Question

Pls help i am on a timer

Answer 1

Answer:

$x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}$

Step-by-step explanation:

Given;

x² - 2x - 1 = 0

Solve by completing the square method;

⇒ take the constant to the right hand side of the equation.

x² - 2x = 1

⇒ take half of coefficient of x = ¹/₂ x -2 = -1

⇒ square half of coefficient of x and add it to the both sides of the equation

$x^2 + (-1)^2 = 1 + (-1)^2$

$(x-1)^2 = 1 + 1\\\\(x-1)^2 = 2$

⇒ take the square root of both sides;

$x-1 = +/- \ \ \sqrt{2} \\\\x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}$

Therefore, option B is the right solution.