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# Radical and rational equations and expressions on the SAT

## Studying radical and rational equations and expressions

On the SAT test radical and rational equations and expressions are part of passport to advanced mathematics subscore that includes 9 advanced topics (see the full topics list on the top menu).

Radical and rational equations and expressions topic is the seventh topic of passport to advances mathematics subscore. Start learning passport to advances mathematics subscore with its first six topics (the first topic is quadratic equations and quadratic functions).

Radical and rational equations and expressions topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other passport to advances mathematics subscore topic. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding passport to advanced mathematics subscore topics).

## Radical and rational equations and expressions - summary

Solving radicals and rational exponents requires creating equations that are different from the original equations. This may result in getting extraneous solutions, meaning that the solutions to the new equations don’t satisfy the original equations and therefore are not correct.

Rational expressions:
A rational expression is an expression containing at least one fraction with a variable in the denominator. The variables in the nominator and the denominator can be a quadratic or a higher order polynomial. For example: 3x/(x2-2) is a rational expression with linear variable in the nominator and a quadratic variable in the denominator.

Solving rational expressions includes simplifying, performing 4 operations with fractions (adding, subtracting, dividing, and multiplying) and cancelling common factors.

Rewriting a rational expression as a quotient and a remainder can be done with 2 methods: long division and grouping the numerator.

Rational equations:
A rational equation is an equation containing at least one fraction with a variable in the denominator. For example: 2/(x+2)=1.

Solving rational equations is done by multiplying both sides of the equation by the least common denominator. Since rational expressions contain a variable in the denominator, we need to exclude an extraneous solution for which the denominator equals to zero (we can’t divide by 0).

Radicals are rational exponents that are written with roots. For example √x. The symbol of a radical is √ and it represents a square root (instead of writing 2√x we write only √x).

A radical equation is an equation in which a variable appears under a radical. For example: √(x+1)=1 is a radical equation and √(4+12)=x is not a radical equation.

Solving radical equation is done by squaring both of its sides, this action cancels the radical sign and results in a linear or quadratic equation that we can solve. Extraneous solution is a solution that we get after solving a squared equation that is not a solution to the original equation. We need to exclude extraneous solutions since they are incorrect.

Previous knowledge for this topic includes solving quadratic equations, operations with exponents and solving a linear equation with fractions topics.

## Rational expressions on the SAT

A rational expression is an expression containing at least one fraction with a variable in the denominator. Rational expression can also contain variables in the nominator. The variables in the nominator and the denominator can be a quadratic or a higher order polynomial. For example:
2/(x+2) is a rational expression with a variable in the denominator (the minimal requirement for a rational expression).
2x/(x+3) is a rational expression with linear variables in the nominator and the denominator.
3x/(x2-2) is a rational expression with linear variable in the nominator and a quadratic variable in the denominator.

### Simplifying a rational expression

Before performing operations with two or more rational expressions, you should check if it is possible to simplify each expression by cancelling out common factors (factors that appear in both the nominator and the denominator). Making different operations on simplified expressions will be much easier.

The steps for simplifying one rational expression:

Step 1- Check if you can factor out common factors from the nominator and the denominator. Cancel common factors that appear both in the numerator and the denominator.

Step 2- Given a quadratic expression, check if you can rewrite it to its factored form or apply special factoring. Cancel common factors that appear both in the numerator and the denominator.
For detailed explanations learn the factored form of a quadratic function subject and the special factoring subject.

Consider the following example:

Simplify the expression (x3-4x)/(x2-2x).

Step 1- Factoring out common factors:
We can factor out x in the nominator and the denominator getting (x3-4x)/(x2-2x)=x(x2-4)/x(x-2)
Cancelling common factors: x(x2-4)/x(x-2)=(x2-4)/(x-2).

Step 2- Applying special factoring:
According to the difference of squares formula a2-b2=(a+b)(a-b). We can identify this formula in the nominator of the expression (x2-4)/(x-2) since x2 is a2 and 4 is b2, therefore x2-4=(x+2)(x-2).

Cancelling common factors: (x2-4)/(x-2)=( x+2)(x-2)/(x-2)=x+2.

### Solving rational expressions

After checking simplifying each rational expression as explained before, you can start performing operations with two or more rational expressions.

