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Rational and Radical Expressions and Equations

35 questions

All types35Multiple choice19Numeric7Fill in the blank9
Question 1Question →

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

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Answer: x+2x + 2

Answer

The expression is equivalent to x+2x + 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.

Step-by-Step Solution

1
Combine the two rational terms since they already share a common denominator of xx.
(x+1)2−1x\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(x+1)^2.
x2+2x+1−1x\frac{x^2 + 2x + 1 - 1}{x}
The square of a binomial is given by (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2.
3
Simplify the numerator by combining like terms (1−1=01 - 1 = 0).
x2+2xx\frac{x^2 + 2x}{x}
Combine the constant terms in the numerator.
4
Factor xx out of each term in the numerator and divide by the denominator xx.
x+2x + 2
Since x≠0x \neq 0, we can cancel the common factor of xx from both the numerator and the denominator.

Key Concept

Simplifying rational expressions by combining fractions and expanding binomials
Estimated Time:45s
Question 2Question →

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

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Answer: 22 only

Answer

The correct answer is 22 only.
The correct answer is 22 only. Isolating the radical in the original equation gives 4x+28=x+4\sqrt{4x+28} = x+4. Squaring both sides yields 4x+28=x2+8x+164x+28 = x^2+8x+16. Rearranging into standard quadratic form gives x2+4x−12=0x^2+4x-12 = 0. Factoring the quadratic yields (x+6)(x−2)=0(x+6)(x-2) = 0, giving the candidate solutions x=−6x = -6 and x=2x = 2. Testing x=2x = 2 in the original equation gives 36−2=4\sqrt{36} - 2 = 4, which is valid. Testing x=−6x = -6 gives 4−(−6)=8≠4\sqrt{4} - (-6) = 8 \neq 4, meaning −6-6 is extraneous.

Step-by-Step Solution

1
Isolate the radical expression on one side of the equation.
4x+28=x+4\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=x2+8x+164x + 28 = x^2 + 8x + 16
Squaring a square root eliminates the radical, and squaring the binomial (x+4)(x+4) yields x2+8x+16x^2 + 8x + 16.
3
Move all terms to one side of the equation to form a standard quadratic equation.
x2+4x−12=0x^2 + 4x - 12 = 0
Subtracting 4x4x and 2828 from both sides of the equation simplifies it to the form ax2+bx+c=0ax^2 + bx + c = 0.
4
Factor the quadratic equation to find potential solutions.
(x+6)(x−2)=0(x + 6)(x - 2) = 0, which gives potential solutions x=−6x = -6 and x=2x = 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=2x = 2: 4(2)+28−2=6−2=4\sqrt{4(2)+28}-2 = 6-2 = 4, which is true. For x=−6x = -6: 4(−6)+28−(−6)=2+6=8≠4\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.

Key Concept

Solving radical equations by isolating the radical, squaring both sides, and identifying extraneous solutions.
Question 3Question →

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

Show answer & explanation

Answer: x2−3x−9x2−6x+9\frac{x^2 - 3x - 9}{x^2 - 6x + 9}

Answer

x2−3x−9x2−6x+9\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−3x-3 and (x−3)2(x-3)^2, so the least common denominator is (x−3)2(x-3)^2. We multiply the numerator and denominator of the first term by x−3x-3, resulting in x(x−3)(x−3)2−9(x−3)2\frac{x(x-3)}{(x-3)^2} - \frac{9}{(x-3)^2}. Expanding the first numerator gives x2−3xx^2 - 3x. Subtracting the second numerator gives x2−3x−9x^2 - 3x - 9 in the numerator. Expanding the denominator (x−3)2(x-3)^2 yields x2−6x+9x^2 - 6x + 9. Therefore, the simplified equivalent expression is x2−3x−9x2−6x+9\frac{x^2 - 3x - 9}{x^2 - 6x + 9}.

