How do you solve #(x-3)^2 = 5#?

2 Answers
Jun 27, 2018

Answer:

See a solution process below:

Explanation:

First, put the equation in standard quadratic form:

#(x - 3)^2 = 5#

#(x - 3)(x - 3) = 5#

#x^2 - 3x - 3x + 9 = 5#

#x^2 - 6x + 9 = 5#

#x^2 - 6x + 9 - color(red)(5) = 5 - color(red)(5)#

#x^2 - 6x + 4 = 0#

We can now use the quadratic equation to solve this problem:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(1)# for #color(red)(a)#

#color(blue)(-6)# for #color(blue)(b)#

#color(green)(4)# for #color(green)(c)# gives:

#x = (-color(blue)(-6) +- sqrt(color(blue)(-6)^2 - (4 * color(red)(1) * color(green)(4))))/(2 * color(red)(1))#

#x = (6 +- sqrt(36 - 16))/2#

#x = (6 +- sqrt(20))/2#

#x = (6 - sqrt(4 * 5))/2#; #x = (6 + sqrt(4 * 5))/2#

#x = (6 - sqrt(4)sqrt(5))/2#; #x = (6 + sqrt(4)sqrt(5))/2#

#x = (6 - 2sqrt(5))/2#; #x = (6 + 2sqrt(5))/2#

#x = 6/2 - (2sqrt(5))/2#; #x = 6/2 + (2sqrt(5))/2#

#x = 3 - sqrt(5)#; #x = 3 + sqrt(5)#

Jun 27, 2018

Answer:

#x=3+sqrt5# and #x=3-sqrt5#

Explanation:

Let's take the square root of both sides. This gives us

#x-3=+-sqrt5#

We can add #3# to both sides to get

#x=3+sqrt5# and #x=3-sqrt5#

Hope this helps!