What is the missing constant term in the perfect square that starts with x^2 - 16x ?

Question

asked Nov 1, 2020 193k views

2 Answers

Qehgt · Answer 1 · 2020-11-07T13:34:11+0000

ANSWER

The missing constant term is 64

EXPLANATION

The given expression is

${x}^(2) - 16x$

Let the k be the constant term, then

${x}^(2) - 16x + k$

is a perfect square.

This implies that, the discriminant is zero.

The discriminant is

$D= {b}^(2) - 4ac$

${( - 16)}^(2) - 4(1)(k) = 0$

256 - 4k = 0

- 4k = - 256

k = ( - 256)/( - 4)

k = 64