Step-by-Step Solution

1
Identify the least common denominator (LCD) for the two rational expressions.
The LCD of x−3x - 3 and (x−3)2(x - 3)^2 is (x−3)2(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−3x - 3.
x(x−3)(x−3)2=x2−3x(x−3)2\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.
x2−3x−9(x−3)2\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(x - 3)^2 to match the standard polynomial form of the options.
(x−3)2=x2−6x+9(x - 3)^2 = x^2 - 6x + 9, yielding x2−3x−9x2−6x+9\frac{x^2 - 3x - 9}{x^2 - 6x + 9}.
Expanding the denominator allows direct comparison with the polynomial choices provided.

Key Concept

Simplifying rational expressions by finding a common denominator
Question 4Question →

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

x8−x2⋅x3x2(1−x)−(x2−x)2\frac{\sqrt{x^8} - x^2 \cdot x^3}{x^2(1-x) - (x^2 - x)^2}
Show answer & explanation

Answer: xx

Answer

The expression is equivalent to xx.
The correct answer is xx. Simplifying the numerator yields x4−x5=x4(1−x)x^4 - x^5 = x^4(1-x). Simplifying the denominator by expanding both parts yields (x2−x3)−(x4−2x3+x2)=x3−x4=x3(1−x)(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)(1-x) leaves x4x3\frac{x^4}{x^3}, which simplifies to xx.

Step-by-Step Solution

1
Simplify the radical term in the numerator.
x8=x4\sqrt{x^8} = x^4
Since x>1x > 1, we can take the square root of x8x^8 directly by dividing the exponent by 2: (x8)1/2=x8/2=x4(x^8)^{1/2} = x^{8/2} = x^4.
2
Simplify the multiplication of exponential terms in the numerator.
x2⋅x3=x5x^2 \cdot x^3 = x^5
By exponent rules, when multiplying terms with the same base, we add the exponents: x2+3=x5x^{2+3} = x^5.
3
Combine the terms in the numerator and factor out the greatest common factor.
Numerator = x4−x5=x4(1−x)x^4 - x^5 = x^4(1-x)
Factoring out x4x^4 prepares the numerator for potential cancellation with the denominator.
4
Expand and simplify the terms in the denominator.
Denominator = x3(1−x)x^3(1-x)
Expanding the first term gives x2(1−x)=x2−x3x^2(1-x) = x^2 - x^3. Expanding the squared binomial gives (x2−x)2=x4−2x3+x2(x^2-x)^2 = x^4 - 2x^3 + x^2. Subtracting them yields (x2−x3)−(x4−2x3+x2)=x3−x4=x3(1−x)(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.
x4(1−x)x3(1−x)=x\frac{x^4(1-x)}{x^3(1-x)} = x
Since x>1x > 1, the factor (1−x)(1-x) is non-zero, allowing us to cancel it. Finally, dividing the remaining terms gives x4x3=x4−3=x1=x\frac{x^4}{x^3} = x^{4-3} = x^1 = x.

Key Concept

Simplifying rational expressions by factoring and applying exponent rules, including radical simplification and binomial expansion.
Question 5Question →

The rational expression x−3x2−x−12\frac{x - 3}{x^2 - x - 12} is undefined for two real values of xx. 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?

Fill in the blanks below

The smaller value is and the larger value is .
Show answer & explanation

Answer

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 00. For the expression x−3x2−x−12\frac{x - 3}{x^2 - x - 12}, setting the denominator equal to zero gives x2−x−12=0x^2 - x - 12 = 0. Factoring the quadratic expression yields (x−4)(x+3)=0(x - 4)(x + 3) = 0. Setting each factor to zero gives the solutions x=4x = 4 and x=−3x = -3. Thus, the smaller value for which the expression is undefined is −3-3 and the larger value is 44.

Step-by-Step Solution

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.
x2−x−12=0x^2 - x - 12 = 0
This equation will yield the values of xx for which the denominator is zero.
3
Factor the quadratic equation.
(x−4)(x+3)=0(x - 4)(x + 3) = 0
Factoring allows us to find the roots by setting each linear factor to zero.
4
Solve for xx by setting each factor equal to zero.
x=4x = 4 or x=−3x = -3
If x−4=0x - 4 = 0, then x=4x = 4. If x+3=0x + 3 = 0, then x=−3x = -3.
5
Identify the smaller and larger values.
The smaller value is −3-3 and the larger value is 44.
Comparing the two values, −3<4-3 < 4.

