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?
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
Thank you!
You are on the right track, but there is a small error in your calculation. Let's go step by step to find the value of k.
The given expression is 121a^2b^2 + kab + 4b^2.
To determine if it is a perfect square trinomial, we need to check if it can be factored into the square of a binomial.
First, let's factor out the greatest common factor from the three terms. The GCF of 121a^2b^2, kab, and 4b^2 is b^2. So we have:
b^2(121a^2 + ka + 4)
Now, let's focus on the expression inside the parentheses: 121a^2 + ka + 4. In order for it to be a perfect square trinomial, the second term needs to be twice the product of the square roots of the first and third terms.
In this case, the first term is 121a^2, which is the square of 11a, and the third term is 4, which is the square of 2. Therefore, the second term should be:
2 * (√(121a^2) * √(4))
2 * (11a * 2) = 44a
So, the correct value of k is 44a, not just 44 as you calculated.
To summarize, the positive value of k must be 44a, not 44.