Question

Find k so that 9x^2-48x+k is the square of a binomial.

Answer 1

Answer:

Step-by-step explanation:

The discriminant must be zero.

Discriminant = (-48)^2 - 4*9k = 2304 - 36K

36k = 2304

k = 64

Answer 2

Using the square of a subtraction, it is found that k = 64.

What is a square of a subtraction?

It is a notable product, given by:

$(ax - b)^2 = a^2x^2 - 2abx + b^2$

In this problem, the expression is:

$9x^2 - 48x + k$

We want it to be the square of a binomial, hence:

$a^2x^2 - 2abx + b^2 = 9x^2 - 48x + k$

Then, comparing both sides, first the value of a is found:

$a^2 = 9$

$a = \sqrt{9}$

$a = 3$

Then, with the second term, the value of b is found.

$2ab = 48$

$6b = 48$

$b = \frac{48}{6}$

$b = 8$

Finally, the value of k is found.

$k = b^2 = 8^2 = 64$

You can learn more about the square of a subtraction at https://brainly.com/question/9239489