How to Solve Equations Including Radicals

Solve radical equations by converting the radical to a polynomial.

Radical equations in algebra contain a radical or square root term. Solve radical equations by isolating the radical term and squaring both sides of the equation to transform the radical equation into a polynomial equation. When solving radical equations, remember that squaring both sides of an equation can create false solutions. Always plug your answers into the original equation to verify that they produce a valid equation.

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Instructions

- 1
  
  Move the radical term to one side of the equation and all other terms to the other side of the equation. For example, subtract 1 from both sides of the equation √(2x + 5) + 1 = 2x to get the equation √(2x + 5) = 2x - 1.
- 2
  
  Square both sides of the equation to cancel out the radical and turn the radical equation into a polynomial equation. In the above example, square both sides of the equation to get √(2x + 5) = (2x - 1)^2, which simplifies to 2x + 5 = (2x - 1)^2.
- 3
  
  Use the FOIL method to expand the expression on the other side of the equation if there are multiple terms. FOIL means first, outside, inside, last. In the above example, use FOIL to expand (2x - 1)^2 to 4x^2 - 4x + 1.
- 4
  
  Rewrite the polynomial equation in standard form, Ax^2 + Bx + C = 0, by moving all terms to one side of the equation, combining like terms and ordering terms by descending degree. In the example, the equation 2x + 5 = 4x^2 - 4x + 1 is rewritten as 4x^2 - 6x - 4 = 0 in standard form.
- 5
  
  Solve the polynomial equation using factoring techniques or the quadratic equation. The solutions to the equation 4x^2 - 6x - 4 can be found by factoring the equation to get 2(2x + 1)(x - 2) = 0. The solutions to the equation are the zeros of the factors: x = -1/2 and x = 2.
- 6
  
  Verify the solutions by plugging them back into the original equation. Plugging the value x = 2 yields the equation √(2*2 + 5) = 2*2 - 1, or √9 = 3. Plugging the value x = -1/2 yields the equation √2*(-1/2) + 5 = 2*(-1/2) - 1, or √4 = -2 which is invalid. The solution to the equation is therefore x = 2 only.