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-   -   Solving problems in simplest form (https://www.askmehelpdesk.com/showthread.php?t=267201)

  • Oct 7, 2008, 08:13 PM
    andyhaus1057
    Solving problems in simplest form
    I know the answer to some of these, and the others I'm not sure of. Can you please explain to me how I can work them out in simplest form?

    1) (-5 - √-9^2) answer is 30 + 16i?

    2) x^2 + 5x + 2 = 0 answer is {-5 + √17 / 2, -5 - √17 / 2

    3) √6x + 1 = x - 1 answer is {0, 8}?

    4) 2x + 5 <17 answer is x<6?

    5) 3x^2 - 1 = 47 answer is {-4, 4}

    6) x^2 - 2x + 17 = 0 answer is {1 + 4i, 1 - 4i)

    7) 4x^3 - 12x^2 = 9x - 27 answer is { -3/2, 3, 3/2}

    8) |x| = 6 answer is {6}?

    9) √x + 10 = x - 2 answer is {-1}?

    10) (-7 + 5i) - (-9 -11i) answer is 2 + 16i

    11) (2 + 7i) ( 2 - 7i) answer is 4 - 49i

    12) √-81 - √-144 answer is -3

    13) Solve x^2 - 13x + 36 = 0 answer is {4, 9}

    14) Solve 5x^2 - 20x = 0 answer is {0, 4}
  • Oct 8, 2008, 01:07 AM
    Unknown008

    1) I don't unbderstand where the 'i' comes from iin your answer. What is ? Then solve.

    2)You must know your formulae!!



    3) Use the same formula, but here, you'll have the square root of x instead of x.



    4) Solve normally, i.e. subtract 5 from both sides, then divide both sides by 2. Keep the '<' symbol.

    5) Solve normally, i.e. add 1 to both sides, then divide both sides by 3. Lastly, put square root on both sides. But then, remember that the value of x may be positive or negative, so put both.

    6) Same as 2)

    7) That's more complicated. See my explanation that i gave you a while back

    https://www.askmehelpdesk.com/math-s...ml#post1293230

    8) The two bars surrounding the x (modullus) show that for any value of x, the answer will be positive. As the result is 6, x must be either 6 or -6. Little mistake i think in your given answer

    9) Same as in 3)

    10) Expand normally then simplify. That's relatively easy!

    11) Expand or use your formula:



    12) What is and ?

    13) same as in 2) and 6)

    14) Same as in 2), 6), and 13). But here, you can also factorise by putting the x outside 9better if you take the 5 also since both terms are divisible by 5x):



    or



    Then solve.

    Hope that helped. Just remember your basic formulae, and all will be fine.
  • Oct 8, 2008, 11:49 AM
    andyhaus1057
    Thanks for your help.

    12) The square root of 81 and the square root of 144.. what would that be?
  • Oct 8, 2008, 01:52 PM
    andyhaus1057
    These are the ones I'm not sure of..

    1) Solve √x + 10 = x - 2. I know that it has to be either
    {-1, 6}
    {-1}
    {6}
    or {0}

    2) √6x + 1 = x - 1 I know that it has to be either...
    {9, 18}
    {0, 8}
    {0}
    or {8}

    3) |x| = 6 I know that it has to be either..
    {6}
    {-6, 6}
    {-6}
    or {0, 6}

    4)
    √-81 - √-144 I know it has to be either..
    -3, -3i, 3i, or 21i

    5) x^2 = 11x - 10
    {-10, -1}
    {1, 10}
    {5, 2}
    or {10, -1}
  • Oct 9, 2008, 12:41 AM
    Unknown008

    I'll give you an easy away to be sure of your answers. REPLACE the answers you got into your question. By the way, i haven't done the square root of negatives yet... if you could provide me with the rule, i could perhaps help you more of the numbers requiring such formulae.

    1)





    Are they correct? Check it. Anyway, at first galnce, that cannot be 0 since it will be



    which is not correct.

    2) Check it as for the 1)

    3) Replace your answers.

    |x| = 6

    |6| = 6 √

    |-6| = 6 √

    |0| = 6 X

    4) Give me that formula, asap.

    5)





    So, x = 10 or 1

    the other answers are wrong. You can check the others also.

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