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Radical and Rational Equations

43 questions

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Question 1Question →

If x+5x−1=3\frac{x + 5}{x - 1} = 3, what is the value of x+1x + 1?

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Answer: 5

Answer

5
The correct answer is 55. To solve the rational equation x+5x−1=3\frac{x + 5}{x - 1} = 3, we first multiply both sides of the equation by x−1x - 1, which gives x+5=3(x−1)x + 5 = 3(x - 1). Distributing the 33 on the right side yields x+5=3x−3x + 5 = 3x - 3. Subtracting xx from both sides gives 5=2x−35 = 2x - 3. Adding 33 to both sides yields 8=2x8 = 2x. Dividing both sides by 22 gives x=4x = 4. Substituting this value back into the expression x+1x + 1 yields 4+1=54 + 1 = 5.

Step-by-Step Solution

1
Multiply both sides of the equation by x−1x - 1 to clear the fraction.
x+5=3(x−1)x + 5 = 3(x - 1)
Multiplying both sides by the denominator clears the fraction and allows us to solve the equation linearly.
2
Distribute the 33 to the terms inside the parentheses on the right side of the equation.
x+5=3x−3x + 5 = 3x - 3
This step simplifies the equation to prepare for isolating the variable xx.
3
Isolate the variable xx by subtracting xx from both sides and then adding 33 to both sides.
8=2x8 = 2x
Grouping like terms on opposite sides of the equation simplifies the process of finding the value of the variable.
4
Divide both sides of the equation by 22 to find the value of xx.
x=4x = 4
Dividing by the coefficient of xx isolates the variable completely.
5
Substitute the value of xx into the expression x+1x + 1 to find the final answer.
4+1=54 + 1 = 5
The question asks for the value of the expression x+1x + 1, not just the value of xx.

Key Concept

Solving rational equations that simplify to linear equations
Question 2Question →

What is the sum of all real values of xx that satisfy the equation 3x+40−x=4\sqrt{3x + 40} - x = 4?

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Answer: 3

Answer

The sum of all valid real solutions is 3.
Isolating the radical gives 3x+40=x+4\sqrt{3x + 40} = x + 4. Squaring both sides and simplifying results in x2+5x−24=0x^2 + 5x - 24 = 0. Factoring gives (x+8)(x−3)=0(x + 8)(x - 3) = 0, leading to candidate solutions of 3 and -8. Substituting these back shows that only 3 satisfies the original equation, as -8 results in an extraneous solution. Therefore, the sum of all valid solutions is 3.

Step-by-Step Solution

1
Isolate the radical expression on one side of the equation.
3x+40=x+4\sqrt{3x + 40} = x + 4
To prepare both sides of the equation for squaring.
2
Square both sides of the equation to eliminate the radical.
3x+40=(x+4)23x + 40 = (x + 4)^2
Squaring a square root cancels the radical.
3
Expand the right side and move all terms to one side to form a quadratic equation.
x2+5x−24=0x^2 + 5x - 24 = 0
Expanding (x+4)2(x+4)^2 yields x2+8x+16x^2 + 8x + 16. Subtracting 3x3x and 4040 from both sides sets the quadratic equation to zero.
4
Factor the quadratic equation.
(x+8)(x−3)=0(x + 8)(x - 3) = 0
The factors of -24 that add up to 5 are 8 and -3.
5
Solve for the potential values of xx.
x=−8x = -8 or x=3x = 3
Setting each factor to zero yields the candidate solutions.
6
Check each candidate solution in the original equation to identify any extraneous solutions.
For x=3x = 3, 3(3)+40−3=7−3=4\sqrt{3(3)+40}-3 = 7-3 = 4 (valid). For x=−8x = -8, 3(−8)+40−(−8)=4+8=12≠4\sqrt{3(-8)+40}-(-8) = 4+8 = 12 \neq 4 (extraneous).
Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Solving radical equations and checking for extraneous solutions.
Question 3Question →

If 8x−3=2\frac{8}{x - 3} = 2, what is the value of xx?

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Answer: 7

Answer

7
To solve the equation 8x−3=2\frac{8}{x - 3} = 2, we clear the denominator by multiplying both sides of the equation by x−3x - 3. This gives 8=2(x−3)8 = 2(x - 3). Distributing the 22 results in 8=2x−68 = 2x - 6. Adding 66 to both sides yields 14=2x14 = 2x. Dividing both sides by 22 gives the solution x=7x = 7.