Key Concept

Identifying values that make a rational expression undefined by finding the roots of the denominator.
Estimated Time:1m 15s
Question 6Question →
To find the domain of the rational expression f(x)=x+3x2−1÷x2−9x2−3x−4f(x) = \frac{x+3}{x^2 - 1} \div \frac{x^2 - 9}{x^2 - 3x - 4} a student must identify all values of xx for which the expression is undefined. The expression is undefined for exactly five real values of xx. What are these five values in increasing order?

Fill in the blanks below

From least to greatest, the values of xx that make the expression undefined are:
1. x=x =
2. x=x =
3. x=x =
4. x=x =
5. x=x =
Show answer & explanation

Answer

The five real values of xx that make the expression undefined, from least to greatest, are −3-3, −1-1, 11, 33, and 44.
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 (x2−1)=(x−1)(x+1)(x^2 - 1) = (x-1)(x+1), (x2−3x−4)=(x−4)(x+1)(x^2 - 3x - 4) = (x-4)(x+1), and (x2−9)=(x−3)(x+3)(x^2 - 9) = (x-3)(x+3). The values that make any of these equal to zero are −3,−1,1,3,4-3, -1, 1, 3, 4.

Step-by-Step Solution

1
Identify values that make the first denominator zero.
x=1x = 1 and x=−1x = -1
The denominator of the first rational expression is x2−1x^2 - 1. Setting x2−1=0x^2 - 1 = 0 yields (x−1)(x+1)=0(x - 1)(x + 1) = 0, which gives x=1x = 1 and x=−1x = -1.
2
Identify values that make the second denominator zero.
x=4x = 4 and x=−1x = -1
The denominator of the second rational expression (the divisor) is x2−3x−4x^2 - 3x - 4. Setting x2−3x−4=0x^2 - 3x - 4 = 0 yields (x−4)(x+1)=0(x - 4)(x + 1) = 0, which gives x=4x = 4 and x=−1x = -1.
3
Identify values that make the divisor equal to zero.
x=3x = 3 and x=−3x = -3
Since the operation is division, dividing by zero is undefined. The divisor x2−9x2−3x−4\frac{x^2 - 9}{x^2 - 3x - 4} is equal to zero when its numerator is zero. Setting x2−9=0x^2 - 9 = 0 yields (x−3)(x+3)=0(x - 3)(x + 3) = 0, which gives x=3x = 3 and x=−3x = -3.
4
Combine all unique restricted values and sort them in ascending order.
−3,−1,1,3,4-3, -1, 1, 3, 4
Combining the restricted values from steps 1, 2, and 3 gives the set {−3,−1,1,3,4}\{-3, -1, 1, 3, 4\}. Sorting these from least to greatest yields −3,−1,1,3,4-3, -1, 1, 3, 4.

Key Concept

Domain restrictions in division of rational expressions
Estimated Time:2m 30s
Question 7Question →

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

Fill in the blanks below

The rational expression 2x+1x−7\frac{2x + 1}{x - 7} is undefined when $x = .
Show answer & explanation

Answer

The rational expression is undefined when the variable x equals 7.
A rational expression is undefined when its denominator is equal to 00. For the expression 2x+1x−7\frac{2x + 1}{x - 7}, setting the denominator x−7=0x - 7 = 0 and solving for xx yields x=7x = 7.

Step-by-Step Solution

1
Set the denominator of the rational expression equal to 0.
x−7=0x - 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 xx by adding 7 to both sides.
x=7x = 7
Isolating the variable xx gives the value that makes the denominator zero.

Key Concept

Identifying values that make a rational expression undefined
Question 8Question →

For what real values of xx is the rational expression g(x)=x3−3x22x−2−x+4x2−4g(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.

