1. **State the problem:** Solve the equation $6\sqrt{2x+1} = x^2 - 6x + 26$ for $x$. 2. **Recall the formula and rules:** To solve equations involving square roots, isolate the root term and then square both sides to eliminate the square root. Remember to check for extraneous solutions after squaring. 3. **Isolate the square root:** The root is already isolated on the left side: $6\sqrt{2x+1} = x^2 - 6x + 26$. 4. **Square both sides:** $$\left(6\sqrt{2x+1}\right)^2 = \left(x^2 - 6x + 26\right)^2$$ $$36(2x+1) = (x^2 - 6x + 26)^2$$ 5. **Simplify the left side:** $$72x + 36 = (x^2 - 6x + 26)^2$$ 6. **Expand the right side:** Let $y = x^2 - 6x + 26$, then $$y^2 = (x^2 - 6x + 26)^2$$ Expand: $$= (x^2)^2 - 2 \cdot x^2 \cdot 6x + 2 \cdot x^2 \cdot 26 + (6x)^2 - 2 \cdot 6x \cdot 26 + 26^2$$ More precisely, expand fully: $$(x^2 - 6x + 26)^2 = x^4 - 12x^3 + 88x^2 - 312x + 676$$ 7. **Rewrite the equation:** $$72x + 36 = x^4 - 12x^3 + 88x^2 - 312x + 676$$ 8. **Bring all terms to one side:** $$0 = x^4 - 12x^3 + 88x^2 - 312x + 676 - 72x - 36$$ $$0 = x^4 - 12x^3 + 88x^2 - 384x + 640$$ 9. **Solve the quartic equation:** Try to find rational roots using the Rational Root Theorem. Test $x=4$: $$4^4 - 12(4)^3 + 88(4)^2 - 384(4) + 640 = 256 - 768 + 1408 - 1536 + 640 = 0$$ So, $x=4$ is a root. 10. **Divide polynomial by $(x-4)$:** Using synthetic division or polynomial division, the quotient is: $$x^3 - 8x^2 + 56x - 160$$ 11. **Solve cubic $x^3 - 8x^2 + 56x - 160 = 0$:** Try $x=4$ again: $$64 - 128 + 224 - 160 = 0$$ So, $x=4$ is a root again. 12. **Divide cubic by $(x-4)$:** Quotient: $$x^2 - 4x + 40$$ 13. **Solve quadratic $x^2 - 4x + 40 = 0$:** Discriminant: $$\Delta = (-4)^2 - 4 \cdot 1 \cdot 40 = 16 - 160 = -144 < 0$$ No real roots here. 14. **Real roots found:** $x=4$ (double root). 15. **Check for extraneous solutions:** Plug $x=4$ into original equation: Left side: $$6\sqrt{2(4)+1} = 6\sqrt{9} = 6 \times 3 = 18$$ Right side: $$4^2 - 6 \times 4 + 26 = 16 - 24 + 26 = 18$$ Equal, so $x=4$ is a valid solution. **Final answer:** $$\boxed{4}$$