Question

solve the following equation.$left(x^{2}+4x+6
ight)^{\frac{3}{2}} + 1 = 0$select the correct answer below, and if necessary, fill in the answer box toa. the solution(s) is/are x = . (use a comma to separate answers asb. there are no solutions.

Explanation:

Step1: Isolate the rational exponent term

$(x^2 + 4x + 6)^{\frac{3}{2}} = -1$

Step2: Rewrite the exponent form

Recall that $a^{\frac{3}{2}} = (\sqrt{a})^3$. So:
$(\sqrt{x^2 + 4x + 6})^3 = -1$

Step3: Analyze the radicand

First, complete the square for $x^2 + 4x + 6$:
$x^2 + 4x + 6 = (x+2)^2 + 2$
Since $(x+2)^2 \geq 0$ for all real $x$, $(x+2)^2 + 2 \geq 2 > 0$. Thus $\sqrt{x^2 + 4x + 6}$ is a positive real number for all real $x$.

Step4: Evaluate the cube of a positive number

The cube of a positive real number is always positive, so $(\sqrt{x^2 + 4x + 6})^3 > 0$. It can never equal $-1$.

Answer:

B. There are no solutions.