Fill in the blanks below

The values of xx for which g(x)g(x) is undefined, ordered from least to greatest, are x=x = , x=x = , and x=x = .
Show answer & explanation

Answer

The values of xx for which the expression is undefined, ordered from least to greatest, are −2-2, 00, and 22.
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−2x-2 and x2−4x^2-4, which are zero when x=2x = 2 or x=−2x = -2. The entire main denominator simplifies to x(x−2)(x+2)\frac{x}{(x-2)(x+2)}, which is zero when x=0x = 0. Therefore, the three values of xx that make the expression undefined are −2-2, 00, and 22.

Step-by-Step Solution

1
Identify the values of xx that make the denominators of the individual rational terms equal to zero.
The denominators are x−2x - 2 and x2−4x^2 - 4. Setting these to zero gives x=2x = 2 and x=±2x = \pm 2. Thus, x=2x = 2 and x=−2x = -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 xx that make the overall denominator of the main fraction equal to zero by first simplifying it.
The main denominator is 2x−2−x+4x2−4\frac{2}{x-2} - \frac{x+4}{x^2-4}. Finding a common denominator yields 2(x+2)−(x+4)(x−2)(x+2)=x(x−2)(x+2)\frac{2(x+2) - (x+4)}{(x-2)(x+2)} = \frac{x}{(x-2)(x+2)}. Setting this equal to zero gives x=0x = 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=−2x = -2, x=0x = 0, and x=2x = 2.
These are all the distinct real numbers that cause any division by zero in the original expression.

Key Concept

Finding the domain of nested rational expressions
Estimated Time:2m 0s
Question 9Question →

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

Show answer & explanation

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

Answer

5x+3x(x+1)\frac{5x + 3}{x(x + 1)}
To add the rational expressions, find the common denominator x(x+1)x(x + 1). Multiplying the first term by x+1x+1\frac{x + 1}{x + 1} and the second term by xx\frac{x}{x} gives 3(x+1)+2xx(x+1)\frac{3(x + 1) + 2x}{x(x + 1)}. Expanding the numerator yields 3x+3+2x3x + 3 + 2x, which simplifies to 5x+35x + 3. Thus, the equivalent expression is 5x+3x(x+1)\frac{5x + 3}{x(x + 1)}.

Step-by-Step Solution

1
Identify a common denominator for the two rational expressions.
The common denominator for the denominators xx and x+1x + 1 is x(x+1)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)(x + 1) and the second term by xx.
\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.

Key Concept

Adding rational expressions by finding a common denominator
Estimated Time:1m 0s
Question 10Question →

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

Show answer & explanation

Answer: 9

Answer

The only real solution to the equation is 9.
Isolating the radical yields 2x+7=x−4\sqrt{2x + 7} = x - 4. Squaring both sides results in the quadratic equation 2x+7=x2−8x+162x + 7 = x^2 - 8x + 16, which simplifies to x2−10x+9=0x^2 - 10x + 9 = 0. Factoring gives (x−9)(x−1)=0(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.

Step-by-Step Solution

1
Isolate the radical on one side of the equation.
2x+7=x−4\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)22x + 7 = (x - 4)^2
Squaring both sides eliminates the square root.
3
Expand the squared binomial.
2x+7=x2−8x+162x + 7 = x^2 - 8x + 16
Applying the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 is necessary to write the equation in polynomial form.
4
Set the quadratic equation to zero.
x2−10x+9=0x^2 - 10x + 9 = 0
Subtracting 2x2x and 77 from both sides allows us to solve the quadratic equation.
5
Factor the quadratic equation.
(x−9)(x−1)=0(x - 9)(x - 1) = 0, giving potential solutions x=9x = 9 or x=1x = 1.
Finding the roots of the quadratic equation provides the candidate solutions.
6
Verify candidates in the original equation.
The solution x=9x = 9 is valid, while x=1x = 1 is extraneous.
Squaring both sides can introduce extraneous solutions, so each candidate must be checked in the original equation.