Step-by-Step Solution

1
Multiply both sides of the equation by x−3x - 3 to eliminate the fraction.
8=2(x−3)8 = 2(x - 3)
To clear the denominator and simplify the equation.
2
Distribute the 2 to the terms inside the parentheses.
8=2x−68 = 2x - 6
To expand the right-hand side of the equation.
3
Add 6 to both sides of the equation.
14=2x14 = 2x
To isolate the term with the variable xx.
4
Divide both sides by 2 to solve for xx.
x=7x = 7
To find the final value of the variable.

Key Concept

Solving a rational equation that simplifies to a linear equation by multiplying by the common denominator.
Question 4Question →
What is the set of all real solutions to the equation below?
3x+10=x+2\sqrt{3x+10} = x+2
Show answer & explanation

Answer: {2}\{2\}

Answer

The set containing only 2
The correct answer is the set containing only 2. Squaring both sides of the equation yields 3x+10=x2+4x+43x + 10 = x^2 + 4x + 4. Rearranging the terms to set the equation to zero gives x2+x−6=0x^2 + x - 6 = 0. Factoring the quadratic expression gives (x+3)(x−2)=0(x + 3)(x - 2) = 0, indicating potential solutions of x=−3x = -3 and x=2x = 2. Substituting x=2x = 2 back into the original equation gives 16=4\sqrt{16} = 4, which is valid. Substituting x=−3x = -3 back into the original equation gives 1=−1\sqrt{1} = -1, which is invalid because the principal square root is non-negative. Thus, x=−3x = -3 is extraneous and must be discarded.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical.
3x+10=(x+2)23x + 10 = (x + 2)^2, which expands to 3x+10=x2+4x+43x + 10 = x^2 + 4x + 4.
Squaring both sides is the standard algebraic method to isolate the term under a square root.
2
Move all terms to one side to form a quadratic equation equal to zero.
x2+x−6=0x^2 + x - 6 = 0.
Setting the quadratic expression to zero allows us to solve it by factoring.
3
Factor the quadratic equation.
(x+3)(x−2)=0(x + 3)(x - 2) = 0, which gives the potential solutions x=−3x = -3 and x=2x = 2.
Finding the factors helps us determine the roots of the equation.
4
Substitute both potential solutions back into the original equation to check for extraneous solutions.
For x=2x = 2: 3(2)+10=16=4\sqrt{3(2)+10} = \sqrt{16} = 4 and 2+2=42+2 = 4, which is true. For x=−3x = -3: 3(−3)+10=1=1\sqrt{3(-3)+10} = \sqrt{1} = 1 and −3+2=−1-3+2 = -1, which is false.
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Identifying extraneous solutions in radical equations
Estimated Time:1m 30s
Question 5Question →

If x>2x > 2 and satisfies the equation

x−8x2−2x=5−4x−2x - \frac{8}{x^2 - 2x} = 5 - \frac{4}{x - 2}

what is the value of xx?

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Answer: 4

Answer

The correct answer is 4.
To solve the rational equation, we first determine the domain restrictions: x≠0x \neq 0 and x≠2x \neq 2. We then clear the denominators by multiplying the entire equation by the least common denominator x(x−2)x(x - 2), which simplifies the equation to the cubic form x3−7x2+14x−8=0x^3 - 7x^2 + 14x - 8 = 0. Factoring the cubic polynomial yields (x−1)(x−2)(x−4)=0(x - 1)(x - 2)(x - 4) = 0. The potential solutions are x=1x = 1, x=2x = 2, and x=4x = 4. Since x=2x = 2 is an extraneous solution and the constraint requires x>2x > 2, x=4x = 4 is the only valid solution.

