1:
x + 2 = \(\sqrt{3x+10}\)
Square both sides of the equation.
(x + 2)2 = 3x + 10
(x + 2)(x + 2) = 3x + 10
x2 + 4x + 4 = 3x + 10
Subtract 3x from both sides and subtract 10 from both sides.
x2 + x - 6 = 0
Factor the left side.
(x + 3)(x - 2) = 0
Set each factor equal to zero and solve for x .
x = -3 or x = 2
Plug in -3 into the original equation to see if it makes it true.
-3 + 2 = \(\sqrt{3(-3)+10}\) ?
-1 = \(\sqrt{1}\) ?
-1 = 1 False.
Plug in 2 into the original equation to see if it makes it true.
2 + 2 = \(\sqrt{3(2)+10}\) ?
4 = \(\sqrt{16}\) ?
4 = 4 True.
So....
(a) 2 is the solution.
(b) -3 is the extraneous solution, because it does not make the given equation true.
(c) And this can probably explain what an extraneous solution is better than I can.
First we can see that one of the lines has the equation y = x , and the shade is below this line,
so y ≤ x is one of the inequalities.
Then the dotted line has the equation y = -2x - 2 , and the shade is above this line
so one of the inequalities must be y > -2x - 2
The only option that has those two is the last one. The last option is the correct one.
Here's a graph: https://www.desmos.com/calculator/bjnoosya26