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

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

do i do it on a graph?

step-by-step explanation: