Question

Q. If a+2b=3 and ab= -5 then find the value of
(i) a²+4b²
(ii) a³+8b³
I need the answers fast

Answer 1

hope it helps...thanks

Answer 2

Answer:

given a + 2b = 3 and ab = - 5  

b = -5/ a

so a + 2b = a + 2×-5/a

             3 = (a² -10) /a

            3a = a² -10

            a²-3a-10=0

            a = (3 + √((-3)²-(4×1×-10))  )/ 2   or 3 -√((-3)²-(4×1×-10))  )/ 2

               = (3 +√49 ) /2 or (3-√49)/2

               =10/2  or -4/2

              =5 or-2

so b= -5/a

     = -5/5 or -5/-2

     = -1 or 5/2

(i) a² + 4 b²

 ( a+ 2b)²=3²

  a²+ 4ab+4b² =9

  a² + 4b² = 9 - 4ab

                = 9 - 4×-5

               = 9+ 20    

              = 29 (answer)

         

(ii) a³ + 8 b³

(a+ 2b)³ = 3³

a³ + 3×a² ×2b +3a×4b² +(2b)³ = 27

a³ + 8b³ = 27 - 6a²b - 12ab²

            = 27-6×ab ×a -12×ab×b

            =27 -6×(-5)×5 -12× -5×-1

            =27 + 150 -60

            = 117 (answer)

         OR

JUST SUBSTITUTE a and b in the question

(i) a² +4b² = 5² + 4×(-1)²

               = 25+ 4

              = 29 (answer)

(ii) a³ +8b³ = 5³ + 8× (-1)³

            = 125 -8

             = 117(answer)