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    The roots of quadratic equation x² - 4x - 1 = 0 are,
    Question

    The roots of quadratic equation x² - 4x - 1 = 0 are,

    A.

    2±√5

    B.

    2, -2

    C.

    5±√2

    D.

    5, -5

    Correct option is A

    Concept/Formula Used:

    For a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0​ , the roots are given by the quadratic formula:

    x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    where:
    - a is the coefficient of x2x^2​ ,
    - b is the coefficient of x ,
    - c is the constant term.

    Given:

    The equation is:
    x24x1=0x^2 - 4x - 1 = 0
    Here:
    a=1, b=4, c=1a = 1, \ b = -4, \ c = -1

    Formula Used:

    Using the quadratic formula:
    x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    Solution:

    Substitute a = 1 , b = -4 , and c = -1 into the formula:

    1. Compute b24acb^2 - 4ac​ :

    b^2 - 4ac = (-4)^2 - 4(1)(-1) = 16 + 4 = 20


    2. Substitute into the quadratic formula:

    x=(4)±202(1)x = \frac{-(-4) \pm \sqrt{20}}{2(1)}


    x=4±202x = \frac{4 \pm \sqrt{20}}{2}


    3. Simplify 20\sqrt{20}​ :

    20=4×5=25\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}


    4. Simplify the expression:

    x=4±252x = \frac{4 \pm 2\sqrt{5}}{2}


    x=2±5x = 2 \pm \sqrt{5}

    Correct Answer:

    (A)2±5(A) 2 \pm \sqrt{5}

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