+13 If $a$ is a constant such that $9x^2 + 24x + a$ is the square of a binomial, then what is $a$?
+13 If 9x^2+24x+a is the square of a binomial, then the binomial has the form for some number , because . So, we compare to . Expanding gives
(3x+b)^2=(3x)^2+2(3x)(b)+b^2=9x+6bx+b^2
Equating the linear term of this to the linear term of 9x^2+24x+a, we have 6bx=24x, so b=4. Equating the constant term of 9x^2+6bx+b^2 to that of 9x^2+24x+a gives us a=b^2=16.
+36915 9 (x^2 + 8/3 ) complete the square
9 ( x+ 8/6)^2 now put '9' back into the parentheses expand to find 'a'
(3x+4)2 expand to see what 'a' is
9x^2 +24x + (16) a = 16
