How do you solve using the completing the square method #x^2-8x+15=0#?

Question

Eli Assouline · Accepted Answer

The solutions are :
#color(green)(x = 5# or # color(green)(x = 3# #x^2 - 8x +15= 0# #x^2 - 8x = - 15# To write the Left Hand Side as a Perfect Square, we add 16 to both sides: #x^2 - 8x + 16 = -15+ 16# #x^2 - 2*x*4 + 4^2 = 1# Using the Identity #color(blue)((a-b)^2 = a^2 - 2ab + b^2#, we get #(x-4)^2 = 1# #x - 4 = sqrt1# or #x -4 = -sqrt1# #color(green)(x = 1 + 4 = 5# or # color(green)(x = - 1 + 4 = 3#

Savannah Anderson · Answer

undefined To solve the quadratic equation $ x^2 - 8x + 15 = 0 $ using the completing the square method, follow these steps:

1. Move the constant term to the other side of the equation:
$$ x^2 - 8x = -15 $$

2. To complete the square, take half of the coefficient of $ x $ (which is $ -8/2 = -4 $) and square it ($ (-4)^2 = 16 $). Add and subtract this value inside the parentheses:
$$ x^2 - 8x + 16 - 16 = -15 $$

3. Rewrite the expression as a perfect square trinomial and simplify:
$$ (x - 4)^2 - 16 = -15 $$

4. Add 16 to both sides to isolate the perfect square trinomial:
$$ (x - 4)^2 = 1 $$

5. Take the square root of both sides:
$$ x - 4 = \pm \sqrt{1} $$

6. Solve for $ x $:
$$ x - 4 = \pm 1 $$
$$ x = 4 \pm 1 $$

7. The solutions are:
$$ x = 4 + 1 = 5 $$ or $$ x = 4 - 1 = 3 $$