Step-by-Step Solution

1
Determine the domain restrictions of the rational terms in the equation.
x≠0x \neq 0 and x≠2x \neq 2
Division by zero is undefined, so the denominators x(x−2)x(x-2) and x−2x-2 cannot be zero.
2
Multiply the entire equation by the least common denominator x(x−2)x(x - 2) to eliminate the fractions.
x2(x−2)−8=5x(x−2)−4xx^2(x - 2) - 8 = 5x(x - 2) - 4x
Multiplying by the common denominator converts the rational equation into an equivalent polynomial equation within the defined domain.
3
Expand the terms and collect them on one side to form a standard polynomial equation.
x3−7x2+14x−8=0x^3 - 7x^2 + 14x - 8 = 0
Simplifying the expressions allows us to find the roots of the polynomial.
4
Factor the cubic polynomial using the rational root theorem or synthetic division.
(x−1)(x−2)(x−4)=0(x - 1)(x - 2)(x - 4) = 0
Since x=1x = 1 makes the polynomial zero, we factor out (x−1)(x-1) to get the remaining quadratic factor x2−6x+8x^2 - 6x + 8, which factors into (x−2)(x−4)(x - 2)(x - 4).
5
Identify the valid solution based on the domain restriction and the given inequality constraint.
x=4x = 4
The value x=2x = 2 is extraneous because it makes the original equation undefined. The value x=1x = 1 is rejected because the problem specifies x>2x > 2.

Key Concept

Solving rational equations by clearing denominators, factoring polynomials, identifying extraneous solutions, and applying inequality constraints.
Question 6Question →
What is the value of the real solution to the equation below?
3x−2−x+3=1\sqrt{3x - 2} - \sqrt{x + 3} = 1
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Answer: 6

Answer

The only real solution to the equation is 6.
Substituting x=6x = 6 into the original equation yields 3(6)−2−6+3=16−9=4−3=1\sqrt{3(6) - 2} - \sqrt{6 + 3} = \sqrt{16} - \sqrt{9} = 4 - 3 = 1, which is a true statement. Therefore, the only real solution is 6.

Step-by-Step Solution

1
Isolate one of the radical terms
3x−2=x+3+1\sqrt{3x - 2} = \sqrt{x + 3} + 1
Preparing the equation to square both sides.
2
Square both sides of the equation
3x−2=x+4+2x+33x - 2 = x + 4 + 2\sqrt{x + 3}
Eliminating one of the square root radicals.
3
Isolate the remaining radical term and simplify
x−3=x+3x - 3 = \sqrt{x + 3}
Simplifying the equation by isolating the second radical and dividing both sides by 2.
4
Square both sides again to eliminate the second radical
x2−6x+9=x+3x^2 - 6x + 9 = x + 3
Converting the radical equation into a polynomial equation.
5
Write in standard quadratic form and factor
(x−6)(x−1)=0(x - 6)(x - 1) = 0
Setting the quadratic equation to zero and factoring to find potential solutions.
6
Verify solutions in the original equation to check for extraneous solutions
x=6x = 6 is valid; x=1x = 1 is extraneous
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Solving equations containing radical expressions and identifying extraneous solutions.
Estimated Time:2m 30s
Question 7Question →

What is the solution to the equation 3x+13=x+3\sqrt{3x + 13} = x + 3?

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Answer: 1

Answer

1
The value 11 is the correct answer because substituting it back into the original equation results in a true mathematical statement: 3(1)+13=1+3\sqrt{3(1) + 13} = 1 + 3, which simplifies to 4=44 = 4.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical.
3x+13=(x+3)23x + 13 = (x + 3)^2
Squaring both sides is the standard method to solve equations containing square roots.
2
Expand the right side of the equation.
3x+13=x2+6x+93x + 13 = x^2 + 6x + 9
Expanding the squared binomial gives a quadratic expression.
3
Move all terms to one side to set the quadratic equation to zero.
x2+3x−4=0x^2 + 3x - 4 = 0
Setting the quadratic equation to zero allows us to solve for xx by factoring.
4
Factor the quadratic equation.
(x+4)(x−1)=0(x + 4)(x - 1) = 0
Finding two numbers that multiply to −4-4 and add to 33 gives the binomial factors.
5
Solve for the potential values of xx.
x=−4x = -4 or x=1x = 1
Setting each factor equal to zero yields the potential solutions.
6
Substitute each potential solution back into the original equation to check for extraneous solutions.
For x=1x = 1: 3(1)+13=1+3⇒4=4\sqrt{3(1) + 13} = 1 + 3 \Rightarrow 4 = 4 (valid). For x=−4x = -4: 3(−4)+13=−4+3⇒1=−1\sqrt{3(-4) + 13} = -4 + 3 \Rightarrow 1 = -1 (invalid). Thus, the only solution is 11.
Squaring both sides can introduce extraneous solutions, so checking all potential solutions in the original equation is necessary to verify their validity.

