Rational and Radical Expressions and Equations Alıştırma Soruları — ACT

Soru 1Soru →

For all non-zero real numbers $x$ , which of the following is equivalent to the expression $\frac{(x+1)^2}{x} - \frac{1}{x}$ ?

x + 2

x

x^2 + 2x

x^2 + 2

x^3 + 2

Cevabı ve açıklamayı göster

Cevap: $x + 2$

Cevap

The expression is equivalent to

x + 2

.

Combining the fractions gives the combined numerator over the denominator. Expanding the binomial in the numerator yields the sum of terms, which simplifies after subtracting one. Factoring out the variable in the numerator and dividing by the same variable simplifies the expression to a first-degree binomial.

Adım Adım Çözüm

1

Combine the two rational terms since they already share a common denominator of

x

.

\frac{(x+1)^2 - 1}{x}

When subtracting fractions with the same denominator, subtract the numerators and keep the denominator.

2

Expand the squared binomial

(x+1)^2

.

\frac{x^2 + 2x + 1 - 1}{x}

The square of a binomial is given by

(a+b)^2 = a^2 + 2ab + b^2

.

3

Simplify the numerator by combining like terms (

1 - 1 = 0

).

\frac{x^2 + 2x}{x}

Combine the constant terms in the numerator.

4

Factor

x

out of each term in the numerator and divide by the denominator

x

.

x + 2

Since

x \neq 0

, we can cancel the common factor of

x

from both the numerator and the denominator.

Anahtar Kavram

Simplifying rational expressions by combining fractions and expanding binomials

Tahmini Süre:45s

Soru 2Soru →

Which of the following represents the complete set of real solutions to the equation $\sqrt{4x + 28} - x = 4$ ?

-6

and

2

-2

and

6

2

only

6

only

-6

only

Cevabı ve açıklamayı göster

Cevap: $2$ only

Cevap

The correct answer is

2

only.

The correct answer is

2

only. Isolating the radical in the original equation gives

\sqrt{4x+28} = x+4

. Squaring both sides yields

4x+28 = x^2+8x+16

. Rearranging into standard quadratic form gives

x^2+4x-12 = 0

. Factoring the quadratic yields

(x+6)(x-2) = 0

, giving the candidate solutions

x = -6

and

x = 2

. Testing

x = 2

in the original equation gives

\sqrt{36} - 2 = 4

, which is valid. Testing

x = -6

gives

\sqrt{4} - (-6) = 8 \neq 4

, meaning

-6

is extraneous.

Adım Adım Çözüm

1

Isolate the radical expression on one side of the equation.

\sqrt{4x + 28} = x + 4

Isolating the radical allows us to square both sides directly to eliminate the square root.

2

Square both sides of the equation.

4x + 28 = x^2 + 8x + 16

Squaring a square root eliminates the radical, and squaring the binomial

(x+4)

yields

x^2 + 8x + 16

.

3

Move all terms to one side of the equation to form a standard quadratic equation.

x^2 + 4x - 12 = 0

Subtracting

4x

and

28

from both sides of the equation simplifies it to the form

ax^2 + bx + c = 0

.

4

Factor the quadratic equation to find potential solutions.

(x + 6)(x - 2) = 0

, which gives potential solutions

x = -6

and

x = 2

.

Factoring is a standard method to solve quadratic equations of this form.

5

Substitute the potential solutions back into the original equation to check for extraneous solutions.

For

x = 2

:

\sqrt{4(2)+28}-2 = 6-2 = 4

, which is true. For

x = -6

:

\sqrt{4(-6)+28}-(-6) = 2+6 = 8 \neq 4

, which is false.

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original radical equation.

Anahtar Kavram

Solving radical equations by isolating the radical, squaring both sides, and identifying extraneous solutions.

Soru 3Soru →

For all real values of $x$ such that $x \neq 3$ , which of the following expressions is equivalent to $\frac{x}{x - 3} - \frac{9}{(x - 3)^2}$ ?

\frac{x^2 - 3x - 9}{x^2 + 9}

\frac{x - 9}{x - 3}

\frac{x^2 - 3x - 9}{x^2 - 6x + 9}

\frac{x - 9}{x^2 - 7x + 12}

\frac{x^2 - 3x - 9}{x^2 - 9}

Cevabı ve açıklamayı göster

Cevap: $\frac{x^2 - 3x - 9}{x^2 - 6x + 9}$

Cevap

\frac{x^2 - 3x - 9}{x^2 - 6x + 9}

To subtract the rational expressions, we must first find a common denominator. The denominators are

x-3

and

(x-3)^2

, so the least common denominator is

(x-3)^2

. We multiply the numerator and denominator of the first term by

x-3

, resulting in

\frac{x(x-3)}{(x-3)^2} - \frac{9}{(x-3)^2}

. Expanding the first numerator gives

x^2 - 3x

. Subtracting the second numerator gives

x^2 - 3x - 9

in the numerator. Expanding the denominator

(x-3)^2

yields

x^2 - 6x + 9

. Therefore, the simplified equivalent expression is

\frac{x^2 - 3x - 9}{x^2 - 6x + 9}

.

Adım Adım Çözüm

1

Identify the least common denominator (LCD) for the two rational expressions.

The LCD of

x - 3

and

(x - 3)^2

is

(x - 3)^2

.

To perform subtraction between two rational expressions, they must have a common denominator.

2

Rewrite the first expression with the common denominator by multiplying its numerator and denominator by

x - 3

.

\frac{x(x - 3)}{(x - 3)^2} = \frac{x^2 - 3x}{(x - 3)^2}

.

Multiplying the numerator and denominator by the same non-zero quantity preserves the value of the expression.

3

Subtract the numerators of the two expressions while keeping the common denominator.

\frac{x^2 - 3x - 9}{(x - 3)^2}

.

Once denominators are identical, subtract the numerators directly over the common denominator.

4

Expand the binomial in the denominator

(x - 3)^2

to match the standard polynomial form of the options.

(x - 3)^2 = x^2 - 6x + 9

, yielding

\frac{x^2 - 3x - 9}{x^2 - 6x + 9}

.

Expanding the denominator allows direct comparison with the polynomial choices provided.

Anahtar Kavram

Simplifying rational expressions by finding a common denominator

Soru 4Soru →

For all real numbers $x > 1$ , the expression below is equivalent to which of the following?

\frac{\sqrt{x^8} - x^2 \cdot x^3}{x^2(1-x) - (x^2 - x)^2}

-x^2(x+1)

0

x

x^2 + x

\frac{x^2}{x+2}

Cevabı ve açıklamayı göster

Cevap: $x$

Cevap

The expression is equivalent to

x

.

The correct answer is

x

. Simplifying the numerator yields

x^4 - x^5 = x^4(1-x)

. Simplifying the denominator by expanding both parts yields

(x^2 - x^3) - (x^4 - 2x^3 + x^2) = x^3 - x^4 = x^3(1-x)

. Dividing the numerator by the denominator and canceling the common non-zero term

(1-x)

leaves

\frac{x^4}{x^3}

, which simplifies to

x

.

Adım Adım Çözüm

1

Simplify the radical term in the numerator.

\sqrt{x^8} = x^4

Since

x > 1

, we can take the square root of

x^8

directly by dividing the exponent by 2:

(x^8)^{1/2} = x^{8/2} = x^4

.

2

Simplify the multiplication of exponential terms in the numerator.

x^2 \cdot x^3 = x^5

By exponent rules, when multiplying terms with the same base, we add the exponents:

x^{2+3} = x^5

.

3

Combine the terms in the numerator and factor out the greatest common factor.

Numerator =

x^4 - x^5 = x^4(1-x)

Factoring out

x^4

prepares the numerator for potential cancellation with the denominator.

4

Expand and simplify the terms in the denominator.

Denominator =

x^3(1-x)

Expanding the first term gives

x^2(1-x) = x^2 - x^3

. Expanding the squared binomial gives

(x^2-x)^2 = x^4 - 2x^3 + x^2

. Subtracting them yields

(x^2 - x^3) - (x^4 - 2x^3 + x^2) = x^3 - x^4 = x^3(1-x)

.

5

Divide the simplified numerator by the simplified denominator.

\frac{x^4(1-x)}{x^3(1-x)} = x

Since

x > 1

, the factor

(1-x)

is non-zero, allowing us to cancel it. Finally, dividing the remaining terms gives

\frac{x^4}{x^3} = x^{4-3} = x^1 = x

.

Anahtar Kavram

Simplifying rational expressions by factoring and applying exponent rules, including radical simplification and binomial expansion.

Soru 5Soru →

The rational expression $\frac{x - 3}{x^2 - x - 12}$ is undefined for two real values of $x$ . What are these two values, where the smaller value is entered in the first blank and the larger value is entered in the second blank?

Aşağıdaki boşlukları doldurun

The smaller value is and the larger value is .

Cevabı ve açıklamayı göster

Cevap

The expression is undefined when the smaller value is -3 and the larger value is 4.

A rational expression is undefined when the denominator is equal to

0

. For the expression

\frac{x - 3}{x^2 - x - 12}

, setting the denominator equal to zero gives

x^2 - x - 12 = 0

. Factoring the quadratic expression yields

(x - 4)(x + 3) = 0

. Setting each factor to zero gives the solutions

x = 4

and

x = -3

. Thus, the smaller value for which the expression is undefined is

-3

and the larger value is

4

.

Adım Adım Çözüm

1

Identify the condition that makes a rational expression undefined.

A rational expression is undefined when its denominator is equal to zero.

Division by zero is undefined in real numbers.

