How do you solve #sqrt( 11x+3) -2x =0#?

2 Answers
Jun 6, 2018

Answer:

#x=-1/4# and #x=3#

Explanation:

#sqrt( 11x+3) -2x =0#

Isolate the radical:

#sqrt( 11x+3)=2x#

#(sqrt( 11x+3))^2=(2x)^2#

#11x+3=4x^2#

#0=4x^2 -11x-3#

Factor:

#(4 x + 1) (x - 3) =0#

#x=-1/4# and #x=3#

Jun 6, 2018

Answer:

#x = (13 + sqrt217)/8 or (13 - sqrt217)/8#

Explanation:

#sqrt(11x + 3) - 2x = 0#

Add #2x# to both sides;

#sqrt(11x + 3) - 2x + 2x = 0 + 2x#

#sqrt(11x+3) = 2x#

Square both sides;

#sqrt(11x + 3)^2 = (2x)^2#

#11x + 3 = 4x^2#

Rearranging the equation;

#4x^2 - 11x - 3 = 0#

Using the Quadratic Formula;

#x= (-b+-sqrt(b^2 - 4ac))/(2a)#

Where;

#a= 4#

#b= -11#

#c =-3#

Substituting the values into the equation..

#x = (-(-13) +- sqrt((-13)^2- 4(4)(-3)))/(2(4))#

#x= (13 +- sqrt(169 + 48))/8#

#x = (13 +- sqrt217)/8#

#x = (13 + sqrt217)/8 or (13 - sqrt217)/8#