What are the values that make x^2 + bx + 64 a perfect square?

b = ±16

Step-by-step explanation:

Normally to find the number we add to make it a perfect square

We take the coefficient of x

b

Divide by 2

b/2

Then square it

(b/2) ^2

In this case, we are adding 64

(b/2) ^2 = 64

Take the square root of each side

sqrt((b/2) ^2) = sqrt(64)

b/2 = ±8

Multiply each side by 2

b/2*2 =  ±8 *2

b = ±16

±16

Step-by-step explanation:

What is the square root of 64?  8.  But -8 is also a square root of 64.  Squaring either 8 or -8 results in ±16.

The following (the square of a binomial) is a "special product" or "perfect square."

(a + b)^2 = a^2 + 2ab + b^2

Compare this pattern to the given:  1 x^2 + bx + 64:

                                                            a^2 + 2ab + b^2

Here a = 1 and b^2 = 64.  Thus, b must be either +8 or -8.

Then the given expression becomes 1x^2 ± 16x + 64; that is:

b = ±16

it should be a and d. lmk if i'm wrong xd

the barter