How do you solve the quadratic equation by completing the square: #x^2-2x-5=0#?

Question

Isaiah Anthony · Accepted Answer

#x_1 = 1 + sqrt(6)#, #x_2 = 1-sqrt(6)#

Write your quadratic in the form first.

#x^2 + b/ax = -c/a#

In your case, #a=1# to begin with, so you have

#x^2 - 2x = 5#

You must write the left side of the equation as the square of a binomial by adding a term to both sides in order to solve this quadratic by completing the square.

You can determine what this term must be by dividing the coefficient of the #x#-term by 2, then squaring the result.

As for you, you've

#(-2)/2 = (-1)#, then

#(-1)^2 = 1#

Accordingly, the quadratic becomes

#x^2 -2x + 1 = 5 + 1#

This allows us to write the left side of the equation as

#x^2 - 2x + 1 = x^2 + 2 * (-1) + (-1)^2 = (x-1)^2#

This indicates that you've

#(x-1)^2 = 6#

To find, take the square root of each side.

#sqrt((x-1)^2) = sqrt(6)#

#x-1 = +-sqrt(6) => x_(1,2) = 1 +- sqrt(6)#

Thus, your quadratic will have two solutions:

#x_1 = color(green)(1 + sqrt(6))# and #x_2 = color(green)(1 - sqrt(6))#

Savannah Anderson · Answer

undefined To solve the quadratic equation $x^2 - 2x - 5 = 0$ by completing the square:

1. Move the constant term to the other side of the equation:
   $x^2 - 2x = 5$

2. To complete the square, take half of the coefficient of $x$, square it, and add it to both sides of the equation:
   $$x^2 - 2x + (-2/2)^2 = 5 + (-2/2)^2$$

3. Simplify both sides:
   $$x^2 - 2x + 1 = 5 + 1$$
   $$x^2 - 2x + 1 = 6$$

4. Rewrite the left side as a perfect square:
   $$(x - 1)^2 = 6$$

5. Take the square root of both sides:
   $$x - 1 = \pm \sqrt{6}$$

6. Solve for $x$:
   $$x = 1 \pm \sqrt{6}$$

Therefore, the solutions to the quadratic equation $x^2 - 2x - 5 = 0$ by completing the square are $x = 1 + \sqrt{6}$ and $x = 1 - \sqrt{6}$.