Key Concept

Solving radical equations by squaring both sides and checking for extraneous solutions.
Question 8Question →

What is the set of all solutions to the equation x+12=x\sqrt{x + 12} = x?

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Answer: 44 only

Answer

The solution set containing only 44
The correct answer is the set containing only 44. Squaring both sides of the equation x+12=x\sqrt{x + 12} = x produces x+12=x2x + 12 = x^2. Rearranging this into standard quadratic form gives x2−x−12=0x^2 - x - 12 = 0. Factoring the quadratic gives (x−4)(x+3)=0(x - 4)(x + 3) = 0, which yields potential solutions of x=4x = 4 and x=−3x = -3. Testing x=4x = 4 in the original equation yields 4+12=16=4\sqrt{4 + 12} = \sqrt{16} = 4, which is a true statement. Testing x=−3x = -3 yields −3+12=9=3\sqrt{-3 + 12} = \sqrt{9} = 3, which does not equal −3-3. Thus, x=−3x = -3 is extraneous, and 44 is the only valid solution.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical.
x+12=x2x + 12 = x^2
Squaring both sides of x+12=x\sqrt{x + 12} = x removes the square root.
2
Rewrite the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−x−12=0x^2 - x - 12 = 0
Subtracting xx and 1212 from both sides moves all terms to one side of the equation.
3
Factor the quadratic equation.
(x−4)(x+3)=0(x - 4)(x + 3) = 0
We search for two numbers that multiply to −12-12 and add to −1-1. These numbers are −4-4 and 33.
4
Solve for the potential solutions.
x=4x = 4 or x=−3x = -3
Setting each factor equal to zero yields the prospective values of xx.
5
Check both potential solutions in the original equation to identify extraneous solutions.
Substituting x=4x = 4: 4+12=16=4\sqrt{4 + 12} = \sqrt{16} = 4 (valid). Substituting x=−3x = -3: −3+12=9=3≠−3\sqrt{-3 + 12} = \sqrt{9} = 3 \neq -3 (extraneous).
Squaring both sides of an equation can introduce extraneous roots that must be discarded.

Key Concept

Solving radical equations and verifying solutions to eliminate extraneous roots
Question 9Question →

If 20x−3=4\frac{20}{x - 3} = 4, what is the value of x+2x + 2?

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Answer: 10

Answer

10
To solve the equation 20x−3=4\frac{20}{x - 3} = 4, multiply both sides by x−3x - 3 to get 20=4(x−3)20 = 4(x - 3). Distributing the 4 gives 20=4x−1220 = 4x - 12. Adding 12 to both sides yields 32=4x32 = 4x, and dividing by 4 gives x=8x = 8. Substituting x=8x = 8 into the expression x+2x + 2 gives 8+2=108 + 2 = 10.

Step-by-Step Solution

1
Multiply both sides of the equation by the denominator x−3x - 3.
20=4(x−3)20 = 4(x - 3)
To eliminate the fraction and rewrite the rational equation in linear form.
2
Distribute the constant on the right side of the equation.
20=4x−1220 = 4x - 12
To remove the parentheses by multiplying 4 by both xx and −3-3.
3
Add 12 to both sides of the equation.
32=4x32 = 4x
To isolate the term with the variable xx on one side of the equation.
4
Divide both sides of the equation by 4.
x=8x = 8
To solve for the variable xx.
5
Substitute the value of xx into the expression x+2x + 2.
1010
The question asks for the value of the expression x+2x + 2 rather than just the variable xx.

Key Concept

Solving rational equations by clearing the denominator and isolating the variable to evaluate algebraic expressions.
Question 10Question →

If 2x+15−6=x\sqrt{2x + 15} - 6 = x, what is the value of x+5x + 5?

