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Solving Radical Equations

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Presentation on theme: "Solving Radical Equations"— Presentation transcript:

1 Solving Radical Equations

2 Extraneous Solutions Power Rule (text only talks about squaring, but applies to other powers, as well). If both sides of an equation are raised to the same power, solutions of the new equation contain all the solutions of the original equation, but might also contain additional solutions. A proposed solution of the new equation that is NOT a solution of the original equation is an extraneous solution.

3 Solving Radical Equations
Steps for Solving Radical Equations Isolate one radical on one side of equal sign. Raise each side of the equation to a power equal to the index of the isolated radical, and simplify. (With square roots, the index is 2, so square both sides.) If equation still contains a radical, repeat steps 1 and 2. If not, solve equation. Check proposed solutions in the original equation.

4 that contains a radical.
A radical equation is an equation that contains a radical.

5 mathematical definitions
Did you ever notice how much sense mathematical definitions usually make?

6 in solving most equations.
The goal in solving radical equations is the same as the goal in solving most equations.

7 We need to isolate the variable.

8 But there is only one way
to move the variable out from under the square root sign.

9 We need to square the radical expression.

10 And, because it is an equation, what we do to one side,

11 we have to do to the other.
And, because it is an equation, what we do to one side, we have to do to the other.

12 And now, we need to simplify:

13 (Even if n is an expression)
Remember, no matter what n is. (Even if n is an expression)

14 So we have:

15 So we have: So we have:

16 Solving Radical Equations
Solve the following radical equation. Substitute into the original equation. true So the solution is x = 24.

17 Solving Radical Equations
Solve the following radical equation. Substitute into the original equation. Does NOT check, since the left side of the equation is asking for the principal square root. So the solution is .


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