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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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