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Answer: 2

Answer

The correct answer is 2.
The correct answer is 2. To solve the equation 2x+15−6=x\sqrt{2x + 15} - 6 = x, we first isolate the radical by adding 66 to both sides, obtaining 2x+15=x+6\sqrt{2x + 15} = x + 6. Squaring both sides yields 2x+15=x2+12x+362x + 15 = x^2 + 12x + 36. Rearranging the terms to form a quadratic equation gives x2+10x+21=0x^2 + 10x + 21 = 0, which factors as (x+7)(x+3)=0(x + 7)(x + 3) = 0. This gives two potential solutions: x=−7x = -7 and x=−3x = -3. We must check both potential solutions in the original equation. For x=−3x = -3, the equation holds true: 2(−3)+15−6=3−6=−3\sqrt{2(-3) + 15} - 6 = 3 - 6 = -3. For x=−7x = -7, it does not: 2(−7)+15−6=1−6=−5≠−7\sqrt{2(-7) + 15} - 6 = 1 - 6 = -5 \neq -7. Thus, x=−3x = -3 is the only valid solution. Finally, substituting this into the expression x+5x + 5 gives −3+5=2-3 + 5 = 2.

Step-by-Step Solution

1
Isolate the radical term on one side of the equation.
2x+15=x+6\sqrt{2x + 15} = x + 6
Before squaring both sides, the radical term must be isolated to avoid cross-terms on that side.
2
Square both sides of the equation to eliminate the radical.
2x+15=(x+6)2  ⟹  2x+15=x2+12x+362x + 15 = (x + 6)^2 \implies 2x + 15 = x^2 + 12x + 36
Squaring a square root cancels the radical, allowing us to solve the resulting algebraic equation.
3
Rearrange the equation into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
x2+10x+21=0x^2 + 10x + 21 = 0
Subtracting 2x2x and 1515 from both sides collects all terms on one side so the quadratic can be factored.
4
Factor the quadratic equation.
(x+7)(x+3)=0  ⟹  x=−7 or x=−3(x + 7)(x + 3) = 0 \implies x = -7 \text{ or } x = -3
Finding two numbers that multiply to 21 and add to 10 allows us to factor the quadratic and find potential solutions.
5
Substitute the potential solutions back into the original equation to check for extraneous solutions.
For x=−3x = -3, 2(−3)+15−6=−3\sqrt{2(-3) + 15} - 6 = -3 is true. For x=−7x = -7, 2(−7)+15−6=−7\sqrt{2(-7) + 15} - 6 = -7 is false (gives −5≠−7-5 \neq -7). Thus, x=−3x = -3 is the only valid solution.
Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.
6
Evaluate the expression x+5x + 5 using the valid solution.
−3+5=2-3 + 5 = 2
The question asks for the value of x+5x + 5, not just the value of xx.

Key Concept

Solving radical equations by isolating the radical, squaring both sides, checking for extraneous solutions, and evaluating algebraic expressions.
Question 11Question →

If xx−2−8x+2=8x2−4\frac{x}{x - 2} - \frac{8}{x + 2} = \frac{8}{x^2 - 4}, what is the value of xx?

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Answer: 4

Answer

4
To solve the rational equation, multiply both sides by the least common denominator, x2−4x^2 - 4. This results in the equation x(x+2)−8(x−2)=8x(x + 2) - 8(x - 2) = 8. Expanding and combining like terms yields the quadratic equation x2−6x+8=0x^2 - 6x + 8 = 0. Factoring this equation gives (x−2)(x−4)=0(x - 2)(x - 4) = 0, which yields the potential solutions x=2x = 2 and x=4x = 4. However, x=2x = 2 must be excluded because it makes the denominators of the original equation equal to zero. Therefore, the only valid solution is x=4x = 4.

Step-by-Step Solution

1
Determine the least common denominator (LCD) for the terms in the equation.
The LCD is (x−2)(x+2)=x2−4(x - 2)(x + 2) = x^2 - 4.
Multiplying the entire equation by the LCD clears all rational expressions.
2
Multiply the entire equation by the LCD to eliminate the denominators.
x(x+2)−8(x−2)=8x(x + 2) - 8(x - 2) = 8
This simplifies the rational equation into a polynomial equation.
3
Expand, simplify, and solve the resulting quadratic equation.
x2−6x+8=0⇒(x−2)(x−4)=0⇒x=2 or x=4x^2 - 6x + 8 = 0 \Rightarrow (x - 2)(x - 4) = 0 \Rightarrow x = 2 \text{ or } x = 4
Factoring the quadratic equation gives the potential solutions.
4
Substitute each potential solution into the original equation to check for extraneous solutions.
Substituting x=2x = 2 leads to division by zero, making the expression undefined. Substituting x=4x = 4 yields a valid identity: 42−86=23\frac{4}{2} - \frac{8}{6} = \frac{2}{3}. Therefore, x=4x = 4 is the only valid solution.
Values that make any denominator in the original equation equal to zero must be excluded.