Key Concept

Solving radical equations and verifying for extraneous solutions
Question 11Question →

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

Show answer & explanation

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

Answer

The expression is equivalent to 2x2−x+15x2−x−6\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)=x2−x−6(x - 3)(x + 2) = x^2 - x - 6. We rewrite each fraction with this denominator, which gives 2x(x+2)x2−x−6=2x2+4xx2−x−6\frac{2x(x + 2)}{x^2 - x - 6} = \frac{2x^2 + 4x}{x^2 - x - 6} and 5(x−3)x2−x−6=5x−15x2−x−6\frac{5(x - 3)}{x^2 - x - 6} = \frac{5x - 15}{x^2 - x - 6}. Subtracting the second numerator from the first gives (2x2+4x)−(5x−15)=2x2−x+15(2x^2 + 4x) - (5x - 15) = 2x^2 - x + 15. Placing this over the common denominator gives the simplified expression 2x2−x+15x2−x−6\frac{2x^2 - x + 15}{x^2 - x - 6}.

Step-by-Step Solution

1
Identify the common denominator.
The common denominator is (x−3)(x+2)=x2−x−6(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 2x(x+2)(x−3)(x+2)=2x2+4xx2−x−6\frac{2x(x + 2)}{(x - 3)(x + 2)} = \frac{2x^2 + 4x}{x^2 - x - 6}, and the second term becomes 5(x−3)(x−3)(x+2)=5x−15x2−x−6\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.
(2x2+4x)−(5x−15)=2x2+4x−5x+15=2x2−x+15(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)(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.

Key Concept

Subtraction of rational expressions involves finding a common denominator, expanding the numerators, and distributing negative signs carefully.
Question 12Question →

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

Show answer & explanation

Answer: 6

Answer

The value of xx is 6.
Squaring both sides of the equation 3x−2=4\sqrt{3x - 2} = 4 eliminates the radical, leading to 3x−2=163x - 2 = 16. Adding 2 to both sides results in 3x=183x = 18. Dividing both sides by 3 gives x=6x = 6. Substituting 6 back into the original equation yields 3(6)−2=18−2=16=4\sqrt{3(6) - 2} = \sqrt{18 - 2} = \sqrt{16} = 4, which verifies that the solution is correct.

Step-by-Step Solution

1
Square both sides of the equation to remove the radical.
3x−2=163x - 2 = 16
Squaring a square root removes the radical sign since (a)2=a(\sqrt{a})^2 = a for non-negative values.
2
Add 2 to both sides of the equation.
3x=183x = 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=6x = 6
Dividing by the coefficient of xx yields the final solution.

Key Concept

Solving basic radical equations by isolating the radical and squaring both sides.
Question 13Question →
For all real values of xx where the expression is defined, consider the equation:
2xx+3=x+3−1\frac{2x}{\sqrt{x + 3}} = \sqrt{x + 3} - 1
Which of the following represents the complete set of real solutions to this equation?
Show answer & explanation

Answer: {1}\{1\}

Answer

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 x+3\sqrt{x+3}, which yields 2x=x+3−x+32x = x + 3 - \sqrt{x+3}. Isolating the radical gives x+3=3−x\sqrt{x+3} = 3-x. Squaring both sides produces the quadratic equation x+3=x2−6x+9x+3 = x^2-6x+9, which simplifies to x2−7x+6=0x^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=24 = 2), making it extraneous. Substituting 1 yields a valid statement (1=11 = 1), meaning the only real solution is 1.