2

Set the denominator of the expression equal to zero.

x^2 - x - 12 = 0

This equation will yield the values of

x

for which the denominator is zero.

3

Factor the quadratic equation.

(x - 4)(x + 3) = 0

Factoring allows us to find the roots by setting each linear factor to zero.

4

Solve for

x

by setting each factor equal to zero.

x = 4

or

x = -3

If

x - 4 = 0

, then

x = 4

. If

x + 3 = 0

, then

x = -3

.

5

Identify the smaller and larger values.

The smaller value is

-3

and the larger value is

4

.

Comparing the two values,

-3 < 4

.

Anahtar Kavram

Identifying values that make a rational expression undefined by finding the roots of the denominator.

Tahmini Süre:1m 15s

Soru 6Soru →

To find the domain of the rational expression

f(x) = \frac{x+3}{x^2 - 1} \div \frac{x^2 - 9}{x^2 - 3x - 4}

a student must identify all values of

x

for which the expression is undefined. The expression is undefined for exactly five real values of

x

. What are these five values in increasing order?

Aşağıdaki boşlukları doldurun

From least to greatest, the values of

x

that make the expression undefined are:
1.

x =

2.

x =

3.

x =

4.

x =

5.

x =

Cevabı ve açıklamayı göster

Cevap

The five real values of

x

that make the expression undefined, from least to greatest, are

-3

,

-1

,

1

,

3

, and

4

.

The rational expression is undefined when the denominator of the first expression is zero, when the denominator of the divisor is zero, or when the numerator of the divisor is zero. Factoring these three parts gives

(x^2 - 1) = (x-1)(x+1)

,

(x^2 - 3x - 4) = (x-4)(x+1)

, and

(x^2 - 9) = (x-3)(x+3)

. The values that make any of these equal to zero are

-3, -1, 1, 3, 4

.

Adım Adım Çözüm

1

Identify values that make the first denominator zero.

x = 1

and

x = -1

The denominator of the first rational expression is

x^2 - 1

. Setting

x^2 - 1 = 0

yields

(x - 1)(x + 1) = 0

, which gives

x = 1

and

x = -1

.

2

Identify values that make the second denominator zero.

x = 4

and

x = -1

The denominator of the second rational expression (the divisor) is

x^2 - 3x - 4

. Setting

x^2 - 3x - 4 = 0

yields

(x - 4)(x + 1) = 0

, which gives

x = 4

and

x = -1

.

3

Identify values that make the divisor equal to zero.

x = 3

and

x = -3

Since the operation is division, dividing by zero is undefined. The divisor

\frac{x^2 - 9}{x^2 - 3x - 4}

is equal to zero when its numerator is zero. Setting

x^2 - 9 = 0

yields

(x - 3)(x + 3) = 0

, which gives

x = 3

and

x = -3

.

4

Combine all unique restricted values and sort them in ascending order.

-3, -1, 1, 3, 4

Combining the restricted values from steps 1, 2, and 3 gives the set

\{-3, -1, 1, 3, 4\}

. Sorting these from least to greatest yields

-3, -1, 1, 3, 4

.

Anahtar Kavram

Domain restrictions in division of rational expressions

Tahmini Süre:2m 30s

Soru 7Soru →

For what real value of $x$ is the rational expression $\frac{2x + 1}{x - 7}$ undefined? Fill in the blank with the correct number.

Aşağıdaki boşlukları doldurun

The rational expression

\frac{2x + 1}{x - 7}

is undefined when $x = .

Cevabı ve açıklamayı göster

Cevap

The rational expression is undefined when the variable x equals 7.

A rational expression is undefined when its denominator is equal to

0

. For the expression

\frac{2x + 1}{x - 7}

, setting the denominator

x - 7 = 0

and solving for

x

yields

x = 7

.

Adım Adım Çözüm

1

Set the denominator of the rational expression equal to 0.

x - 7 = 0

A rational expression is undefined when its denominator is equal to 0 because division by zero is undefined in the real number system.

2

Solve the linear equation for

x

by adding 7 to both sides.

x = 7

Isolating the variable

x

gives the value that makes the denominator zero.

Anahtar Kavram

Identifying values that make a rational expression undefined

Soru 8Soru →

For what real values of $x$ is the rational expression $g(x) = \frac{x^3 - 3x^2}{\frac{2}{x-2} - \frac{x+4}{x^2 - 4}}$ undefined? Determine these three values and enter them in the blanks below in order from least to greatest.

Aşağıdaki boşlukları doldurun

The values of

x

for which

g(x)

is undefined, ordered from least to greatest, are

x =

,

x =

, and

x =

.

Cevabı ve açıklamayı göster

Cevap

The values of

x

for which the expression is undefined, ordered from least to greatest, are

-2

,

0

, and

2

.

