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.
No way. I just did one like this for you.
Damon, that is against the rules!
Pleeease answer my question Damon!
hey damon!
To determine the real values of a for which the quadratic expression x^2 + ax + 25 is the square of a binomial, we need to find the binomial expression that, when squared, results in the given quadratic equation.
Let's set up the expression for the square of a binomial as (x + b)^2 and expand it:
(x + b)^2 = x^2 + 2bx + b^2
Comparing this with the given quadratic equation x^2 + ax + 25, we can see that the constant term in both equations is 25. Therefore, b^2 must be equal to 25, which means b can be ±5.
Substituting b = 5 into the binomial expression:
(x + 5)^2 = x^2 + 10x + 25
Comparing the coefficients of this expression with the original quadratic equation, we see that the quadratic expression x^2 + ax + 25 is equal to (x + 5)^2 when a = 10.
Now, substituting b = -5 into the binomial expression:
(x - 5)^2 = x^2 - 10x + 25
Comparing coefficients again, we see that the quadratic expression x^2 + ax + 25 is equal to (x - 5)^2 when a = -10.
Therefore, the real values for a that make x^2 + ax + 25 the square of a binomial are a = 10 and a = -10. In increasing order, the values of a are -10, 10.