find the value of a?

Question

find the value of a such that

z^2 + 18z + a is a perfect square

Is this the cube root deal? or how do I do this?

Answer 1

guess and check. set z equal to a number such as 0, 1 or 2
then set a equal to a number that would make the expressions value be equal to that of a perfect square.

ex. z=1, then 1+18+a=n^2 (n^2 stands for any square) , n^2-19=a , if the square is 36 then a will be 17.

step 2. plug your value for a into another equation and change the value for z. see if you get another perfect square.

ex. z=2, so 4+36+a=n^2 since 57 is not a perfect square, this statement is not true, so your answer is wrong

also keep in mind that a could be a number in terms of z

try 17z

z^2+18z+17z=n^2

z^2+35z=n^2

let z=6

36+210=n^2

246 is not a square root, so again this is not the right answer

how about 17z^2

if z=3

9+54+153
216, again not a perfect square


if you keep changing the constant that will give u a square in another equation, keep changing the degree of z that a is in terms of, and keep trying different square roots in the original equations that u solve a for, u should eventually get it

Answer 2

z^2+18z+81 is the answer
a=81

18z/2=9
(9)(9)=81 and that equals a.

Answer 3

z^2+18z+81= (z+9)^2
so a=81

Answer 4

apply the formula
(a+b)^2 = a^2+2*a*b+b^2


so z^2+2*9*z+9^2


so value of a = 81 :)

Answer 5

z^2 + 18 z + 81 = (z+9)^2, so a = 81

are asking what the cube root of 81 is? it is 4.32....
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