The rational expression is undefined where any constituent denominator is equal to zero, or where the entire main denominator is equal to zero. The constituent denominators are

x-2

and

x^2-4

, which are zero when

x = 2

or

x = -2

. The entire main denominator simplifies to

\frac{x}{(x-2)(x+2)}

, which is zero when

x = 0

. Therefore, the three values of

x

that make the expression undefined are

-2

,

0

, and

2

.

Adım Adım Çözüm

1

Identify the values of

x

that make the denominators of the individual rational terms equal to zero.

The denominators are

x - 2

and

x^2 - 4

. Setting these to zero gives

x = 2

and

x = \pm 2

. Thus,

x = 2

and

x = -2

are values that make the individual terms undefined.

Any rational term is undefined if its denominator is equal to zero.

2

Find the values of

x

that make the overall denominator of the main fraction equal to zero by first simplifying it.

The main denominator is

\frac{2}{x-2} - \frac{x+4}{x^2-4}

. Finding a common denominator yields

\frac{2(x+2) - (x+4)}{(x-2)(x+2)} = \frac{x}{(x-2)(x+2)}

. Setting this equal to zero gives

x = 0

.

The main rational expression is undefined if its entire denominator is equal to zero.

3

Combine all restricted values and list them in order from least to greatest.

The restricted values are

x = -2

,

x = 0

, and

x = 2

.

These are all the distinct real numbers that cause any division by zero in the original expression.

Anahtar Kavram

Finding the domain of nested rational expressions

Tahmini Süre:2m 0s

Soru 9Soru →

For all real numbers $x$ such that $x \neq 0$ and $x \neq -1$ , which of the following expressions is equivalent to $\frac{3}{x} + \frac{2}{x + 1}$ ?

\frac{5}{2x + 1}

\frac{5x + 1}{x(x + 1)}

\frac{5x + 3}{x(x + 1)}

\frac{5x + 3}{2x + 1}

\frac{6}{x(x + 1)}

Cevabı ve açıklamayı göster

Cevap: $\frac{5x + 3}{x(x + 1)}$

Cevap

\frac{5x + 3}{x(x + 1)}

To add the rational expressions, find the common denominator

x(x + 1)

. Multiplying the first term by

\frac{x + 1}{x + 1}

and the second term by

\frac{x}{x}

gives

\frac{3(x + 1) + 2x}{x(x + 1)}

. Expanding the numerator yields

3x + 3 + 2x

, which simplifies to

5x + 3

. Thus, the equivalent expression is

\frac{5x + 3}{x(x + 1)}

.

Adım Adım Çözüm

1

Identify a common denominator for the two rational expressions.

The common denominator for the denominators

x

and

x + 1

is

x(x + 1)

.

To add fractions with different denominators, they must be written with a common denominator.

2

Rewrite each fraction with the common denominator by multiplying the numerator and denominator of the first term by

(x + 1)

and the second term by

x

.

\frac{3(x + 1)}{x(x + 1)} + \frac{2x}{x(x + 1)}

This scales the fractions to have matching denominators without changing their value.

3

Add the numerators together over the common denominator and simplify.

\frac{3x + 3 + 2x}{x(x + 1)} = \frac{5x + 3}{x(x + 1)}

Combining like terms simplifies the numerator to its final form.

Anahtar Kavram

Adding rational expressions by finding a common denominator

Tahmini Süre:1m 0s

Soru 10Soru →

Solve the equation $\sqrt{2x + 7} - x = -4$ for $x$ . What is the value of the real solution?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The only real solution to the equation is 9.

Isolating the radical yields

\sqrt{2x + 7} = x - 4

. Squaring both sides results in the quadratic equation

2x + 7 = x^2 - 8x + 16

, which simplifies to

x^2 - 10x + 9 = 0

. Factoring gives

(x - 9)(x - 1) = 0

, yielding potential solutions of 9 and 1. Checking these solutions in the original equation reveals that 9 is valid, while 1 is extraneous. Therefore, the correct real solution is 9.

Adım Adım Çözüm

1

Isolate the radical on one side of the equation.

\sqrt{2x + 7} = x - 4

Before squaring both sides, the radical term must be isolated to simplify the algebraic manipulation.

2

Square both sides of the equation.

2x + 7 = (x - 4)^2

Squaring both sides eliminates the square root.

3

Expand the squared binomial.

2x + 7 = x^2 - 8x + 16

Applying the identity

(a - b)^2 = a^2 - 2ab + b^2

is necessary to write the equation in polynomial form.

4

Set the quadratic equation to zero.

x^2 - 10x + 9 = 0

Subtracting

2x

and

7

from both sides allows us to solve the quadratic equation.

5

Factor the quadratic equation.

(x - 9)(x - 1) = 0

, giving potential solutions

x = 9

or

x = 1

.

Finding the roots of the quadratic equation provides the candidate solutions.

6

Verify candidates in the original equation.

The solution

x = 9

is valid, while

x = 1

is extraneous.

Squaring both sides can introduce extraneous solutions, so each candidate must be checked in the original equation.