The steps for solving rational expressions:

Step 1- Check if you can simplify each expression:
As said before, prior to making operations with two or more rational expressions, you should first try to simplify each expression.

Step 2- Make the operation between the expressions:
Solving rational expressions questions requires performing 4 operations with fractions (adding, subtracting, dividing, and multiplying).
For detailed explanations learn the solving a linear equation with fractions topic.

Step 3- Cancel common factors that appear both in the numerator and the denominator.

#### Multiplying rational expressions

Consider the following example:

What is the product of (x2+2x)/(x-1) and (x3-x)/x2?

Step 1- Simplify each expression:
We can simplify the first expression by factoring out x in the nominator: (x2+2x)=x(x+2).
We can simplify the second expression by factoring out x in the nominator and then cancelling it: (x3-x)/x2=x(x2-1)/x*x=(x2-1)/x.

Step 2- Making operation between the expressions:
The multiplication between fractions formula is ab*cd=a*cb*d.
In the question [x(x+2)/(x-1)] * [(x2-1)/x] we have a=x(x+2), b=(x-1), c=(x2-1) and d=x, therefore:
[x(x+2)/(x-1)] * [(x2-1)/x] = [x(x+2)*(x2-1)] / [(x-1)*x]

Step 3- Cancelling common factors that appear both in the numerator and the denominator:
We can cancel x and x-1 getting [x(x+2)*(x2-1)] / [(x-1)*x]=(x+2)*(x-1).

#### Dividing rational expressions

Consider the following example:

Solve [(x2+4x+3)/(x2+4x+4] / [(x+1)/(x+2)].

Step 1- Simplify each expression:
Rewriting the expression x2+4x+3 to its factored form will give us x2+4x+3=x2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3).
Rewriting the expression x2+4x+4 according to the square of sum formula
(a+b)2=a2+b2+2ab will give us x2+4x+4=(x+2)2.
Therefore the first expression is simplified to (x+1)(x+3)/(x+2)2

Step 2- Making operation between the expressions:
We need to solve [(x+1)(x+3)/(x+2)2] / [(x+1)/(x+2)]
According to fraction division formula ab:cd=ab*dc=a*db*c, in the given question we get a=(x+1)(x+3), b=(x+2)2, c=x+1, d=x+2, therefore:
[(x+1)(x+3)/(x+2)2] / [(x+1)/(x+2)] =[(x+1)(x+3)(x+2)] / [(x+2)2(x+1)]

Step 3- Cancelling common factors that appear both in the numerator and the denominator:
We can cancel x+1 and x+2 getting (x+3)/(x+2).

#### Adding and subtracting rational expressions

Consider the following example:

What is the difference between (5x2-2x)/(x2+x) and 3x/(x-1)?

Step 1- Simplify each expression:
We can simplify the first expression by factoring out x in the nominator and the denominator and then cancelling it.
(5x2-2x)/(x2+x)=[x(5x-2)/x(x+1)]=(5x-2)/(x+1)

Step 2- Making operation between the expressions:
We have to solve (5x-2)/(x+1)-3x/(x-1).
First, we need to find a common denominator of the expressions using the formula for subtraction of fractions with unlike denominators ab±cd=(ad±bc)bd:
(5x-2)/(x+1)-3x/(x-1)=[(5x-2)(x-1)]/[(x+1)(x-1)][3x(x+1)]/[(x-1)(x+1)]

After finding a common denominator we can solve using the formula for addition and subtraction of fractions with like denominators ab±cb=(a±c)b :
[(5x-2)(x-1)-3x(x+1)]/[(x+1)(x-1)]=(5x2-7x+2-3x2-3x)/(x2-1)=(2x2-10x+2)/(x2-1)

We can factor out 2 in the nominator:
(2x2-10x+2)/(x2-1)=2(x2-5x+1)/(x2-1)

### Rewriting a rational expression as a quotient and a remainder

A quotient is the result of dividing one number by another. If the answer is not an integer, then the quotient is only the integer part of the division. The part that is left is called a remainder.
For example:
28/7=4 the quotient of 28 and 7 is 4 (no remainder).
30/7=4+2/7 the quotient of 30 and 7 is 4 and the remainder is 2.

