Question

IF 121a^2b^2 + kab + 4b^2 is a perfect square trinomial, then the positive value of k must be ________.

my work:
121a^2b^2 = 11ab
4b^2 = 2b

Then 2nd term k is twice the product of 11ab & 2b , so the value of k must be: 44

am i right or wrong?

Answer 1

you are correct in the final answer, but I object to your "work"
the statement 121a^2b^2 = 11ab is wrong it should be
121a^2b^2 = (11ab)^2
same for your next line

you could have checked your answer ...

(11ab + 2b)^2
= 121a^2b^2 + 44ab + 4b^2

Answer 2

Thank you!