Anahtar Kavram

Solving radical equations and verifying for extraneous solutions

Soru 11Soru →

For all real numbers $x$ such that $x \neq -2$ and $x \neq 3$ , the expression $\frac{2x}{x - 3} - \frac{5}{x + 2}$ is equivalent to which of the following?

\frac{2x - 5}{2x - 1}

\frac{2x^2 - x - 15}{x^2 - x - 6}

\frac{x + 15}{x^2 - x - 6}

\frac{2x^2 - x + 15}{x^2 - x - 6}

\frac{2x^2 - x + 15}{x^2 - 6}

Cevabı ve açıklamayı göster

Cevap: \frac{2x^2 - x + 15}{x^2 - x - 6}

Cevap

The expression is equivalent to

\frac{2x^2 - x + 15}{x^2 - x - 6}

.

To subtract the rational expressions, we find the common denominator, which is

(x - 3)(x + 2) = x^2 - x - 6

. We rewrite each fraction with this denominator, which gives

\frac{2x(x + 2)}{x^2 - x - 6} = \frac{2x^2 + 4x}{x^2 - x - 6}

and

\frac{5(x - 3)}{x^2 - x - 6} = \frac{5x - 15}{x^2 - x - 6}

. Subtracting the second numerator from the first gives

(2x^2 + 4x) - (5x - 15) = 2x^2 - x + 15

. Placing this over the common denominator gives the simplified expression

\frac{2x^2 - x + 15}{x^2 - x - 6}

.

Adım Adım Çözüm

1

Identify the common denominator.

The common denominator is

(x - 3)(x + 2) = x^2 - x - 6

.

To subtract rational expressions, we need a common denominator.

2

Rewrite each rational expression with the common denominator.

The first term becomes

\frac{2x(x + 2)}{(x - 3)(x + 2)} = \frac{2x^2 + 4x}{x^2 - x - 6}

, and the second term becomes

\frac{5(x - 3)}{(x - 3)(x + 2)} = \frac{5x - 15}{x^2 - x - 6}

.

Multiplying the numerator and denominator of each term by the missing factor keeps the values of the expressions unchanged.

3

Subtract the numerators.

(2x^2 + 4x) - (5x - 15) = 2x^2 + 4x - 5x + 15 = 2x^2 - x + 15

.

Subtracting the second numerator requires distributing the negative sign to both terms of the expression

(5x - 15)

.

4

Combine the result over the common denominator.

\frac{2x^2 - x + 15}{x^2 - x - 6}

Write the simplified numerator over the common denominator.

Anahtar Kavram

Subtraction of rational expressions involves finding a common denominator, expanding the numerators, and distributing negative signs carefully.

Soru 12Soru →

If $\sqrt{3x - 2} = 4$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of

x

is 6.

Squaring both sides of the equation

\sqrt{3x - 2} = 4

eliminates the radical, leading to

3x - 2 = 16

. Adding 2 to both sides results in

3x = 18

. Dividing both sides by 3 gives

x = 6

. Substituting 6 back into the original equation yields

\sqrt{3(6) - 2} = \sqrt{18 - 2} = \sqrt{16} = 4

, which verifies that the solution is correct.

Adım Adım Çözüm

1

Square both sides of the equation to remove the radical.

3x - 2 = 16

Squaring a square root removes the radical sign since

(\sqrt{a})^2 = a

for non-negative values.

2

Add 2 to both sides of the equation.

3x = 18

Adding the constant term moves it to the other side to isolate the term containing the variable.

3

Divide both sides by 3.

x = 6

Dividing by the coefficient of

x

yields the final solution.

Anahtar Kavram

Solving basic radical equations by isolating the radical and squaring both sides.

Soru 13Soru →

For all real values of

x

where the expression is defined, consider the equation:

\frac{2x}{\sqrt{x + 3}} = \sqrt{x + 3} - 1

Which of the following represents the complete set of real solutions to this equation?

\{1\}

\{1, 6\}

\{6\}

\{-3, 4\}

\{2\}

Cevabı ve açıklamayı göster

Cevap: $\{1\}$

Cevap

The set containing only 1

The correct answer is the set containing only 1. To solve the equation, we first multiply both sides by the denominator

\sqrt{x+3}

, which yields

2x = x + 3 - \sqrt{x+3}

. Isolating the radical gives

\sqrt{x+3} = 3-x

. Squaring both sides produces the quadratic equation

x+3 = x^2-6x+9

, which simplifies to

x^2-7x+6=0

. Solving this quadratic gives candidate solutions of 1 and 6. Substituting 6 back into the original equation results in an invalid statement (

4 = 2

), making it extraneous. Substituting 1 yields a valid statement (

1 = 1

), meaning the only real solution is 1.

Adım Adım Çözüm

1

Determine the domain of the equation.

x > -3

The expression inside the square root must be non-negative (

x+3 \ge 0

), and since it is in the denominator, it cannot be zero (

x+3 \neq 0

).

