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

2 Answers
ali ergin
Mar 11, 2016

x={3,12}x={3,12}

Explanation:

sqrt(3x)+8=x+2√3x+8=x+2
sqrt(3x)=x+2-8√3x=x+2−8
sqrt(3x)=x-6√3x=x−6
(sqrt(3x))^2=(x-6)^2(√3x)2=(x−6)2
3x=x^2-12x+363x=x2−12x+36
x^2-12x-3x+36=0x2−12x−3x+36=0
x^2-15x+36=0x2−15x+36=0
(x-12)(x-3)=0(x−12)(x−3)=0
"if (x-12)=0 then x=12"if (x-12)=0 then x=12
"if (x-3)=0 then x=3"if (x-3)=0 then x=3
x={3,12}x={3,12}

Nghi N.
Mar 11, 2016

3 and 12

Explanation:

sqrt(3x) + 8 = x + 2√3x+8=x+2
Isolate the radical term.
sqrt(3x) = x - 6√3x=x−6
Square both sides:
3x = (x - 6)^2 = x^2 - 12x + 363x=(x−6)2=x2−12x+36
x^2 - 15x + 36 = 0x2−15x+36=0
Find 2 numbers (real roots) knowing sum (15 = -b) and product (c = 36). They are: 3 and 12.

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