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.
if it is a square of a binomial, then a has to be 10
(x+5)^2
or a=-10
(x-5)^2
To determine the values of a for which the expression x^2 + ax + 25 is the square of a binomial, we can consider the form of a perfect square binomial.
A perfect square binomial has the form (x + b)^2, where b is a constant.
Expanding (x + b)^2 using the square of a binomial formula yields x^2 + 2bx + b^2.
Comparing this expanded form with the given expression x^2 + ax + 25, we can see that the constant term in the expanded form is b^2, while in the given expression, it is 25. Therefore, we have:
b^2 = 25.
We can find the values of b by taking the square root of both sides of the equation:
√(b^2) = √25
Taking the square root yields:
b = ±5
So, the possible values for b are 5 and -5.
Since (x + b)^2 = x^2 + 2bx + b^2, the given expression x^2 + ax + 25 can be expressed as a perfect square binomial if and only if the coefficient of x is twice the value of b.
Hence, we have:
a = 2b
Substituting the values of b into the equation, we get:
a = 2(5) and a = 2(-5)
This simplifies to:
a = 10 and a = -10.
Therefore, the real values of a for which x^2 + ax + 25 is the square of a binomial are 10 and -10.
In increasing order, the values are: -10, 10.