Step-by-Step Solution

1
Determine the domain of the equation.
x>−3x > -3
The expression inside the square root must be non-negative (x+3≥0x+3 \ge 0), and since it is in the denominator, it cannot be zero (x+3≠0x+3 \neq 0).
2
Clear the denominator by multiplying both sides by x+3\sqrt{x + 3}.
2x=x+3−x+32x = 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.
x+3=3−x\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≤3x \le 3
Since the principal square root on the left side is non-negative, the right side 3−x3 - x must also be non-negative, which restricts any valid solutions to x≤3x \le 3.
5
Square both sides and simplify to form a quadratic equation.
x2−7x+6=0x^2 - 7x + 6 = 0
Squaring both sides eliminates the radical: x+3=(3−x)2  ⟹  x+3=x2−6x+9  ⟹  x2−7x+6=0x + 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=1x = 1 or x=6x = 6
Factoring (x−1)(x−6)=0(x-1)(x-6) = 0 gives the potential solutions.
7
Verify the solutions in the original equation.
x=1x = 1 is valid; x=6x = 6 is extraneous.
Substituting x=6x = 6 yields 123=4\frac{12}{3} = 4 on the left and 9−1=2\sqrt{9}-1 = 2 on the right, which are not equal. Substituting x=1x = 1 yields 22=1\frac{2}{2} = 1 on the left and 4−1=1\sqrt{4}-1 = 1 on the right, which are equal.

Key Concept

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.
Question 14Question →

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

Show answer & explanation

Answer: {9}\{9\}

Answer

{9}\{9\}
The correct option is the set containing only the value 9. To solve the equation, we square both sides to get 5x−9=x2−6x+95x - 9 = x^2 - 6x + 9, which simplifies to the quadratic equation x2−11x+18=0x^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.

Step-by-Step Solution

1
Isolate the radical and square both sides of the equation.
5x−9=(x−3)25x - 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=x2−6x+95x - 9 = x^2 - 6x + 9
The square of a binomial (a−b)2(a - b)^2 is a2−2ab+b2a^2 - 2ab + b^2.
3
Move all terms to one side to set the quadratic equation to zero, then factor.
x2−11x+18=0⇒(x−9)(x−2)=0x^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=9x = 9 (valid) and x=2x = 2 (extraneous)
Checking x=9x = 9 gives 5(9)−9=9−3⇒6=6\sqrt{5(9) - 9} = 9 - 3 \Rightarrow 6 = 6 (true). Checking x=2x = 2 gives 5(2)−9=2−3⇒1=−1\sqrt{5(2) - 9} = 2 - 3 \Rightarrow 1 = -1 (false).

Key Concept

Solving radical equations and verifying for extraneous solutions
Question 15Question →

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

Show answer & explanation

Answer: -3

Answer

The only real solution to the equation is −3-3.
The value −3-3 is the only real number that satisfies the original equation. Substituting −3-3 back into the original equation yields 2(−3)+7+−3+3=1+0=1\sqrt{2(-3) + 7} + \sqrt{-3 + 3} = \sqrt{1} + 0 = 1, which is true.

Step-by-Step Solution

1
Isolate the first radical term.
2x+7=1−x+3\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+4−2x+32x + 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=−2x+3x + 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.
x2+6x+9=4(x+3)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=−3x = -3 and x=1x = 1
Rearranging to x2+2x−3=0x^2 + 2x - 3 = 0 and factoring as (x+3)(x−1)=0(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=−3x = -3.
Substituting x=1x = 1 yields 5=15 = 1 (invalid), while substituting x=−3x = -3 yields 1=11 = 1 (valid).

Key Concept

Solving radical equations by isolating radicals and squaring, then testing for extraneous solutions.
Question 16Question →
What is the sum of all real values of xx that satisfy the equation x2−3xx2−3x+2+x2−3x−1x2−3x−3=113\frac{x^2 - 3x}{x^2 - 3x + 2} + \frac{x^2 - 3x - 1}{x^2 - 3x - 3} = \frac{11}{3}?
Show answer & explanation

Answer: 3

Answer

The sum of all real values of xx that satisfy the equation is 33.
Substituting y=x2−3xy = x^2 - 3x simplifies the original rational equation into the quadratic form y2−y−12=0y^2 - y - 12 = 0. Solving for yy yields the values 44 and −3-3. Substituting back x2−3xx^2 - 3x for yy produces two quadratic equations. The first, x2−3x−4=0x^2 - 3x - 4 = 0, has real solutions of 44 and −1-1. The second, x2−3x+3=0x^2 - 3x + 3 = 0, has a negative discriminant and produces no real solutions. Summing the valid real solutions gives 4+(−1)=34 + (-1) = 3.