We can express a division of 2 polynomial expressions a and b as 2 new polynomials: a polynomial q that represents the quotient and another polynomial r  that represents the remainder:
a(x)/b(x)=q(x)+r(x)/b(x)

Note that:
a(x)>b(x), otherwise we can’t create a quotient.
q(x) has no denominator, therefore it represents the quotient; r(x) is divided by the denominator b(x), therefore r(x) is the remainder.

There are 2 methods to divide polynomials: long division and grouping the numerator. It is enough to know the long division method to solve polynomial division questions.

#### Long division polynomials method

When dividing 2 numbers we can relate to them as a dividend (the number that we are dividing) and a divisor (the number that divides another number). When we start to do the long division, the dividend is the numerator and the divisor is the denominator. A we continue, the dividend changes according to what is left to divide and the divisor stays equal to the denominator.

The rule in long division of polynomials is that we need to look at the highest exponents of the dividend and the divisor. For example: in the expression 5x2-2x the highest exponential term is 5x2.

long division of polynomials a(x)/b(x) steps:

Step 1: Divide the terms with the highest exponents of the dividend (the numerator a(x)) and the divisor (the denominator b(x)) and place the result above the bar.

Step 2: Multiply the result of step 1 by the denominator b(x) and place the result below the dividend adding it a minus sign (the numerator):

Step 3: Subtract the result from step 2 from the dividend a(x).

Step 4: Continue dividing replacing and subtracting the dividend with what is left from the subtraction:

Consider the following example:

Rewrite (5x3+4x2+3)/x2 as a quotient and a remainder.

Step 1: Dividing the terms with the highest exponents of the dividend (the numerator 5x3+4x2+3) and the divisor (the denominator x2) and place the result above the bar:
The term with the highest exponents of the dividend (the numerator) is 5x2.
The term with the highest exponents of the divisor (the numerator) is x2.
We need to divide 5x3/x2 and place the result 5x above the bar.

5x
____________
x2  |  5x3+4x2+3

Step 2: Multiplying the result of step 1 by the denominator x2 and placing the result below the dividend (the numerator) adding it a minus sign (we add a minus because we need to subtract in the nest step):

5x
____________
x2  |  5x3+4x2+3
-5x3

Step 3: Subtracting the result from step 2 (in the previous step we added the minus sign to 5x3) from the dividend 5x3+4x2+3:

5x
____________
x2  |  5x3+4x2+3
-5x3
_______
4x2+3

Step 4: Continuing dividing replacing and subtracting the dividend with what is left from the subtraction:
The new dividend (instead of the numerator a(x)) is now 4x2+3, the denominator b(x) stays as the divisor b(x) which is x2. The answer is 4 and what is left is 4x2+3-4x2=3. Therefore we write -4x2 under the dividend and add 4 to the answer that is written above the bar (getting 5x+4 above the bar):

5x+4
____________
x2  |  5x3+4x2+3
-5x3
_______
4x2+3
-4x2
_______
3

We are left with dividend of 3. We can’t divide 3 by x2, therefore the remainder is 3/ x2.

Step 5: We can check the answer by calculating a common denominator of the answer 5x+4+3/x2:
The common denominator is x2:
(5x+4)*x2/x2+3/x2=(5x3+4x2+3)/x2 this is the original expression.

#### Polynomials division- factoring the numerator

In this technic we split the numerator to a multiplication of the denominator and another expression and add what is left. This will allow to cancel the denominator and get us a quotient that we need, what is left will become the remainder.

Polynomials division a(x)/b(x)- factoring the numerator steps:
Step 1: Split the numerator to a multiplication of the denominator and another expression and another expression.
Step 2: Cancel the denominator and get a quotient and a remainder.

Consider the previous example:

Rewrite (5x3+4x2+3)/x2 as a quotient and a remainder.

