Professor Kate Poirier | D046 | Fall 2023

OpenLab assignment: Test #2 solutions

Individual assignment due Monday, December 12

Choose one question from Test #2 that you did not earn full credit for. Write a complete solution and share your work on the OpenLab (if you are not able to share a photo of your hand-written work, upload the photo to another platform like Dropbox or Google docs and share the link in your OpenLab post).

If you choose a true/false question or a “give an example of…” question, write complete sentences to explain your thinking.

Post title: Test #2 solutions Version [your test version] Question [the number of the question you chose]

Post category: Test #2 solutions

5 Comments

  1. Tyrone

    Test #2 solutions version A, question #1

    (a) There are a minimum of 10 solutions to the equation (x^5 – 3x) (x – 3) = 5. Untrue. There are only n possible solutions for a polynomial equation of degree n. There can be a maximum of 6 solutions to the given equation because it is of degree 6.

    (b) The solution to (x^5 – 3x) (x – 3) = 5 is (b) x = 8. Untrue. We can replace x in the equation with 8 to see if it is true to see if x = 8 is a solution. The result of doing this is: (8^5 – 3*8) (8 – 3) = 5 (32768 – 24). (5) = 5 (163720 = 5) (32744) (5) = 5

    Since this is obviously untrue, x = 8 is not a solution.

    (c) There are no solutions to the equation sqrt x – 3 + 9 = 2. True. Since the square root of a negative number is not defined in the real numbers, this equation is equal to sqrt x = -4, which has no actual solutions.

    (d)There are six potential roots for a polynomial of degree four. Untrue. As portion (a) explains, a 4th degree polynomial can have up to 4 distinct roots.

    (e) There are no solutions for the equation 1/x-3+ 9 = 9 False. x = 3 is the only possible answer to this equation. We can multiply by x – 3 and remove 9 from both sides to find it:

    1/x-3 + 9 = 9 1/x-3 = 0 (x – 3) (1/x-3) = 0 1 = 0

    Only when x – 3 = 0, which suggests x = 3, is this true.

  2. Pavlo

    ?rlkey=dabivpwnspvism9zlbn9iyoqy&dl=0″ rel=”nofollow ugc”>?rlkey=dabivpwnspvism9zlbn9iyoqy&dl=0

  3. Pavlo

    https://imgur.com/a/9BFtSm1

  4. Amanda Padilla

    1A: The equation (x^5 – 3x)(x-3) = 5 has at least 10 solutions. It is false, because for the equation to have ten solutions, it would need to be a polynomial equation of degree 10. However, the given equation may not necessarily lead to such a polynomial equation. The degree of the polynomial is determined by the highest power of x in the equation.

  5. Omar Abaza

    View post on imgur.com

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