Solving Radical Equations with Two Radicals

Steps to Solve the Equation:

1. Separate Radicals:
   • Place one radical on each side of the equation.
2. Raise to Power of Index:
   • Square both sides (since it's a square root).
3. Simplify:
   • Simplify the resulting equation.
   • Isolate any remaining radicals on one side.
4. Repeat Raising and Simplifying:
   • If there is still a radical, repeat the process of squaring and simplifying until radicals are eliminated.
5. Solve the Resulting Equation:
   • Solve the equation obtained after eliminating radicals.
6. Verify Solutions:
   • Check potential solutions in the original equation to verify they are not extraneous.

Example Problem:

• Original Equation: ( \sqrt{2x + 6} = \sqrt{x + 4} + 1 )

Solution Process:

• Square Both Sides:

  • Left: ((\sqrt{2x + 6})^2 = 2x + 6)
  • Right: ((\sqrt{x + 4} + 1)^2) requires expansion:
    • (= (\sqrt{x + 4})^2 + 2(\sqrt{x + 4})(1) + 1)
    • Simplifies to (x + 4 + 2\sqrt{x+4} + 1)

• Simplified Equation:

  • (2x + 6 = x + 5 + 2\sqrt{x + 4})
  • Isolate the remaining square root:
    • Subtract (x) and 5 from both sides:
    • (x + 1 = 2\sqrt{x + 4})

• Square Again to Remove Radical:

  • Divide by 2:
    • ((x + 1)/2 = \sqrt{x + 4})
  • Square both sides:
    • (((x + 1)/2)^2 = x + 4)
    • ((x^2 + 2x + 1) / 4 = x + 4)
  • Clear fractions by multiplying by 4:
    • (x^2 + 2x + 1 = 4x + 16)

• Form Quadratic Equation:

  • Subtract (4x + 16) to set it to zero:
    • (x^2 - 2x - 15 = 0)
  • Factorize:
    • ((x - 5)(x + 3) = 0)

• Potential Solutions:

  • (x = 5) and (x = -3)

Verification:

• Check Each Solution:
  • (x = 5):
    • (\sqrt{2(5) + 6} = \sqrt{16} = 4)
    • (\sqrt{5 + 4} + 1 = \sqrt{9} + 1 = 3 + 1 = 4)
    • Solution is valid.
  • (x = -3):
    • (\sqrt{2(-3) + 6} = \sqrt{0} = 0)
    • (\sqrt{-3 + 4} + 1 = \sqrt{1} + 1 = 1 + 1 = 2)
    • Not valid; (x = -3) is extraneous.

Conclusion:

• Final Solution:
  • The valid solution is (x = 5).

This process ensures that all potential solutions are verified against the original equation to eliminate any extraneous solutions.