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basic algebra

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1) 2x²-7x-15
2) 9x²+24x+16
Can you show me all the steps...i don't know how to do it...Thankyou!!:)

  • basic algebra - ,

    first: add 7x to each side. Then divide both sides by 9

    second: subtract 9x+16 from each side. Then divide both sides by 15

  • basic algebra - ,

    we have to factor not solve

  • basic algebra - ,

    1. Y = 2x^2-7x-15 = 0.
    Use the AC method of factoring:
    A*C = 2*-15 = -30 = 1*-30 = 3*-10.

    Choose the pair of factors whose sum = B(-7): 3, and -10.

    2x^2 + (3x-10x) - 15 = 0.
    Group the 4 terms into 2 factorable pairs:
    (2x^2-10x) + (3x-15)
    2x(x-5) + 3(x-5)
    (x-5) is common to both terms. So we factor out(x-5):
    (x-5)(2x+3) = 0

    x-5 = 0
    X = 5.

    2x+3 = 0
    2x = -3
    X = -1.5.

    Solution set: X = 5, and -1.5.

    2. 9x^2+24x+16 = 0.
    The given expression is a perfect square: 9x^2+24x+16 = (3x+4)^2 = 0.

    (3x+4)^2 = 0
    Take sgrt of both sides:
    +-(3x+4) = 0

    3x+4 = 0
    3x = -4
    X = -1 1/3 = -1.33333.

    -3x-4 = 0
    -3x = 4
    X = -1 1/3 = -1.33333.

    Solution X = -1.33333, and -1.33333.

    Note: The given expression was set to zero, because the solutions are x-intercepts and Y = 0 at the x-intercepts.

  • basic algebra - ,

    9x^2+24x+16. Factor only.

    9x^2+24x+16 = (3x+4)^2 = (3x+4)(3x+4).

  • basic algebra - ,

    Thankyou so much!!

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