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1 May, 06:53

What are the x-intercepts of 4x^ + 8x - 5

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  1. 1 May, 07:38
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    Assuming the exponent is supposed to be "^2" your equation will read:

    4x² + 8x - 5

    It must set equal to y to be a valid function and the y must be set equal to zero to find x-intercepts.

    4x² + 8x - 5 = 0

    4x² - 2x + 10x - 5 = 0

    2x (2x - 1) + 5 (2x - 1) = 0

    (2x + 5) (2x - 1) = 0

    Set each binomial equal to zero.

    2x + 5 = 0

    2x = 0 - 5

    2x = - 5

    Divide both sides by 2

    x = - 5/2

    2x - 1 = 0

    2x = 0 + 1

    2x = 1

    x = 1/2

    Your x-intercepts are x = - 5/2, 1/2 or ( - 5/2, 0) and (1/2, 0)
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