In this example the denominator is composed from only one component x2, therefore we can factor it out:

Step 1: Splitting the numerator to a multiplication of the denominator and another expression and another expression:
We need to split the numerator 5x3+4x2+3 x2 into the denominator x2 multiplied by another expression and some other expression. We see that we can factor out a common factor of x2 from 5x3+4x2 leaving 3, getting 5x3+4x2+3=x2(5x+4)+3.
The original expression was (5x3+4x2+3)/x2, combining that (5x3+4x2+3)/x2= [x2(5x+4)+3]/x2

Step 2: Cancelling the denominator and getting a quotient and a remainder:
[x2(5x+4)+3]/x2=5x+4+3/x2 the quotient is 5x+4 and the remainder is 3.

Step 3: We can check the answer by calculating a common denominator of the answer 5x+4+3/x2:
The common denominator is x2:
(5x+4)*x2/x2+3/x2=(5x3+4x2+3)/x2 this is the original expression.

Consider the following example:

Rewrite (5x3+4x2+3)/x2 as a quotient and a remainder.

In this example the denominator is composed from an expression x-1, therefore we need to factor out x-1 from the nominator. For detailed explanations learn the factored form of a quadratic function subject.

Step 1: Splitting the numerator to a multiplication of the denominator and another expression and another expression:

We have x2+3 and we need to split it to a multiplication of x-1 by another expression and add what is left: x2+0x+3=(x-1)(x+?)+??

We need 2 numbers that add to 0 and we know that one of them is -1. Therefore the other number is 1. We get (x-1)(x+1)=x2-x+x-1= x2-1, the full numerator was x2+3 therefore we are left with 4 that we couldn’t factor: x2+3=(x-1)(x+1)+4.

Step 2: Cancelling the denominator and getting a quotient and a remainder:
(x2+3)/(x-1)=[(x-1)(x+1)+4]/(x-1)= [(x-1)(x+1)]/(x-1)+4/(x-1)=x+1+4/(x-1) the quotient is x+1 and the remainder is 4.

Step 3: We can check the answer by calculating a common denominator of the answer x+1+4/(x-1):
The common denominator is x-1:
x+1+4/(x-1)=(x+1)(x-1)/(x-1)+4/(x-1)=[(x+1)(x-1)+4]/(x-1)=(x2-x+x-1+4)/(x-1)=(x2+3)/(x-1) this is the original expression.

## Rational equations on the SAT

### Solving rational equations steps

Step 1: Reaching a common denominator to cancel the denominators: Reaching a common denominator is done by multiplying by the smallest common multiple of the denominators of the fractions.

Step 2: Solving the equation: We can solve the new equation since it does not include denominators, it can be a linear equation or a quadratic equation.

Consider the following example:

Solve the equation 2/(x-1)=3/x.

2/(x-1)=3/x

The common denominator is x(x-1):
2x(x-1)/(x-1)=3x(x-1)/x
2x=3(x-1)
2x=3x-3
-x=-3
x=3

2/(x-1)=3/x
2/2=3/3
1=1 is a true statement therefore the answer x=3 is correct.

Step 3: Excluding extraneous solutions:

Plug the solutions of the equation without denominators into the original equation to make sure that they satisfy the original equation.

Sometimes the answers of the equation without denominators don’t satisfy the original equation. See the next paragraph for further explanation.

### Extraneous solutions of rational equations

Extraneous solution is a solution that we get after solving a squared equation that is not a solution to the original equation. We need to exclude extraneous solutions since they are incorrect. To find if the solution is extraneous check the solutions by plugging it into the original equation (the solution is extraneous if it does not satisfy the equation).

When do we get extraneous solutions? Extraneous solution happens when a solution for the equation without denominators causes a division by 0 in the original equation.

Consider the following example:

Solve 4/(x-3)=(x+1)/(x-3).

4/(x-3)=(x+1)/(x-3)

The common denominator is (x-3):
4(x-3)/(x-3)=(x+1)(x-3)/(x-3)
4=x+1
x=3

4/(3-3)=(x+1)/(3-3) the denominator on both sides of the equation is 3=3=0, we cannot divide by 0 therefore x=3 is an extraneous solution and there is no solution to the equation.

## Radical equations on the SAT

To solve the radical equation, we need to cancel the radical. This can be done by raising both sides of the equation to the index of the radical. Usually, the radical is a square root, therefore we need to raise it to a second exponent. Squaring both sides of the equation leads us to a new equation that we need to solve.

