For what real values of a is x^2 + ax + 25 the square of a binomial? If you find more than one, then list your values in increasing order, separated by commas.
Answer please!!!
Damon did a whole bunch of those for you yesterday.
The idea is to use a previous solution and apply it to a new question.
x^2 + ax + 25 = (x + ???)^2
Gee what do you think we need so when squared we will get 25?
And what will be the middle term if we expand that?
To find the real values of "a" for which the expression x^2 + ax + 25 is a perfect square of a binomial, we need to utilize a few properties of perfect squares.
Let's suppose the given expression is the square of a binomial in the form (x+b)^2, where b is a constant.
Expanding (x+b)^2, we get:
x^2 + 2bx + b^2
Comparing this with the given expression x^2 + ax + 25, we can deduce that:
2bx = ax and b^2 = 25
From the first equation, we have:
2bx = ax
2b = a
Now, substituting this value of "a" in the second equation, we get:
(2b)^2 = 25
4b^2 = 25
Taking the square root of both sides, we have:
2b = ±√25
2b = ±5
Dividing by 2, we get:
b = ±5/2
Thus, the possible values of "a" are:
a = 2b = 2(±5/2)
a = ±5
Hence, the real values of "a" for which x^2 + ax + 25 is the square of a binomial are -5 and 5.