x^2+18x+?=?^2 what numbers makes this true ?

Question

anonymous · Answer

did you copy the correct prob here?

anonymous · Answer

Yes , I have to find what number fits in thee blanks . I dont get it at all .

anonymous · Answer

x^2+18x+?=?^2 try completing the squares
x^2+18x+(18/2)^2=y^2+(18/2)^2
(x+9)^2=y^2+81

anonymous · Answer

Let me see if I can first write the correct form of the problem:
$$x^2+18x+a=a^2$$
which implies:
$$x^2+18x+a-a^2=0$$

Now, we use the quadratic formula: $$x=(1/2a)(-b+or-\sqrt{b^2-4ac})$$

Set a=1, b=18, and c= a-a^2=a(1-a)

Using the these values for the quadratic formula from above we see that x must be one of two values:

1.$$(1/(2a(1-a))(-18+\sqrt{324-4a(1-a})$$   or
2.$$(1/(2a(1-a))(-18-\sqrt{324-4a(1-a})$$

Plug these values in for the "x" variable in the original quadratic equation and you will obtain the criteria that a must satisfy in order for the statement to be true.

anonymous · Answer

*sorry, at the end I said quadratic equation and it could easily be confused with the quadratic formula.  Plug the two values into the original polynomial