#### Step 1- Isolate the radical expression on one side of the equation:

If we have an expression that includes other components in addition to the radical we can’t cancel the radical in the next step.
For example: in the equation √x+1=1 the radical is not isolated in one side of the equation therefore we cannot cancel the radical by squaring both parts of the equation:
√x+1=1
Squaring both sides: (√x+1)2=12
Remember the formula (a+b)2=a2+b2+2ab therefore (√x+1)2=x+1+2√x.
The new squared equation is x+1+2√x=1, the radical sign still exists in the equation.
Isolating the radical on one side of the equation at first gives us:
√x+1=1
√x=1-1
√x=0
In this arrangement of the equation (the radical was isolated on one side of the equation before squaring), squaring both parts of the equations will cancel the radical:
√x=0
(√x)2=0
x=0

#### Step 2: Square both sides of the equation:

If we raise the radical of a square root to a second exponent, we get only the expression under the radical, since (√x)2= x1/2*2= x2/2=x1=x. Since there is no radical in the equation, we can solve it in the next step.

#### Step 3: Solve the new squared equation:

We can solve the new equation since it does not include radicals. It can be a linear equation or a quadratic equation.

We will get a linear equation if on the other side of the equation there is a constant (squaring a constant will give us another constant). For example:
√(2x+4)=4 (4 is a constant)
(√(2x+4))2=42 (42 is also a constant)
2x+4=16 is a linear equation
2x=12
x=6

We will get a squared equation if there is a squared variable under the radical. For example: √(x2-2x)=-1.
Another case in which we will get a squared equation is when on the other side of the original equation there is a variable or an expression with variables (squaring a variable will give us a squared variable). For example:
√(5x-4)=x (x is a variable)
(√(5x-4))2=x2
5x-4=x2 is a squared equation

Solving the squared equation ax2+bx+c=0 is done using the quadratic formula
x=(-b±√(b2-4ac))/2a,    x1=(-b+√(b2-4ac))/2a,    x2=(-b-√(b2-4ac))/2a
5x-4=x2
-x2+5x-4=0
x1=(-5+√(25-4*4))/-2=(-5+√9)/-2=-2/-2=1
x2=(-5-√(25-4*4))/-2=(-5-√9)/-2=-8/-2=4

For x=1 we get 5-4=1, 1=1
For x=4 we get 20-4=16, 16=16

#### Step 4: Excluding extraneous solutions:

Plug the solutions of the squared equation into the original equation to make sure that they satisfy the original equation.

Sometimes the answers of the squared equation don’t satisfy the original equation. See the next paragraph for further explanation.

### Extraneous solutions of a radical equations

Extraneous solution is a solution that we get after solving a squared equation that is not a solution to the original equation. We need to exclude extraneous solutions since they are incorrect. To find if the solution is extraneous check the solutions by plugging it into the original equation (the solution is extraneous if it does not satisfy the equation).

Why do we get extraneous solutions? The reason is that the squared equation is different from the original equation, as a result it has more solutions.

For example: the original equation is x=2 (the solution is 2), its squared equation is x2=4 (the solution is 2 or -2). After squaring the original equation x=2 to a new equation x2=4 we get 2 solutions 2 or -2, whole the only solution for the original equation x=2 is 2. Squaring got us another answer x=-2 that is an answer for the squared equation but not an answer for the original equation. Since our goal was to solve the original equation, we must exclude the extraneous answer x=-2.

Consider the following example:

What are the solutions to the equation √(-3x-2)=x?

√(-3x-2)=x

Squaring both sides of the equation: (√(-3x-2))2=x2
-3x-2=x2
x2+3x+2=0
x1=(-3+√(9-4*2))/2=(-3+1)/2=-1
x2=(-3-√(9-4*2))/2=(-3-1)/2=-2

Checking the answers by plugging them into the original equation:
For x=-1 we get √(-3*-1-2)=-1, √1=-1, 1=-1 is a false statement therefore the solution x=-1 is extraneous and we need to exclude it.
For x=-2 we get √(-3*-2-2)=-2, √4=-2, 2=-2 is a false statement therefore the solution x=-1 is extraneous and we need to exclude it.
The original equation √(-3x-2)=x has no solutions while the quadratic equation had 2 solutions.

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