2

Clear the denominator by multiplying both sides by

\sqrt{x + 3}

.

2x = x + 3 - \sqrt{x + 3}

Multiplying both sides by the denominator simplifies the rational expression into a form where we can isolate the radical.

3

Isolate the radical term.

\sqrt{x + 3} = 3 - x

Grouping all non-radical terms on one side prepares the equation for squaring to eliminate the radical.

4

Analyze constraints on the variable.

x \le 3

Since the principal square root on the left side is non-negative, the right side

3 - x

must also be non-negative, which restricts any valid solutions to

x \le 3

.

5

Square both sides and simplify to form a quadratic equation.

x^2 - 7x + 6 = 0

Squaring both sides eliminates the radical:

x + 3 = (3 - x)^2 \implies x + 3 = x^2 - 6x + 9 \implies x^2 - 7x + 6 = 0

.

6

Solve the quadratic equation by factoring.

x = 1

or

x = 6

Factoring

(x-1)(x-6) = 0

gives the potential solutions.

7

Verify the solutions in the original equation.

x = 1

is valid;

x = 6

is extraneous.

Substituting

x = 6

yields

\frac{12}{3} = 4

on the left and

\sqrt{9}-1 = 2

on the right, which are not equal. Substituting

x = 1

yields

\frac{2}{2} = 1

on the left and

\sqrt{4}-1 = 1

on the right, which are equal.

Anahtar Kavram

Solving equations containing both rational and radical expressions requires clearing denominators, isolating the radical, squaring both sides, and verifying candidate solutions to eliminate extraneous solutions.

Soru 14Soru →

Which of the following is the complete set of real solutions to the equation $\sqrt{5x - 9} = x - 3$ ?

\{9\}

\{2, 9\}

\{0, 5\}

\{1.5, 2\}

\{2\}

Cevabı ve açıklamayı göster

Cevap: $\{9\}$

Cevap

\{9\}

The correct option is the set containing only the value 9. To solve the equation, we square both sides to get

5x - 9 = x^2 - 6x + 9

, which simplifies to the quadratic equation

x^2 - 11x + 18 = 0

. Factoring this yields the potential solutions 9 and 2. Substituting these back into the original equation, we find that 9 satisfies the equation while 2 results in a contradiction, making it extraneous.

Adım Adım Çözüm

1

Isolate the radical and square both sides of the equation.

5x - 9 = (x - 3)^2

Squaring both sides eliminates the square root to allow solving for the variable.

2

Expand the right-hand side using the binomial squaring rule.

5x - 9 = x^2 - 6x + 9

The square of a binomial

(a - b)^2

is

a^2 - 2ab + b^2

.

3

Move all terms to one side to set the quadratic equation to zero, then factor.

x^2 - 11x + 18 = 0 \Rightarrow (x - 9)(x - 2) = 0

Rearranging terms simplifies the equation into a standard quadratic form that can be factored.

4

Solve for the potential roots and check for extraneous solutions in the original equation.

x = 9

(valid) and

x = 2

(extraneous)

Checking

x = 9

gives

\sqrt{5(9) - 9} = 9 - 3 \Rightarrow 6 = 6

(true). Checking

x = 2

gives

\sqrt{5(2) - 9} = 2 - 3 \Rightarrow 1 = -1

(false).

Anahtar Kavram

Solving radical equations and verifying for extraneous solutions

Soru 15Soru →

If $x$ is a real number that satisfies the equation $\sqrt{2x + 7} + \sqrt{x + 3} = 1$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: -3

Cevap

The only real solution to the equation is

-3

.

The value

-3

is the only real number that satisfies the original equation. Substituting

-3

back into the original equation yields

\sqrt{2(-3) + 7} + \sqrt{-3 + 3} = \sqrt{1} + 0 = 1

, which is true.

Adım Adım Çözüm

1

Isolate the first radical term.

\sqrt{2x + 7} = 1 - \sqrt{x + 3}

This allows for squaring both sides to eliminate one radical.

2

Square both sides and simplify.

2x + 7 = x + 4 - 2\sqrt{x + 3}

Squaring removes the radical on the left side, though it creates a middle term on the right side.

3

Isolate the remaining radical term.

x + 3 = -2\sqrt{x + 3}

Grouping the non-radical terms on one side prepares the equation for a second squaring step.

4

Square both sides again to eliminate the remaining radical.

x^2 + 6x + 9 = 4(x + 3)

Squaring both sides eliminates the radical completely, converting the expression into a polynomial equation.

5

Solve the quadratic equation.

x = -3

and

x = 1

Rearranging to

x^2 + 2x - 3 = 0

and factoring as

(x + 3)(x - 1) = 0

gives the candidate solutions.

6

Substitute candidates back into the original equation to check for extraneous solutions.

The only valid solution is

x = -3

.

