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Friday 1st of July Home Square Roots and Radical Expressions Solving Radical Equations Simplifying Radical Expressions Irrational Numbers in General and Square Roots in Particular Roots of Polynomials Simplifying Radical Expressions Exponents and Radicals Products and Quotients Involving Radicals Roots of Quadratic Equations Radical Expressions Radicals and Rational Exponents Find Square Roots and Compare Real Numbers Radicals Radicals and Rational Exponents Theorems on the Roots of Polynomial Equations SYNTHETIC DIVISION AND BOUNDS ON ROOTS Simplifying Radical Expressions Exponents and Radicals Properties of Exponents and Square Roots Solving Radical Equations Rational Exponents and Radicals,Rationalizing Denominators Rational Exponents and Radicals,Rationalizing Denominators Quadratic Roots Exponents and Roots Multiplying Radical Expressions Exponents and Radicals Solving Radical Equations Solving Quadratic Equations by Factoring and Extracting Roots Newton's Method for Finding Roots Roots of Quadratic Equations Studio Roots, Radicals, and Root Functions Review division factoring and Root Finding Radicals Simplifying Radical Expressions Multiplying and Simplifying Radical Expressions LIKE RADICALS Multiplication and Division of Radicals Radical Equations BOUNDING ROOTS OF POLYNOMIALS
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A "radical" equation is an equation in which the variable is hiding inside  To solve radical equations: 1. Isolate the radical (or one of the radicals) to one side of the equal sign. 2. If the radical is a square root, square each side of the equation. (If the radical is not a square root, raise each side to a power equal to the index of the root.) 3. Solve the resulting equation. 4. Check your answer(s) to avoid extraneous roots.

BEWARE!!

When working with radical equations (that are square roots), ...

1. you must square sides, not terms. Consider: 2. you must check your answers. The process of squaring the sides of an equation creates a
"derived" equation which may not be equivalent to the original radical equation. Consequently,
solving this new derived equation may create solutions that never previously existed. These
"extra" roots that are not true solutions of the original radical equation are called extraneous
roots and are rejected as answers. Consider:

 The first statement is false, but when each side is squared, the concluding statement is true. The first statement is false, but when each side is squared, the concluding statement is true. We examine ways to solve equations such as . In solving any equation, we
always “undo” the operation so that we can get x isolated. In this case we have to square
both sides of the equation. The examples cited below describe some of the methods
needed to solve equations which include radicals.

#1. To solve we first square both sides of the equation.
The result is x - 2 = 81. This equation is simple to solve. We have x = 83.

#2. A more complicated situation is .
In this case we still begin by squaring both sides of the equation. The result is .
To finish solving this requires us to set all terms equal to zero and either factor or use the
quadratic formula. We get .
This factors: and the solutions are x = 2 or x = - 1.

We must check each of these solutions in the original equation to see if the value of x gives
a solution.

x = 2 gives which is correct.
x = - 1 gives which is impossible since for n > 0 is always
positive.
Therefore the only solution is x = 2.
Keeping in mind the basic idea of squaring both sides of the equation in order to solve for x
when there is a radical in the equation, we will now show some additional examples.

#3. Solve for x if .
Squaring both sides of the equation gives us .
Setting terms equal to zero gives .
The expression factors: .
Setting each factor equal to 0 gives two possible answers: x = 3 or x = – 1.
We check each answer in the original equation:
If x = 3 we have which is correct.
If x = – 1 we have which is impossible since the square root cannot be
negative.
Therefore the only answer is x = 3.
#4. Suppose .
We begin again by squaring both sides of the equation. The result is 3x – 2 = x + 7.
We solve for x getting 2x = 9 or .
We check this possible answer in the original equation.
If we have .
This is the same as which is correct.
Therefore the solution is .

#5. Solve for x if .
We square both sides. This gives x + 2 = 2 – x.
Solving for x gives 2x = 0. The only solution is x = 0.
Checking this in the original equation gives which is correct.
Therefore the solution is x = 0.

#6. Let .
Our first step is to separate the radicals. The result is .
Now we square both sides. The result is .
We next isolate the radical term on the right side. The result is .
We square both sides again. The result is .
We set the terms equal to zero. The result is .
We solve this by factoring or using the quadratic formula. This does factor: (x – 5)(x + 3) =
0. This gives two possible solutions: x = 5 or x = – 3.
We check these possible solutions in the original equation.
If x = 5 we have . This is the same as 4 – 3 = 1.
If x = – 3 we have . This is impossible since .
Therefore the only solution is x = 5.