How do you solve #sqrt(3x)+8=x+2#?

2 Answers
Mar 11, 2016

Answer:

#x={3,12}#

Explanation:

#sqrt(3x)+8=x+2#
#sqrt(3x)=x+2-8#
#sqrt(3x)=x-6#
#(sqrt(3x))^2=(x-6)^2#
#3x=x^2-12x+36#
#x^2-12x-3x+36=0#
#x^2-15x+36=0#
#(x-12)(x-3)=0#
#"if (x-12)=0 then x=12"#
#"if (x-3)=0 then x=3"#
#x={3,12}#

Mar 11, 2016

Answer:

3 and 12

Explanation:

#sqrt(3x) + 8 = x + 2#
Isolate the radical term.
#sqrt(3x) = x - 6#
Square both sides:
#3x = (x - 6)^2 = x^2 - 12x + 36#
#x^2 - 15x + 36 = 0#
Find 2 numbers (real roots) knowing sum (15 = -b) and product (c = 36). They are: 3 and 12.