What positive value of b makes 9^2-bx+4 a perfect square trinomial? How would you solve this problem? Thanks
9x^2-bx+4
First of all to be a perfect square, the first and last terms should be perfect squares, and they are!
so it looks like we had
(3x ± 2)^2
for (3x+2)^2 we would get 9x^2 + 12x + 4
so -bx = 12x
b = -12
for (3x - 2)^2 we would get 9x^2 - 12x + 4
so -bx = -12x
b = 12
So b = ± 12
Thank you!
Welcome
To solve this problem, we need to determine the value of "b" that makes the trinomial 9^2 - bx + 4 a perfect square trinomial.
A perfect square trinomial can be written as the square of a binomial. So, we need to find values for "b" that will allow us to rewrite the trinomial in the form (a ± b)^2, where "a" is a constant and "b" is a variable.
Let's start by expanding (a ± b)^2:
(a ± b)^2 = a^2 ± 2ab + b^2
In our case, the trinomial is 9^2 - bx + 4. To match the form of a perfect square trinomial, we need to equate the terms from the expansion of (a ± b)^2 with the terms from the given trinomial.
Comparing the terms, we have:
a^2 = 9^2 (which means a = 9)
2ab = -bx
b^2 = 4
Solving for "b" in the second equation, we get:
2ab = -bx
2(9)(b) = -bx
18b = -bx
Dividing both sides by "x", we get:
18b/x = -b
Since a perfect square trinomial does not have any "x" term, the coefficient of "x" in our equation should be zero. Therefore:
-18b/x = 0
This implies that either b = 0 or x = 0.
However, since we are interested in finding a positive value for "b", we can eliminate x = 0 from our consideration.
Thus, the only positive value for "b" that makes 9^2 - bx + 4 a perfect square trinomial is b = 0.
Therefore, when b = 0, the trinomial 9^2 - bx + 4 becomes a perfect square trinomial.