How do you solve by completing the square #x^2- 4x-11=0#?

Question

Avery Alexander · Accepted Answer

#2+sqrt(15), #x=2-sqrt(15)# This equation should be rewritten using a full square, giving us: #x^2-4x+c=0#. I don't want to delve any further, so let's end this section with #c=4#. Thus: x^2 -4x + 4#= #(x-2)^2 Let's now enter #(x-2)^2# in the provided equation. x^2 -4x-11=0#, (x-2)^2+n=x^2-4x-11#, and n=-15# are the given values. Thus: #abs(x-2)=sqrt(15)#
#x-2=-sqrt(15), sqrt(15)#
#x=2-sqrt(15),2+sqrt(15)#
#(x-2)^2-15=0#
#sqrt((x-2)^2) = 15#

Isaiah Anthony · Answer

undefined To solve the equation $x^2 - 4x - 11 = 0$ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $x^2 - 4x = 11$

2. Add the square of half the coefficient of $x$ to both sides of the equation:
   $x^2 - 4x + (-4/2)^2 = 11 + (-4/2)^2$

3. Simplify both sides:
   $x^2 - 4x + 4 = 11 + 4$

4. Rewrite the left side as a perfect square:
   $(x - 2)^2 = 15$

5. Take the square root of both sides:
   $x - 2 = \pm \sqrt{15}$

6. Solve for $x$:
   $x = 2 \pm \sqrt{15}$

So, the solutions are $x = 2 + \sqrt{15}$ and $x = 2 - \sqrt{15}$.