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Posted by on Friday, December 2, 2011 at 11:04am.

How would you find the x intercepts of 2x^2+17=0. The original equation is 2(x-3)^2-1.

  • Algebra 2 - , Friday, December 2, 2011 at 11:12am

    To have x-intercepts you must have a function,
    I bet your equation was
    y = 2(x-3)^2 - 1 or f(x) = 2(x-3)^2 -1
    then
    y = 2(x^2 - 6x + 9) - 1
    y = 2x^2 - 12x + 17 , your expansion was incorrect

    so at the x-intercept, y = 0 and
    2x^2 - 12x + 17 = 0
    x = (12 ± √8)/4
    = (12 ± 2√2)/4
    x = (6+√2)/2 or x = (6 - √2)/2

  • Algebra 2 - , Friday, December 2, 2011 at 12:06pm

    thanks so much!

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