs, 2Ds and 1E?a) 70 b)71 c) 15(C)5 d) 15(P)5 e) None of theseQ.3 - Find the no. of ways of selecting n articles out of 3n+1, out of which n are identical.a) 2^(2n-1)b) [ (3n+1)(C)n ] / n!c) [ (3n+1)(P)n ] / n!d) 2^(2n)e) None of these
@sa1m007 said:@Subhashdec2 It means find ........ the number or ..... how many times digit 6 comes in the multiplication !! ....... It came in Elitmus Exam . pLease help !!
can hardly think of anything other than multiplication.............
( may be leave it... IMO)
@jain4444 said:How many pairs of integers (not necessarily positive) are there such that both a^2 + 6b^2 and b^2 + 6a^2 are both squares ?
There exists only one solution and the whole challenge is in proving the fact that there does not exist any other solution. So here we go...
Let's add both these expressions. Then as per the problem,
7(a^2+b^2)=c^2+d^2
Clearly LHS is a multiple of 7 and so RHS must be a multiple of 7 in order for the solution to exist. Now, let's investigate the RHS.
Both c and d can be expressed as 7k, 7k+1, 7k+2,7k+3,7k+4, 7k+5,7k+6.
So, remainder when (c^2+d^2) is divided by 7 can be the same if one among the following is divided by7:
0, (1+2^2), (1+3^2), (1+4^2), (1+5^2), (1+6^2)
(2^2+3^2), (2^2+4^2), (2^2+5^2), (2^2+6^2)
(3^2+4^2), (3^2+5^2), (3^2+6^2)
(4^2+5^2), (4^2+6^2)
(5^2+6^2)
None of the above except 0 is divisible by 7.
Hence there exists only one solution for (a,b) -> (0,0)
Team BV - Vineet
Team BV - Vineet
@ani6 said:Q.2 - In how many ways can a selection be made of 5 letters out of 5As, 4Bs, 3Cs, 2Ds and 1E?a) 70 b)71 c) 15(C)5 d) 15(P)5 e) None of theseQ.3 - Find the no. of ways of selecting n articles out of 3n+1, out of which n are identical.a) 2^(2n-1)b) [ (3n+1)(C)n ] / n!c) [ (3n+1)(P)n ] / n!d) 2^(2n)e) None of these
chk #22833
How many pairs of integers (not necessarily positive) are there such that a^2 + 6b^2 is squares (a