Solving Radical Equations with Multiple Radicals page 3

Radical Equations Overview and Principles

• A radical equation is an equation in which a variable is contained inside a radical expression, such as a square root.

• Solving radical equations requires algebraic techniques to isolate radical terms and eliminate them by raising both sides of the equation to a power corresponding to the index of the radical (e.g., squaring both sides for square roots).

• Extraneous Solutions:

  • Squaring both sides of an equation is a non-reversible operation that can introduce extraneous solutions (solutions that emerge during algebraic manipulation but fail to satisfy the original equation).

  • Testing all candidate solutions in the original equation is an essential step in solving radical equations.

Problem 1: Solving an Equation with Two Radicals

• Problem Statement:

  • Solve the equation containing two radicals:   $\sqrt{3x+7} + \sqrt{x+2} = 1$

• Step 1: Isolate One of the Radical Expressions

  • Subtract $\sqrt{x+2}$ from both sides to isolate $\sqrt{3x+7}$ on the left side:   $\sqrt{3x+7} = 1 - \sqrt{x+2}$

• Step 2: Square Both Sides

  • Square both sides to remove the outer square root on the left side:   $(\sqrt{3x+7})^2 = (1 - \sqrt{x+2})^2$

  • Apply the binomial expansion formula $(a - b)^2 = a^2 - 2ab + b^2$ to the right side:   $3x + 7 = 1^2 - 2(1)(\sqrt{x+2}) + (\sqrt{x+2})^2$   $3x + 7 = 1 - 2\sqrt{x+2} + x + 2$

  • Combine like terms on the right side:   $3x + 7 = x + 3 - 2\sqrt{x+2}$

• Step 3: Isolate the Remaining Radical Term

  • Move non-radical terms to the left side by subtracting $x$ and $3$ from both sides:   $3x + 7 - x - 3 = -2\sqrt{x+2}$   $2x + 4 = -2\sqrt{x+2}$

  • Divide both sides by $2$ to simplify the coefficients:   $x + 2 = -\sqrt{x+2}$

• Step 4: Square Both Sides a Second Time

  • Square both sides to eliminate the remaining square root:   $(x + 2)^2 = (-\sqrt{x+2})^2$

  • Expand the left side using the perfect square trinomial formula $(a + b)^2 = a^2 + 2ab + b^2$:   $x^2 + 4x + 4 = x + 2$

• Step 5: Solve the Resulting Quadratic Equation

  • Rearrange terms to write the equation in standard quadratic form $ax^2 + bx + c = 0$:   $x^2 + 4x + 4 - x - 2 = 0$   $x^2 + 3x + 2 = 0$

  • Factor the quadratic polynomial:   $(x + 1)(x + 2) = 0$

  • Set each factor to zero to solve for $x$:   $x + 1 = 0 \quad \text{or} \quad x + 2 = 0$   $x = -1 \quad \text{or} \quad x = -2$

• Step 6: Check Potential Solutions in the Original Equation

  • Test $x = -1$:

  • Substitute $x = -1$ into $\sqrt{3x+7} + \sqrt{x+2} = 1$:     $\sqrt{3(-1)+7} + \sqrt{-1+2} = \sqrt{-3+7} + \sqrt{1}$     $\sqrt{4} + \sqrt{1} = 2 + 1 = 3$

  • Since $3 \neq 1$, $x = -1$ is an extraneous solution and is rejected.

  • Test $x = -2$:

  • Substitute $x = -2$ into $\sqrt{3x+7} + \sqrt{x+2} = 1$:     $\sqrt{3(-2)+7} + \sqrt{-2+2} = \sqrt{-6+7} + \sqrt{0}$     $\sqrt{1} + 0 = 1 + 0 = 1$

  • Since $1 = 1$, $x = -2$ is a valid solution.

• Conclusion for Problem 1:

  • The solution is $x = -2$.

Problem 2: Solving an Equation with Radicals on Opposite Sides

• Problem Statement:

  • Solve the equation:   $\sqrt{m} + 2 = \sqrt{m+12}$

  • Answer format requirement: Provide the solution as an integer or as a simplified fraction in the form $\frac{A}{B}$.

• Step 1: Square Both Sides of the Equation

  • Square both sides directly, as $\sqrt{m+12}$ is already isolated on the right side:   $(\sqrt{m} + 2)^2 = (\sqrt{m+12})^2$

  • Expand the left side using the binomial expansion formula $(a + b)^2 = a^2 + 2ab + b^2$:   $(\sqrt{m})^2 + 2(\sqrt{m})(2) + 2^2 = m + 12$   $m + 4\sqrt{m} + 4 = m + 12$

• Step 2: Isolate the Radical Term

  • Subtract $m$ from both sides:   $4\sqrt{m} + 4 = 12$

  • Subtract $4$ from both sides:   $4\sqrt{m} = 8$

  • Divide both sides by $4$:   $\sqrt{m} = 2$

• Step 3: Solve for the Variable

  • Square both sides to solve for $m$:   $(\sqrt{m})^2 = 2^2$   $m = 4$

• Step 4: Check the Solution

  • Substitute $m = 4$ into the original equation $\sqrt{m} + 2 = \sqrt{m+12}$:

  • Left-Hand Side (LHS):     $\sqrt{4} + 2 = 2 + 2 = 4$

  • Right-Hand Side (RHS):     $\sqrt{4+12} = \sqrt{16} = 4$

  • Since $\text{LHS} = \text{RHS} = 4$, $m = 4$ is a valid solution.

• Conclusion for Problem 2:

  • The final answer is $m = 4$.