Daniel claimed that the equation x2 +36 = 0 can be solved by using its factored form of (x + 6i)(x – 6i) = 0. He stated that there are two solutions zeros for this %3D function: -6i and 6i. Choice Description Daniel is incorrect. The factored form of x2 +36 = 0 is (x + 6i)2 = 0, A and there is only one zero of 6i. Daniel is incorrect. The factored form of x? +36 = 0 is (x + 6i)2 = 0, %3D and there is only one zero of -6i. Daniel is partially correct. The factored form of a2 +36 = 0 is (x+6i)(x- 6i) = 0, but the zeros are 6 and -6. %3D C %3D Daniel is correct. The factored form of x2 + 36 = 0 is (x + 6i)(x – 6i) = 0, and the zeros are 6i and –6i. %3D O Choice A O Choice B O Choice C O Choice D B.
Daniel claimed that the equation x2 +36 = 0 can be solved by using its factored form of (x + 6i)(x – 6i) = 0. He stated that there are two solutions zeros for this %3D function: -6i and 6i. Choice Description Daniel is incorrect. The factored form of x2 +36 = 0 is (x + 6i)2 = 0, A and there is only one zero of 6i. Daniel is incorrect. The factored form of x? +36 = 0 is (x + 6i)2 = 0, %3D and there is only one zero of -6i. Daniel is partially correct. The factored form of a2 +36 = 0 is (x+6i)(x- 6i) = 0, but the zeros are 6 and -6. %3D C %3D Daniel is correct. The factored form of x2 + 36 = 0 is (x + 6i)(x – 6i) = 0, and the zeros are 6i and –6i. %3D O Choice A O Choice B O Choice C O Choice D B.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
Question 7
![Daniel claimed that the equation x2 +36 = 0 can be solved by using its factored
form of (x + 6i)(x – 6i) = 0. He stated that there are two solutions zeros for this
|3D
--
function: -6i and 6i.
Choice
Description
Daniel is incorrect. The factored form of x2 +36 = 0 is (x + 6i)2 = 0,
%3D
A
and there is only one zero of 6i.
Daniel is incorrect. The factored form of x? +36 = 0 is (x + 6i)2 = 0,
and there is only one zero of -6i.
Daniel is partially correct. The factored form of x2 + 36 = 0 is
(x+6i)(x – 6i) = 0, but the zeros are 6 and -6.
%3D
%3D
Daniel is correct. The factored form of x2 + 36 = 0 is
(x+ 6i)(x – 6i) = 0, and the zeros are 6i and–6i.
%3D
O Choice A
O Choice B
O Choice C
O Choice D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e78f021-af53-433e-b041-f8aa941a99ea%2F94ca7273-8e2d-4137-9b48-a7e7ae1ac445%2Ff0o4uld_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Daniel claimed that the equation x2 +36 = 0 can be solved by using its factored
form of (x + 6i)(x – 6i) = 0. He stated that there are two solutions zeros for this
|3D
--
function: -6i and 6i.
Choice
Description
Daniel is incorrect. The factored form of x2 +36 = 0 is (x + 6i)2 = 0,
%3D
A
and there is only one zero of 6i.
Daniel is incorrect. The factored form of x? +36 = 0 is (x + 6i)2 = 0,
and there is only one zero of -6i.
Daniel is partially correct. The factored form of x2 + 36 = 0 is
(x+6i)(x – 6i) = 0, but the zeros are 6 and -6.
%3D
%3D
Daniel is correct. The factored form of x2 + 36 = 0 is
(x+ 6i)(x – 6i) = 0, and the zeros are 6i and–6i.
%3D
O Choice A
O Choice B
O Choice C
O Choice D
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