Question

Find the solution of 3 times the square root of the quantity of x plus 6 equals negative 12, and determine if it is an extraneous solution.

Answer 1

For this case we have the following equation:
 $3 \sqrt{x+6}=-12$
 Rewriting we have:
 $\sqrt{x+6}= \frac{-12}{3}$
 $\sqrt{x+6}=-4$
 We raise both members of the equation to the square:
 $(\sqrt{x+6})^2=(-4)^2$
 Rewriting we have:
 $x+6=16$
 $x=16-6$
 $x=10$
 It's a extraneous solution because equality is not met by substituting x = 10 in the original equation:
 $3 \sqrt{10+6}=-12$
 $3 \sqrt{16}=-12$
 $3(4)=-12$
 $12=-12$
 Answer:
 The solution is:
 $x=10$
 It's a extraneous solution