How do you solve #sqrt(x + 3) = 2x + 1#?

1 Answer
Aug 10, 2015

Answer:

#x= (-3+-sqrt(41))/8#

Explanation:

Given #sqrt(x+3) = 2x+1#

Square both sides:
#color(white)("XXXX")##x+3 = 4x^2+4x+1#

Subtract #(x+3)# from both sides (and reverse sides)
#color(white)("XXXX")##4x^2+3x-2=0#

Using the quadratic formula (see below)
#color(white)("XXXX")##x= (-3+-sqrt(41))/8#

Quadratic formula:
#color(white)("XXXX")#for any quadratic in the general form:
#color(white)("XXXX")##color(white)("XXXX")##ax^2+bx+c=0#
#color(white)("XXXX")#the solutions are given by the formula:
#color(white)("XXXX")##color(white)("XXXX")##x = (-b+-sqrt(b^2-4ac))/(2a)#