You are not logged in.
The product of the four roots is to be 6.
Hence x=3 is certainly a root if x=6 is not a root( so we can say it can be a root) but it can not be a double root.
However x=1 certainly a double root the other roots being 2 & 3.
x=1 can be a triple root the other root being 6.
For 2. use the formula
P(A/D)= {P(A)*P(D/A)}/ {P(A)*P(D/A)+P(B)*P(D/B)}= {0.7*0.02} /{0.7*0.02+0.3*0.05}
Ask each son to ride other's horse.
This can be disproved. if a and b are integers a+b and a-b are either both odd or both even. Therefore the product a^2-b^2 is either odd or divisible by 4. so if p is only divisible by 2 say 10 it is not possible to find a and b.( I mean 27p can be factorized into 2 odd or 2 even numbers.)
A statue.
Consider an integer x. If we add 30, then the result is a perfect square. If we subtract 30, the result is also a perfect square. How many such integers are there?"
Let x be the number.