Question

How do you find all the solutions for x in the equation x+ <22-3x> =6
(Everything in <> is square rooted)

Answer 1

\[\implies \sqrt{22-3x} = 6-x\]\[\implies 22-3x = (6-x)^2; \qquad x \le \frac{22}{3}\]\[\implies 22 - 3x = 36 - 12x + x^2; \qquad x \le \frac{22}{3}\]\[\implies x^2 - 9x + 14 = 0; \qquad x \le \frac{22}{3}\]\[\implies (x-7)(x-2) = 0; \qquad x \le \frac{22}{3}\]\[\implies x \in \{2,7\}\]

Answer 2

we all want 6-x>0
6>x
or
x<6

Answer 3

You can disregard the x=7 solution though, because if you substitute 7 back in the original problem:
7 + √(22 - 3(7)) = 6
7 + √(22 - 21) = 6
7 + √(1) = 6
7 + 1 = 6

Answer 4

also*

Answer 5

we want 6-x>0 since sqrt(something) has domain something>0

Answer 6

something >=0

Answer 7

6-x>=0
x<=6

Answer 8

so just 2 is what i'm saying

Answer 9

since that is only element that was less than or equal to 6

Answer 10

Oh, good point. Forgot to keep track of that restriction.