Key Concept

Solving rational equations and checking for extraneous solutions.
Question 12Question →

If x+2=x\sqrt{x + 2} = x, what is the value of x−1x - 1?

Show answer & explanation

Answer: 1

Answer

The value of x−1x - 1 is 11.
The valid solution to the radical equation is x=2x = 2. Substituting this into the expression x−1x - 1 yields 2−1=12 - 1 = 1.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical.
x+2=x2x + 2 = x^2
Squaring both sides removes the square root, allowing us to solve the resulting equation.
2
Rewrite the equation in standard quadratic form.
x2−x−2=0x^2 - x - 2 = 0
Moving all terms to one side sets the quadratic equation to zero, which is necessary for factoring.
3
Factor the quadratic equation.
(x−2)(x+1)=0(x - 2)(x + 1) = 0, which gives the potential solutions x=2x = 2 and x=−1x = -1.
Factoring allows us to find the roots of the quadratic equation.
4
Check both potential solutions in the original equation to identify any extraneous solutions.
For x=2x = 2, 2+2=2\sqrt{2 + 2} = 2 is true. For x=−1x = -1, −1+2=−1  ⟹  1=−1\sqrt{-1 + 2} = -1 \implies 1 = -1 is false.
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original radical equation.
5
Calculate the value of x−1x - 1 using the valid solution.
2−1=12 - 1 = 1
The question asks for the value of the expression x−1x - 1 rather than just the variable xx.

Key Concept

Radical Equations and Extraneous Solutions
Estimated Time:1m 0s
Question 13Question →

If xx is a solution to the equation 2x−3x−1−2x−2=1x2−3x+2\frac{2x - 3}{x - 1} - \frac{2}{x - 2} = \frac{1}{x^2 - 3x + 2}, what is the value of 2x−12x - 1?

Show answer & explanation

Answer: 6

Answer

6
The correct answer is the value obtained by solving the rational equation for xx and then calculating the expression 2x−12x-1. Multiplying the equation by the least common denominator (x−1)(x−2)(x-1)(x-2) yields the quadratic equation 2x2−9x+7=02x^2 - 9x + 7 = 0, which factors into (2x−7)(x−1)=0(2x - 7)(x - 1) = 0. The potential solution x=1x = 1 is extraneous because it makes the denominators of the original equation equal to zero. Thus, the only valid solution is x=7/2x = 7/2. Substituting this value into the expression gives 2(7/2)−1=62(7/2) - 1 = 6.

Step-by-Step Solution

1
Identify the least common denominator (LCD) of the rational expressions in the equation.
The LCD is (x−1)(x−2)=x2−3x+2(x - 1)(x - 2) = x^2 - 3x + 2. Since these denominators cannot be zero, we must have x≠1x \neq 1 and x≠2x \neq 2.
Finding the LCD allows us to eliminate the denominators by multiplying both sides of the equation.
2
Multiply both sides of the equation by the LCD, (x−1)(x−2)(x - 1)(x - 2), to clear the fractions.
(2x−3)(x−2)−2(x−1)=1(2x - 3)(x - 2) - 2(x - 1) = 1
This simplifies the rational equation into a polynomial equation.
3
Expand the products and simplify the equation.
(2x2−7x+6)−(2x−2)=1  ⟹  2x2−9x+8=1(2x^2 - 7x + 6) - (2x - 2) = 1 \implies 2x^2 - 9x + 8 = 1
Expanding the terms is necessary to group like terms and form a quadratic equation.
4
Subtract 1 from both sides to write the quadratic equation in standard form.
2x2−9x+7=02x^2 - 9x + 7 = 0
Standard form (ax2+bx+c=0ax^2 + bx + c = 0) is required to factor or apply the quadratic formula.
5
Factor the quadratic equation.
(2x−7)(x−1)=0(2x - 7)(x - 1) = 0, which gives potential solutions x=72x = \frac{7}{2} and x=1x = 1.
Factoring helps find the values of xx that satisfy the polynomial equation.
6
Check the potential solutions against the original denominators to identify any extraneous solutions.
For x=1x = 1, the denominators x−1x - 1 and x2−3x+2x^2 - 3x + 2 become zero, so x=1x = 1 is extraneous. For x=72x = \frac{7}{2}, the denominators are non-zero, so x=72x = \frac{7}{2} is the only valid solution.
Multiplying by variables can introduce extraneous solutions that make the original rational expressions undefined.
7
Calculate the value of the requested expression, 2x−12x - 1, using the valid solution x=72x = \frac{7}{2}.
2(72)−1=7−1=62\left(\frac{7}{2}\right) - 1 = 7 - 1 = 6
The question asks for the value of the expression 2x−12x - 1, not just the value of xx.

