there may be mistake in the qes. not sure.

(a-b)^2=1 & (a+b)^2=25
values of a,b?

i)a is a prime no
ii)a<2

• atq
  value of
  set(a,b)={(-2,-3),(3,-2),(-3,2),(2,3)}
  frm 1stmnt
  
  some value is prime and some is nt..but a should have unique value.
  so i cant satisfied
  frm stmnt 2
  not all value of a
• 10 years agoHelpfull: Yes(4) No(2)
• i is true.
• 10 years agoHelpfull: Yes(1) No(1)
• (a-b)^2=a2-2ab+b2=1
  a2+b2=1+2ab;...........................(1)
  also, (a+b)^2=a2+2ab+b2=25
  a2+b2=25-2ab..........................(2)
  frm (1)& (2)
  ab=6
  so ab can be (1,6) , (6,1) ,(2,3) ,(3,2) and also its -ve values
  
  case 1:- a is prime
  1 is neither a prime nor composite no and -ve no cannot be prime and also 6 is not a prime no, hence we r left with options (2,3) ,(3,2) but we dn't have clear values for a and b
  
  case 2: - a
• 10 years agoHelpfull: Yes(1) No(1)
• with i ans can find
• 10 years agoHelpfull: Yes(0) No(1)
• second condition is:
  |a|>2
• 10 years agoHelpfull: Yes(0) No(0)
• both statements are necessary
  (a-b)^2=(b-a)^2
• 10 years agoHelpfull: Yes(0) No(0)
• (a, b)=(2,3), (3,2), (-2,-3), (-3,-2)
  1) a is prime
  (3,2) (-3,-2)
  2) a
• 10 years agoHelpfull: Yes(0) No(0)
• a-b=1
  a+b=5
  by solving answer for a=3
  so its a prime
  i is trye
• 10 years agoHelpfull: Yes(0) No(0)
• (a) If a is prime no. it cannot be -ve
  no. could be (2,3) & (3,2)
  becoz (2-3)^2=1 & (2+3)^2=25 similarly for (3,2)
  since there are two option we cannot find the value of a,b specifically
  (b) since only possible values are (2,3) & (3,2)
  so a
• 9 years agoHelpfull: Yes(0) No(0)
• It's answer should be (-2,-3) and satisfied to both conditions
• 9 years agoHelpfull: Yes(0) No(0)