1. **State the problem:** Solve the equation $$\sqrt{2x + 5} + 2\sqrt{x + 6} = 5$$ for $x$. 2. **Isolate one of the square root terms:** Let's isolate $$\sqrt{2x + 5}$$: $$\sqrt{2x + 5} = 5 - 2\sqrt{x + 6}$$ 3. **Square both sides to eliminate the square root on the left:** $$\left(\sqrt{2x + 5}\right)^2 = \left(5 - 2\sqrt{x + 6}\right)^2$$ $$2x + 5 = 25 - 20\sqrt{x + 6} + 4(x + 6)$$ 4. **Simplify the right side:** $$2x + 5 = 25 - 20\sqrt{x + 6} + 4x + 24$$ $$2x + 5 = 4x + 49 - 20\sqrt{x + 6}$$ 5. **Rearrange to isolate the square root term:** $$-20\sqrt{x + 6} = 2x + 5 - 4x - 49$$ $$-20\sqrt{x + 6} = -2x - 44$$ 6. **Divide both sides by -20:** $$\sqrt{x + 6} = \frac{-2x - 44}{-20} = \frac{2x + 44}{20}$$ $$\sqrt{x + 6} = \frac{2x + 44}{20}$$ 7. **Square both sides again to eliminate the square root:** $$\left(\sqrt{x + 6}\right)^2 = \left(\frac{2x + 44}{20}\right)^2$$ $$x + 6 = \frac{(2x + 44)^2}{400}$$ 8. **Multiply both sides by 400 to clear the denominator:** $$400(x + 6) = (2x + 44)^2$$ $$400x + 2400 = (2x + 44)^2$$ 9. **Expand the right side:** $$(2x + 44)^2 = (2x)^2 + 2 \times 2x \times 44 + 44^2 = 4x^2 + 176x + 1936$$ 10. **Rewrite the equation:** $$400x + 2400 = 4x^2 + 176x + 1936$$ 11. **Bring all terms to one side:** $$0 = 4x^2 + 176x + 1936 - 400x - 2400$$ $$0 = 4x^2 - 224x - 464$$ 12. **Divide the entire equation by 4 to simplify:** $$0 = x^2 - 56x - 116$$ 13. **Use the quadratic formula to solve for $x$:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-56$, and $c=-116$. 14. **Calculate the discriminant:** $$b^2 - 4ac = (-56)^2 - 4(1)(-116) = 3136 + 464 = 3600$$ 15. **Calculate the roots:** $$x = \frac{56 \pm \sqrt{3600}}{2} = \frac{56 \pm 60}{2}$$ 16. **Find the two possible solutions:** - $$x = \frac{56 + 60}{2} = \frac{116}{2} = 58$$ - $$x = \frac{56 - 60}{2} = \frac{-4}{2} = -2$$ 17. **Check for extraneous solutions by substituting back into the original equation:** - For $x=58$: $$\sqrt{2(58) + 5} + 2\sqrt{58 + 6} = \sqrt{121} + 2\sqrt{64} = 11 + 2 \times 8 = 11 + 16 = 27 \neq 5$$ So $x=58$ is extraneous. - For $x=-2$: $$\sqrt{2(-2) + 5} + 2\sqrt{-2 + 6} = \sqrt{1} + 2\sqrt{4} = 1 + 2 \times 2 = 1 + 4 = 5$$ This satisfies the equation. **Final answer:** $$x = -2$$