Substituting

x = 1

yields

5 = 1

(invalid), while substituting

x = -3

yields

1 = 1

(valid).

Anahtar Kavram

Solving radical equations by isolating radicals and squaring, then testing for extraneous solutions.

Soru 16Soru →

What is the sum of all real values of

x

that satisfy the equation

\frac{x^2 - 3x}{x^2 - 3x + 2} + \frac{x^2 - 3x - 1}{x^2 - 3x - 3} = \frac{11}{3}

?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The sum of all real values of

x

that satisfy the equation is

3

.

Substituting

y = x^2 - 3x

simplifies the original rational equation into the quadratic form

y^2 - y - 12 = 0

. Solving for

y

yields the values

4

and

-3

. Substituting back

x^2 - 3x

for

y

produces two quadratic equations. The first,

x^2 - 3x - 4 = 0

, has real solutions of

4

and

-1

. The second,

x^2 - 3x + 3 = 0

, has a negative discriminant and produces no real solutions. Summing the valid real solutions gives

4 + (-1) = 3

.

Adım Adım Çözüm

1

Substitute

y = x^2 - 3x

into the equation to simplify the rational terms.

\frac{y}{y+2} + \frac{y-1}{y-3} = \frac{11}{3}

Using a temporary variable simplifies the algebraic manipulation of the rational expressions.

2

Multiply the entire equation by the least common denominator,

3(y+2)(y-3)

, to eliminate all fractions.

3y(y-3) + 3(y-1)(y+2) = 11(y+2)(y-3)

, where

y \neq -2

and

y \neq 3

This clears the denominators so the equation can be solved as a polynomial.

3

Expand the terms on both sides of the equation.

(3y^2 - 9y) + (3y^2 + 3y - 6) = 11y^2 - 11y - 66

Expanding allows for combining like terms.

4

Combine like terms and move all terms to one side of the equation to set it equal to zero.

5y^2 - 5y - 60 = 0

This sets up the expression in the standard quadratic form

ay^2 + by + c = 0

.

5

Divide the entire quadratic equation by its greatest common factor,

5

, and factor the resulting expression.

y^2 - y - 12 = 0 \implies (y-4)(y+3) = 0 \implies y = 4 \text{ or } y = -3

Factoring solves for the possible values of the substituted variable

y

.

6

Substitute back

y = x^2 - 3x

for each case and solve the resulting quadratic equations for

x

.

For

y = 4

:

x^2 - 3x - 4 = 0 \implies (x-4)(x+1) = 0 \implies x = 4 \text{ or } x = -1

.
For

y = -3

:

x^2 - 3x + 3 = 0

. The discriminant is

(-3)^2 - 4(1)(3) = -3 < 0

, which means there are no real solutions.

This determines the real values of

x

that solve the original equation.

7

Verify that neither solution makes the original denominators zero, and add the valid real solutions.

4 + (-1) = 3

Since both

x = 4

and

x = -1

result in non-zero denominators, both are valid real solutions. Their sum is

3

.

Anahtar Kavram

Solving rational equations by utilizing algebraic substitution to reduce complexity and analyzing quadratic equations for real solutions.

Soru 17Soru →

If $k$ is a positive real number such that $\sqrt{k} + \sqrt{4k} = 15$ , what is the value of $k$ ?

3

5

9

25

45

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

25

The correct answer is

25

. To find this, we simplify

\sqrt{4k}

to

2\sqrt{k}

. Substituting this into the equation gives

\sqrt{k} + 2\sqrt{k} = 15

. Combining like terms yields

3\sqrt{k} = 15

. Dividing both sides by

3

gives

\sqrt{k} = 5

. Squaring both sides of the equation results in

k = 25

. Checking the solution:

\sqrt{25} + \sqrt{4(25)} = 5 + 10 = 15

, which is true.

Adım Adım Çözüm

1

Simplify the radical term

\sqrt{4k}

using the product property of radicals.

\sqrt{4k} = \sqrt{4} \cdot \sqrt{k} = 2\sqrt{k}

This allows us to express both terms in the equation using the same radical base,

\sqrt{k}

.

2

Substitute the simplified term back into the equation and combine like terms.

\sqrt{k} + 2\sqrt{k} = 15 \implies 3\sqrt{k} = 15

Combining like radical terms simplifies the equation to a single radical term.

3

Isolate the radical by dividing both sides of the equation by

3

.

\sqrt{k} = 5

Isolating the radical term is necessary before squaring both sides to solve for the variable.

4

Square both sides of the equation to solve for

k

.

k = 25

Squaring is the inverse operation of taking the square root, which isolates

k

.

Anahtar Kavram

Solving radical equations by simplifying and combining like radical terms

Soru 18Soru →

When solving the radical equation $\sqrt{2x^2 - 4x - 6} = x - 3$ for all real values of $x$ , one of the solutions obtained from the squared equation is extraneous. What is the value of this extraneous solution?

