1. Solve the equation $x + \sqrt{x} = \frac{6}{25}$. First, let $y = \sqrt{x}$, then $x = y^2$. Substitute: $$y^2 + y = \frac{6}{25}$$ Rearranged: $$y^2 + y - \frac{6}{25} = 0$$ Use quadratic formula: $$y = \frac{-1 \pm \sqrt{1 + 4 \times \frac{6}{25}}}{2} = \frac{-1 \pm \sqrt{1 + \frac{24}{25}}}{2} = \frac{-1 \pm \sqrt{\frac{49}{25}}}{2} = \frac{-1 \pm \frac{7}{5}}{2}$$ Two solutions for $y$: $$y_1 = \frac{-1 + \frac{7}{5}}{2} = \frac{-\frac{5}{5} + \frac{7}{5}}{2} = \frac{\frac{2}{5}}{2} = \frac{1}{5}$$ $$y_2 = \frac{-1 - \frac{7}{5}}{2} = \frac{-\frac{5}{5} - \frac{7}{5}}{2} = \frac{-\frac{12}{5}}{2} = -\frac{6}{5}$$ Since $y=\sqrt{x} \geq 0$, discard negative solution. So $y = \frac{1}{5}$, then $$x = y^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25}$$ 2. Solve $x^2 - 6x + 9 = 4\sqrt{x^2 - 6x + 6}$. Notice $x^2 - 6x + 9 = (x-3)^2$ and $x^2 - 6x + 6$ inside the root. Let $A = x^2 - 6x + 6$. Rewrite: $$(x-3)^2 = 4\sqrt{A}$$ Square both sides: $$((x-3)^2)^2 = 16A$$ $$ (x-3)^4 = 16(x^2 - 6x + 6)$$ Expand and simplify both sides (lengthy process): Alternatively, check candidates from the root expression. Consider $\sqrt{A} = t$, so original equation is $(x-3)^2 = 4 t$ and $t^2 = x^2 - 6x + 6$. Substitute $t = \frac{(x-3)^2}{4}$ into $t^2 = x^2 - 6x + 6$: $$\left(\frac{(x-3)^2}{4}\right)^2 = x^2 - 6x + 6$$ $$\frac{(x-3)^4}{16} = x^2 - 6x + 6$$ Multiply both sides by 16: $$(x-3)^4 = 16x^2 - 96x + 96$$ Expand $(x-3)^4$ using binomial expansion: $$(x-3)^2 = x^2 - 6x + 9$$ $$(x-3)^4 = (x^2 - 6x + 9)^2 = x^4 - 12x^3 + 54x^2 - 108x + 81$$ Set equal: $$x^4 - 12x^3 + 54x^2 - 108x + 81 = 16x^2 - 96x + 96$$ Bring all terms to one side: $$x^4 - 12x^3 + 54x^2 - 108x + 81 - 16x^2 + 96x - 96 = 0$$ $$x^4 - 12x^3 + (54 - 16)x^2 + (-108 + 96)x + (81 - 96) = 0$$ $$x^4 - 12x^3 + 38x^2 - 12x - 15 = 0$$ Solve quartic: Try rational roots: candidates ±1, ±3, ±5, ±15 Test $x = 3$: $$81 - 324 + 342 - 36 - 15 = 48 \neq 0$$ Test $x = 1$: $$1 - 12 + 38 - 12 - 15 = 0$$ So $x=1$ is root. Divide polynomial by $(x-1)$: Using synthetic division gives: $$x^3 - 11x^2 + 27x + 15 = 0$$ Try roots again: $x=3$: $27 - 99 + 81 + 15 = 24 \neq 0$ $x=5$: $125 - 275 + 135 + 15 = 0$ So $x=5$ is root. Divide further by $(x-5)$: $$x^2 - 6x - 3= 0$$ Solve quadratic: $$x = \frac{6 \pm \sqrt{36 + 12}}{2} = \frac{6 \pm \sqrt{48}}{2} = 3 \pm 2\sqrt{3}$$ Check original equation validity for roots (presence of square root in original problem requires non-negativity inside radical), keep valid roots $x=1, 5, 3+2\sqrt{3}$. 3. Solve $x^2 - 4x - 12\sqrt{x^2 - 4x + 19} + 51 = 0$. Let $t = \sqrt{x^2 - 4x + 19}$. Rewrite: $$x^2 - 4x + 51 = 12t$$ Square both sides: $$(x^2 - 4x + 51)^2 = 144 (x^2 - 4x + 19)$$ Expanding left: $$x^4 - 8x^3 + 106x^2 - 408x + 2601 = 144x^2 - 576x + 2736$$ Bring all to one side: $$x^4 - 8x^3 + 106x^2 - 408x + 2601 - 144x^2 + 576x - 2736 = 0$$ $$x^4 - 8x^3 - 38x^2 + 168x - 135 = 0$$ Try rational root test ±1, ±3, ±5, ±9,... Test $x=3$: $$81 - 216 - 342 + 504 - 135 = -108 \neq 0$$ Test $x=5$: $$625 - 1000 - 950 + 840 -135 = -620 \neq 0$$ Try $x=9$: $$6561 - 5832 - 3078 + 1512 - 135 = 1028 \neq 0$$ Try to solve via substitution or numerical methods. This is a quartic; approximate solutions are appropriate. 4. Solve $\frac{\sqrt{12 - x}}{5} = \frac{3}{2 + \sqrt{12 - x}}$. Cross multiply: $$(2 + \sqrt{12 - x}) \sqrt{12 - x} = 15$$ Let $t = \sqrt{12 - x}$, then: $$(2 + t) t = 15$$ $$2t + t^2 = 15$$ Rearranged: $$t^2 + 2t - 15 = 0$$ Solve quadratic: $$t = \frac{-2 \pm \sqrt{4 + 60}}{2} = \frac{-2 \pm 8}{2}$$ Solutions: $$t = 3 \quad \text{or} \quad t = -5$$ Since $t = \sqrt{12 - x} \geq 0$, discard $t = -5$. So $t = 3$, then $$\sqrt{12 - x} = 3 \implies 12 - x = 9 \implies x = 3$$ 5. Solve $x^{10} - 33x^5 + 32 = 0$. Let $y = x^5$, rewriting: $$y^2 - 33y + 32 = 0$$ Solve quadratic: $$y = \frac{33 \pm \sqrt{1089 - 128}}{2} = \frac{33 \pm \sqrt{961}}{2} = \frac{33 \pm 31}{2}$$ Two solutions: $$y_1 = \frac{33 + 31}{2} = 32, \, y_2 = \frac{33 - 31}{2} = 1$$ Solve $x^5 = 32$: $$x = 2$$ Solve $x^5 = 1$: $$x = 1$$ **Final answers:** 1. $x = \frac{1}{25}$ 2. $x = 1, 5, 3 + 2\sqrt{3}$ 3. The quartic $x^4 - 8x^3 - 38x^2 + 168x - 135 = 0$ requires numerical methods for roots. 4. $x = 3$ 5. $x = 1, 2$