What are the zeros of the quadratic function? y = x^2 – 2x – 15

Question

What are the zeros of the quadratic function?
y = x^2 – 2x – 15

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Mộc Miên 3 months 2021-08-22T11:07:29+00:00 1 Answers 2 views 0

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    2021-08-22T11:09:12+00:00

    Answer:

    x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}

    Step-by-step explanation:

    -> 0=2x^2-2x-15

    -zeros means the solution aka x intercept, meaning y=0

    x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:2\left(-15\right)}}{2\cdot \:2}

    => according to quadratic formula

    x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

    where ax^2+bx+c=0

    => x_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{31}}{2\cdot \:2}

    simplify

    x_1=\frac{-\left(-2\right)+2\sqrt{31}}{2\cdot \:2},\:x_2=\frac{-\left(-2\right)-2\sqrt{31}}{2\cdot \:2}

    x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}

    Cannot simplify farther

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )