+0  
 
0
89
2
avatar+13 

If $a$ is a constant such that $9x^2 + 24x + a$ is the square of a binomial, then what is $a$?

 Jul 20, 2020
 #1
avatar+13 
+1

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.

 Jul 20, 2020
 #2
avatar+26767 
+1

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

 Jul 20, 2020

12 Online Users

avatar