-5

3

1

-1/2

5

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The value of the extraneous solution is

-5

.

The correct answer is

-5

. Squaring both sides of the equation

\sqrt{2x^2 - 4x - 6} = x - 3

yields

2x^2 - 4x - 6 = x^2 - 6x + 9

, which simplifies to

x^2 + 2x - 15 = 0

. Factoring this quadratic gives

(x + 5)(x - 3) = 0

, leading to potential solutions of

3

and

-5

. Substituting

-5

back into the original equation results in the left side simplifying to

8

and the right side simplifying to

-8

. Since

8 \neq -8

, the value

-5

is an extraneous solution.

Adım Adım Çözüm

1

Square both sides of the original radical equation to eliminate the square root.

2x^2 - 4x - 6 = (x - 3)^2

Squaring both sides removes the radical on the left side.

2

Expand the right side and move all terms to the left side to set the quadratic equation to zero.

x^2 + 2x - 15 = 0

Expanding

(x - 3)^2

gives

x^2 - 6x + 9

. Subtracting this expression from both sides yields the simplified quadratic equation.

3

Factor the quadratic equation to find the potential solutions.

(x + 5)(x - 3) = 0

, which gives

x = -5

and

x = 3

.

The factors of

-15

that add up to

2

are

5

and

-3

.

4

Substitute each potential solution back into the original equation to check for extraneous roots.

For

x = 3

,

\sqrt{2(3)^2 - 4(3) - 6} = 3 - 3

simplifies to

0 = 0

(valid). For

x = -5

,

\sqrt{2(-5)^2 - 4(-5) - 6} = -5 - 3

simplifies to

8 = -8

(invalid).

Extraneous solutions satisfy the squared equation but do not satisfy the original radical equation due to the sign difference introduced by squaring.

Anahtar Kavram

Solving radical equations and identifying extraneous solutions

Tahmini Süre:2m 0s

Soru 19Soru →

Determine the value of $x$ for which the given rational expression is undefined.

Aşağıdaki boşlukları doldurun

The rational expression

\frac{x + 7}{2x - 10}

is undefined when

x =

.

Cevabı ve açıklamayı göster

Cevap

5

The rational expression is undefined when the denominator,

2x - 10

, equals

0

. Solving

2x - 10 = 0

gives

2x = 10

, which simplifies to

x = 5

.

Adım Adım Çözüm

1

Identify the denominator of the rational expression and set it equal to zero.

2x - 10 = 0

A rational expression is undefined when its denominator is equal to zero, as division by zero is not defined in the real number system.

2

Solve the linear equation for

x

by adding 10 to both sides and then dividing by 2.

2x = 10 \Rightarrow x = 5

Isolating

x

yields the value that makes the denominator zero.

Anahtar Kavram

Rational expressions are undefined when the denominator is equal to zero.

Tahmini Süre:45s

Soru 20Soru →

For all real numbers $x$ such that $x \neq 0$ , the expression $\left(\frac{1}{x} + 1\right)^2 - \frac{1}{x^2}$ is equivalent to which of the following?

\frac{x+2}{x}

1

\frac{3}{x^2}

\frac{x+2}{2x}

2x

Cevabı ve açıklamayı göster

Cevap: $\frac{x+2}{x}$

Cevap

\frac{x+2}{x}

Expanding the squared term yields

\frac{1}{x^2} + \frac{2}{x} + 1

. Subtracting

\frac{1}{x^2}

from this leaves

\frac{2}{x} + 1

. Finding a common denominator of

x

to add these terms results in

\frac{x+2}{x}

.

Adım Adım Çözüm

1

Expand the binomial term

\left(\frac{1}{x} + 1\right)^2

.

\frac{1}{x^2} + \frac{2}{x} + 1

Using the algebraic identity

(a+b)^2 = a^2 + 2ab + b^2

where

a = \frac{1}{x}

and

b = 1

.

2

Substitute this expansion back into the original expression and subtract

\frac{1}{x^2}

.

\frac{2}{x} + 1

The positive and negative

\frac{1}{x^2}

terms cancel each other out.

3

Write the expression as a single fraction over the common denominator

x

.

\frac{x+2}{x}

Convert

1

to

\frac{x}{x}

and add the numerators.

Anahtar Kavram

Simplifying rational expressions by expanding binomials and finding common denominators.

Alternatif Yöntem

Use the difference of squares factorization:

A^2 - B^2 = (A-B)(A+B)

. Let

A = \frac{1}{x} + 1

and

B = \frac{1}{x}

. Then the expression simplifies to

\left(\frac{1}{x} + 1 - \frac{1}{x}\right)\left(\frac{1}{x} + 1 + \frac{1}{x}\right) = 1 \cdot \left(\frac{2}{x} + 1\right) = \frac{x+2}{x}

.

Tahmini Süre:1m 0s

Rational and Radical Expressions and Equations

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