how many roots does this has? x^2+(2√5x)+2x=-10​ find Discriminant

Question

how many roots does this has?
x^2+(2√5x)+2x=-10​
find Discriminant

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Mộc Miên 1 month 2021-08-31T12:21:35+00:00 1 Answers 0 views 0

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    2021-08-31T12:22:52+00:00

    Given:

    The equation is

    x^2+(2\sqrt{5})+2x=-10

    To find:

    The number of roots and discriminant of the given equation.

    Solution:

    We have,

    x^2+(2\sqrt{5})x+2x=-10

    The highest degree of given equation is 2. So, the number of roots is also 2.

    It can be written as

    x^2+(2\sqrt{5}+2)x+10=0

    Here, a=1, b=(2\sqrt{5}+2), c=10.

    Discriminant of the given equation is

    D=b^2-4ac

    D=(2\sqrt{5}+2)^2-4(1)(10)

    D=20+8\sqrt{5}+4-40

    D=8\sqrt{5}-16

    D\approx 1.89>0

    Since discriminant is 8\sqrt{5}-16\approx 1.89, which is greater than 0, therefore, the given equation has two distinct real roots.

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