Step-by-Step Solution

1
Substitute y=x2−3xy = x^2 - 3x into the equation to simplify the rational terms.
yy+2+y−1y−3=113\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)3(y+2)(y-3), to eliminate all fractions.
3y(y−3)+3(y−1)(y+2)=11(y+2)(y−3)3y(y-3) + 3(y-1)(y+2) = 11(y+2)(y-3), where y≠−2y \neq -2 and y≠3y \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.
(3y2−9y)+(3y2+3y−6)=11y2−11y−66(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.
5y2−5y−60=05y^2 - 5y - 60 = 0
This sets up the expression in the standard quadratic form ay2+by+c=0ay^2 + by + c = 0.
5
Divide the entire quadratic equation by its greatest common factor, 55, and factor the resulting expression.
y2−y−12=0  ⟹  (y−4)(y+3)=0  ⟹  y=4 or y=−3y^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 yy.
6
Substitute back y=x2−3xy = x^2 - 3x for each case and solve the resulting quadratic equations for xx.
For y=4y = 4: x2−3x−4=0  ⟹  (x−4)(x+1)=0  ⟹  x=4 or x=−1x^2 - 3x - 4 = 0 \implies (x-4)(x+1) = 0 \implies x = 4 \text{ or } x = -1.
For y=−3y = -3: x2−3x+3=0x^2 - 3x + 3 = 0. The discriminant is (−3)2−4(1)(3)=−3<0(-3)^2 - 4(1)(3) = -3 < 0, which means there are no real solutions.
This determines the real values of xx that solve the original equation.
7
Verify that neither solution makes the original denominators zero, and add the valid real solutions.
4+(−1)=34 + (-1) = 3
Since both x=4x = 4 and x=−1x = -1 result in non-zero denominators, both are valid real solutions. Their sum is 33.

Key Concept

Solving rational equations by utilizing algebraic substitution to reduce complexity and analyzing quadratic equations for real solutions.
Question 17Question →

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

Show answer & explanation

Answer: 25

Answer

25
The correct answer is 2525. To find this, we simplify 4k\sqrt{4k} to 2k2\sqrt{k}. Substituting this into the equation gives k+2k=15\sqrt{k} + 2\sqrt{k} = 15. Combining like terms yields 3k=153\sqrt{k} = 15. Dividing both sides by 33 gives k=5\sqrt{k} = 5. Squaring both sides of the equation results in k=25k = 25. Checking the solution: 25+4(25)=5+10=15\sqrt{25} + \sqrt{4(25)} = 5 + 10 = 15, which is true.

Step-by-Step Solution

1
Simplify the radical term 4k\sqrt{4k} using the product property of radicals.
4k=4⋅k=2k\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, k\sqrt{k}.
2
Substitute the simplified term back into the equation and combine like terms.
k+2k=15  ⟹  3k=15\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 33.
k=5\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 kk.
k=25k = 25
Squaring is the inverse operation of taking the square root, which isolates kk.

Key Concept

Solving radical equations by simplifying and combining like radical terms
Question 18Question →

When solving the radical equation 2x2−4x−6=x−3\sqrt{2x^2 - 4x - 6} = x - 3 for all real values of xx, one of the solutions obtained from the squared equation is extraneous. What is the value of this extraneous solution?

Show answer & explanation

Answer: -5

Answer

The value of the extraneous solution is −5-5.
The correct answer is −5-5. Squaring both sides of the equation 2x2−4x−6=x−3\sqrt{2x^2 - 4x - 6} = x - 3 yields 2x2−4x−6=x2−6x+92x^2 - 4x - 6 = x^2 - 6x + 9, which simplifies to x2+2x−15=0x^2 + 2x - 15 = 0. Factoring this quadratic gives (x+5)(x−3)=0(x + 5)(x - 3) = 0, leading to potential solutions of 33 and −5-5. Substituting −5-5 back into the original equation results in the left side simplifying to 88 and the right side simplifying to −8-8. Since 8≠−88 \neq -8, the value −5-5 is an extraneous solution.

