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6 March, 06:09

Find all real zeros of f (x) = 2x^2 + 5x-18 algebraically.

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Answers (2)
  1. 6 March, 06:57
    0
    The zeros are {-4.5, 2}.

    Step-by-step explanation:

    2x^2 + 5x - 18 = 0

    Use the 'ac' method of solution.

    2 * - 18 = - 36; we need 2 numbers whose product is - 36 and whose sum is + 5.

    +9 and - 4 look good so:

    2x^2 + 5x - 18 = 0 Replacing the + 5x by - 4x + 9x:

    2x^2 - 4x + 9x - 18 = 0

    2x (x - 2) + 9 (x - 2) = 0

    x - 2 is common to both parts, so:

    (2x + 9) (x - 2) = 0

    2x + 9 = 0 gives x = - 4.5 and

    x - 2 = 0 gives x = 2.
  2. 6 March, 09:32
    0
    2x2-5x-18=0

    Two solutions were found:

    x = - 2

    x = 9/2 = 4.500

    Step by step solution:

    Step 1:

    Equation at the end of step 1:

    (2x2 - 5x) - 18 = 0

    Step 2:

    Trying to factor by splitting the middle term

    2.1 Factoring 2x2-5x-18

    The first term is, 2x2 its coefficient is 2.

    The middle term is, - 5x its coefficient is - 5.

    The last term, "the constant", is - 18

    Step-1 : Multiply the coefficient of the first term by the constant 2 • - 18 = - 36

    Step-2 : Find two factors of - 36 whose sum equals the coefficient of the middle term, which is - 5.

    -36 + 1 = - 35

    -18 + 2 = - 16

    -12 + 3 = - 9

    -9 + 4 = - 5 That's it

    Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, - 9 and 4

    2x2 - 9x + 4x - 18

    Step-4 : Add up the first 2 terms, pulling out like factors:

    x • (2x-9)

    Add up the last 2 terms, pulling out common factors:

    2 • (2x-9)

    Step-5 : Add up the four terms of step 4:

    (x+2) • (2x-9)

    Which is the desired factorization

    Equation at the end of step 2:

    (2x - 9) • (x + 2) = 0

    Step 3:

    Theory - Roots of a product:

    3.1 A product of several terms equals zero.

    When a product of two or more terms equals zero, then at least one of the terms must be zero.

    We shall now solve each term = 0 separately

    In other words, we are going to solve as many equations as there are terms in the product

    Any solution of term = 0 solves product = 0 as well.
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