How do you solve $\sqrt{3 x} + 8 = x + 2$ $sqrt(3x)+8=x+2$ ?

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How do you solve $\sqrt{3 x} + 8 = x + 2$ $sqrt(3x)+8=x+2$ ?

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Answer 1 · 2016-03-11T15:17:24

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

Explanation:

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

Answer 2 · 2016-03-11T15:19:06

3 and 12

Explanation:

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

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How do you solve $\sqrt{3 x} + 8 = x + 2$ $sqrt(3x)+8=x+2$ ?

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