Step-by-Step Solution

1
Square both sides of the original radical equation to eliminate the square root.
2x2−4x−6=(x−3)22x^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.
x2+2x−15=0x^2 + 2x - 15 = 0
Expanding (x−3)2(x - 3)^2 gives x2−6x+9x^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(x + 5)(x - 3) = 0, which gives x=−5x = -5 and x=3x = 3.
The factors of −15-15 that add up to 22 are 55 and −3-3.
4
Substitute each potential solution back into the original equation to check for extraneous roots.
For x=3x = 3, 2(3)2−4(3)−6=3−3\sqrt{2(3)^2 - 4(3) - 6} = 3 - 3 simplifies to 0=00 = 0 (valid). For x=−5x = -5, 2(−5)2−4(−5)−6=−5−3\sqrt{2(-5)^2 - 4(-5) - 6} = -5 - 3 simplifies to 8=−88 = -8 (invalid).
Extraneous solutions satisfy the squared equation but do not satisfy the original radical equation due to the sign difference introduced by squaring.

Key Concept

Solving radical equations and identifying extraneous solutions
Estimated Time:2m 0s
Question 19Question →

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

Fill in the blanks below

The rational expression x+72x−10\frac{x + 7}{2x - 10} is undefined when x=x = .
Show answer & explanation

Answer

5
The rational expression is undefined when the denominator, 2x−102x - 10, equals 00. Solving 2x−10=02x - 10 = 0 gives 2x=102x = 10, which simplifies to x=5x = 5.

Step-by-Step Solution

1
Identify the denominator of the rational expression and set it equal to zero.
2x−10=02x - 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 xx by adding 10 to both sides and then dividing by 2.
2x=10⇒x=52x = 10 \Rightarrow x = 5
Isolating xx yields the value that makes the denominator zero.

Key Concept

Rational expressions are undefined when the denominator is equal to zero.
Estimated Time:45s
Question 20Question →

For all real numbers xx such that x≠0x \neq 0, the expression (1x+1)2−1x2\left(\frac{1}{x} + 1\right)^2 - \frac{1}{x^2} is equivalent to which of the following?

Show answer & explanation

Answer: x+2x\frac{x+2}{x}

Answer

x+2x\frac{x+2}{x}
Expanding the squared term yields 1x2+2x+1\frac{1}{x^2} + \frac{2}{x} + 1. Subtracting 1x2\frac{1}{x^2} from this leaves 2x+1\frac{2}{x} + 1. Finding a common denominator of xx to add these terms results in x+2x\frac{x+2}{x}.

Step-by-Step Solution

1
Expand the binomial term (1x+1)2\left(\frac{1}{x} + 1\right)^2.
1x2+2x+1\frac{1}{x^2} + \frac{2}{x} + 1
Using the algebraic identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 where a=1xa = \frac{1}{x} and b=1b = 1.
2
Substitute this expansion back into the original expression and subtract 1x2\frac{1}{x^2}.
2x+1\frac{2}{x} + 1
The positive and negative 1x2\frac{1}{x^2} terms cancel each other out.
3
Write the expression as a single fraction over the common denominator xx.
x+2x\frac{x+2}{x}
Convert 11 to xx\frac{x}{x} and add the numerators.

Key Concept

Simplifying rational expressions by expanding binomials and finding common denominators.

Alternative Method

Use the difference of squares factorization: A2−B2=(A−B)(A+B)A^2 - B^2 = (A-B)(A+B). Let A=1x+1A = \frac{1}{x} + 1 and B=1xB = \frac{1}{x}. Then the expression simplifies to (1x+1−1x)(1x+1+1x)=1⋅(2x+1)=x+2x\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}.
Estimated Time:1m 0s
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Rational and Radical Expressions and Equations Practice Questions — ACT | Examkin