Question

コ

OUTLINE

TOOLS

You notice that the following quadratic equation cannot be factored, so you solve it by completing the square.

What value of c would make the left-hand side of this equation a perfect square trinomial?

z28+ c = 13

(1 point)

0-8

0-4

<

Ο 16

ON

ACK

Ο 64

04

Answer 1

What's the answer

Answer 2

To make the left-hand side of the equation \( z^2 + c = 13 \) a perfect square trinomial, we need to rewrite it in the form of \( (z + k)^2 \).

The general form of a perfect square trinomial is given by:

\[
(z + k)^2 = z^2 + 2kz + k^2
\]

For our equation, we have:

\[
z^2 + c = 13
\]

This means we want the expression \( z^2 + c \) to match the form \( z^2 + 2kz + k^2 \).

Notice that our expression lacks the linear term \( 2kz \). This indicates that we will set \( 2k = 0 \) (which implies \( k = 0 \)). Thus, \( c \) needs to be \( k^2 \) where \( k = 0 \).

Since \( k = 0 \):

\[
k^2 = 0^2 = 0
\]

Thus, to make the left-hand side of the equation a perfect square trinomial, \( c \) must be \( 0 \).

Therefore, the value of \( c \) is \( 0 \).

Answer 3

The value of \( c \) that would make the left-hand side of the equation \( z^2 + c = 13 \) a perfect square trinomial is **0**.