Question:

**Step 1: Factor the expressions under radicals** The expressions under the radicals are factored as follows: $$x^2 + 4x - 21 = (x + 7)(x - 3)$$ $$x^2 - x - 6 = (x - 3)(x + 2)$$ $$6x^2 - 5x - 39 = (x - 3)(6x + 13)$$ Substituting these into the original equation yields: $$\sqrt{(x + 7)(x - 3)} + \sqrt{(x - 3)(x + 2)} = \sqrt{(x - 3)(6x + 13)}$$ **Step 2: Determine domain restrictions** For the square roots to be defined, all radicands must be non-negative: 1. $(x + 7)(x - 3) \ge 0 \implies x \le -7 \text{ or } x \ge 3$ 2. $(x - 3)(x + 2) \ge 0 \implies x \le -2 \text{ or } x \ge 3$ 3. $(x - 3)(6x + 13) \ge 0 \implies x \le -\frac{13}{6} \text{ or } x \ge 3$ The intersection of these three conditions defines the domain of the equation: $$x \le -7 \text{ or } x \ge 3$$ **Step 3: Solve the equation** The equation is $\sqrt{(x + 7)(x - 3)} + \sqrt{(x - 3)(x + 2)} = \sqrt{(x - 3)(6x + 13)}$. We can factor out $\sqrt{|x-3|}$ from each term. Case 1: $x - 3 = 0$ If $x = 3$, the equation becomes: $$\sqrt{(3 + 7)(0)} + \sqrt{(0)(3 + 2)} = \sqrt{(0)(6(3) + 13)}$$ $$\sqrt{0} + \sqrt{0} = \sqrt{0}$$ $$0 = 0$$ Thus, $x = 3$ is a solution. This value is within the determined domain ($x \ge 3$). Case 2: $x - 3 > 0$ (i.e., $x > 3$) In this case, $\sqrt{x-3}$ is a real, non-zero term. We can divide both sides by $\sqrt{x-3}$: $$\sqrt{x+7} + \sqrt{x+2} = \sqrt{6x+13}$$ Square both sides: $$(x+7) + (x+2) + 2\sqrt{(x+7)(x+2)} = 6x+13$$ $$2x+9 + 2\sqrt{x^2+9x+14} = 6x+13$$ Isolate the radical term: $$2\sqrt{x^2+9x+14} = 4x+4$$ $$\sqrt{x^2+9x+14} = 2x+2$$ For this equation to hold, we must have $2x+2 \ge 0$, which implies $x \ge -1$. Since we are in Case 2 where $x > 3$, this condition is satisfied. Square both sides again: $$x^2+9x+14 = (2x+2)^2$$ $$x^2+9x+14 = 4x^2+8x+4$$ Rearrange into a quadratic equation: $$3x^2 - x - 10 = 0$$ Solve using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-10)}}{2(3)}$$ $$x = \frac{1 \pm \sqrt{1 + 120}}{6}$$ $$x = \frac{1 \pm \sqrt{121}}{6}$$ $$x = \frac{1 \pm 11}{6}$$ This yields two potential solutions: $$x_1 = \frac{1+11}{6} = 2$$ $$x_2 = \frac{1-11}{6} = -\frac{5}{3}$$ Neither $x=2$ nor $x=-5/3$ satisfy the condition for Case 2 ($x > 3$). Therefore, there are no solutions from Case 2. Case 3: $x - 3 < 0$ (i.e., $x < 3$) From the domain restrictions, we must have $x \le -7$. In this case, $x-3 < 0$, $x+7 \le 0$, $x+2 < 0$, and $6x+13 < 0$. Let $x-3 = -A$, $x+7 = -B$, $x+2 = -C$, and $6x+13 = -D$, where $A, B, C, D > 0$. The equation becomes: $$\sqrt{(-B)(-A)} + \sqrt{(-A)(-C)} = \sqrt{(-A)(-D)}$$ $$\sqrt{AB} + \sqrt{AC} = \sqrt{AD}$$ Since $A = -(x-3) > 0$, we can divide by $\sqrt{A}$: $$\sqrt{B} + \sqrt{C} = \sqrt{D}$$ Substitute back: $$\sqrt{-(x+7)} + \sqrt{-(x+2)} = \sqrt{-(6x+13)}$$ Let $y = -x$. Since $x \le -7$, we have $y \ge 7$. The equation transforms to: $$\sqrt{y-7} + \sqrt{y-2} = \sqrt{6y-13}$$ Square both sides: $$(y-7) + (y-2) + 2\sqrt{(y-7)(y-2)} = 6y-13$$ $$2y-9 + 2\sqrt{y^2-9y+14} = 6y-13$$ Isolate the radical term: $$2\sqrt{y^2-9y+14} = 4y-4$$ $$\sqrt{y^2-9y+14} = 2y-2$$ For this equation to hold, we must have $2y-2 \ge 0$, which implies $y \ge 1$. This condition is satisfied since $y \ge 7$. Square both sides again: $$y^2-9y+14 = (2y-2)^2$$ $$y^2-9y+14 = 4y^2-8y+4$$ Rearrange into a quadratic equation: $$3y^2 + y - 10 = 0$$ Solve using the quadratic formula: $$y = \frac{-1 \pm \sqrt{1^2 - 4(3)(-10)}}{2(3)}$$ $$y = \frac{-1 \pm \sqrt{1 + 120}}{6}$$ $$y = \frac{-1 \pm 11}{6}$$ This yields two potential solutions for $y$: $$y_1 = \frac{-1+11}{6} = \frac{10}{6} = \frac{5}{3}$$ $$y_2 = \frac{-1-11}{6} = -2$$ Neither $y=5/3$ nor $y=-2$ satisfy the condition for Case 3 ($y \ge 7$). Therefore, there are no solutions from Case 3. Combining all cases, the only solution is $x=3$. **Step 4: Verify the solution** Substitute $x=3$ into the original equation: $$\sqrt{3^2+4(3)-21}+\sqrt{3^2-3-6}=\sqrt{6(3)^2-5(3)-39}$$ $$\sqrt{9+12-21}+\sqrt{9-3-6}=\sqrt{54-15-39}$$ $$\sqrt{0}+\sqrt{0}=\sqrt{0}$$ $$0+0=0$$ $$0=0$$ The solution $x=3$ is verified.