How do you solve #sqrt (3x + 4) + x = 8#?

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Answer 1 · 2018-03-23T11:38:45

#color(blue)(x=16)#

#color(blue)(x=4)#

#sqrt(3x+4)+x=8#

Subtract #x# from both sides:

#sqrt(3x+4)=8-x#

Square both sides:

#3x+4=(8-x)^2#

Expand the bracket:

#3x+4=64-16x+x^2#

Collect like terms:

#x^2-19x+60=0#

Factor:

#x^2-15x-4x+60#

#x(x-15)-4(x-15)#

#(x-15){x-4}#

#(x-16)(x-4)=0#

#x-16=0=>color(blue)(x=16)#

#x-4=0=>color(blue)(x=4)#

Answer 2 · 2018-03-23T11:52:37

Solution: #x=4 , x=15#

#sqrt(3x+4)+x =8 or sqrt(3x+4) =8-x # Squaring both sides

we get, #3x+4 = (8-x)^2 or 3x+4 = x^2-16x+64# or

# x^2-16x+64-3x-4=0 or x^2-19x+60=0 # or

# x^2-15x-4x+60=0 # or

#x(x-15)-4(x-15)=0 or (x-15)(x-4)=0 :.#

Either #x-15=0 :. x=15# , or #x-4=0 :. x=4#

Check : #sqrt(3*4+4)+4 =8 or sqrt16+4=8or 8=8 #

#sqrt(3*15+4)+15 =8 or sqrt49+15=8 or -7+15=8 #

Solution: #x=4 , x=15# [Ans]