1. **State the problem:** Solve the equation $$\sqrt{2x} = x - 4$$ for $x$. 2. **Understand the domain:** Since the left side is a square root, the expression inside must be non-negative: $$2x \geq 0 \implies x \geq 0$$. Also, the right side $x - 4$ must be non-negative because a square root is always non-negative: $$x - 4 \geq 0 \implies x \geq 4$$. So the domain is $$x \geq 4$$. 3. **Square both sides to eliminate the square root:** $$\left(\sqrt{2x}\right)^2 = (x - 4)^2$$ $$2x = (x - 4)^2$$ 4. **Expand the right side:** $$2x = x^2 - 8x + 16$$ 5. **Bring all terms to one side to form a quadratic equation:** $$0 = x^2 - 8x + 16 - 2x$$ $$0 = x^2 - 10x + 16$$ 6. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-10$, $c=16$. Calculate the discriminant: $$\Delta = (-10)^2 - 4 \times 1 \times 16 = 100 - 64 = 36$$ Calculate the roots: $$x = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2}$$ So, $$x_1 = \frac{10 + 6}{2} = \frac{16}{2} = 8$$ $$x_2 = \frac{10 - 6}{2} = \frac{4}{2} = 2$$ 7. **Check the solutions against the domain $x \geq 4$:** - $x=8$ satisfies $x \geq 4$. - $x=2$ does not satisfy $x \geq 4$, so discard $x=2$. 8. **Verify the solution $x=8$ in the original equation:** $$\sqrt{2 \times 8} = \sqrt{16} = 4$$ $$8 - 4 = 4$$ Both sides equal 4, so $x=8$ is a valid solution. **Final answer:** $$x = 8$$