SOLUTION: 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.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: 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.      Log On


   

Question 1085961: 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.
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52760) About Me  (Show Source):
You can put this solution on YOUR website!
.
There are two solutions: a = 10   and  a = -10.


x%5E2+%2B+10x+%2B+25 = %28x%2B5%29%5E2,


x%5E2+-+10x+%2B+25 = %28x-5%29%5E2.



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2+%2B+ax+%2B+25+
Looking at the quadratic formula, when the
discriminant is zero, there is a double root
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
The discriminant is:
++b%5E2-4%2Aa%2Ac+
+b%5E2++-+4%2Aa%2Ac+=+0+
+b%5E2+=+4%2Aa%2Ac+
For the given quadratic,
+a+=+1+
+b+=+a+
+c+=+25+
+a%5E2+=+4%2A1%2A25+
+a%5E2+=+100+
+a+=+10+
and also,
+a+=+-10+
---------------------
+a+=+10+
+x%5E2+%2B+10x+%2B+25+=+%28+x+%2B+5+%29%5E2++
and
+a+=+-10+
+x%5E2+-+10x+%2B+25+=+%28+x+-+5+%29%5E2++