Key Concept

Solving rational equations by clearing denominators, factoring the resulting quadratic equation, and checking for extraneous solutions.
Question 14Question →

If x8−2x=0\frac{x}{8} - \frac{2}{x} = 0 and x>0x > 0, what is the value of xx?

Show answer & explanation

Answer: 4

Answer

4
The correct answer is 4. Adding the term 2x\frac{2}{x} to both sides of the equation yields x8=2x\frac{x}{8} = \frac{2}{x}. Cross-multiplying the terms gives x2=16x^2 = 16. Solving for xx by taking the square root of both sides gives x=4x = 4 or x=−4x = -4. Since the question specifies that x>0x > 0, the negative value is discarded, leaving 4 as the only valid solution.

Step-by-Step Solution

1
Add 2x\frac{2}{x} to both sides of the equation.
x8=2x\frac{x}{8} = \frac{2}{x}
To isolate the rational terms on opposite sides of the equation.
2
Cross-multiply the terms.
x2=16x^2 = 16
To eliminate the denominators and form a quadratic equation.
3
Solve for xx by taking the square root of both sides.
x=4x = 4 or x=−4x = -4
Taking the square root of 16 yields both positive and negative solutions.
4
Apply the constraint x>0x > 0.
x=4x = 4
The question specifies that xx must be greater than 0, which excludes the negative solution.

Key Concept

Solving rational equations by isolating terms, cross-multiplying, and applying given constraints.
Question 15Question →

If xx is the real solution to the equation 5x+6−x+3=3\sqrt{5x + 6} - \sqrt{x + 3} = 3, what is the value of xx?

Show answer & explanation

Answer: 6

Answer

6
The correct answer is 66. Isolating the first radical gives 5x+6=x+3+3\sqrt{5x+6} = \sqrt{x+3} + 3. Squaring both sides yields 5x+6=x+12+6x+35x+6 = x+12+6\sqrt{x+3}, which simplifies to 4x−6=6x+34x-6 = 6\sqrt{x+3}. Dividing by 22 and squaring both sides again gives (2x−3)2=9(x+3)(2x-3)^2 = 9(x+3), leading to the quadratic equation 4x2−21x−18=04x^2 - 21x - 18 = 0. Factoring this equation yields (4x+3)(x−6)=0(4x+3)(x-6)=0, which gives the potential solutions x=6x = 6 and x=−34x = -\frac{3}{4}. Checking both solutions in the original equation shows that x=6x = 6 is a valid solution because 36−9=6−3=3\sqrt{36} - \sqrt{9} = 6 - 3 = 3. The solution x=−34x = -\frac{3}{4} is extraneous because 2.25−2.25=0≠3\sqrt{2.25} - \sqrt{2.25} = 0 \neq 3.

Step-by-Step Solution

1
Isolate the first radical term and square both sides.
5x+6=x+12+6x+35x + 6 = x + 12 + 6\sqrt{x + 3}
Isolating one radical term simplifies the squaring process to eliminate that radical.
2
Isolate the remaining radical term and simplify the expression.
2x−3=3x+32x - 3 = 3\sqrt{x + 3}
Simplifying and isolating the remaining radical term prepares the equation for a second squaring step.
3
Square both sides to eliminate the remaining radical and form a quadratic equation.
4x2−21x−18=04x^2 - 21x - 18 = 0
Squaring both sides removes the last radical, turning the radical equation into a polynomial equation.
4
Factor the quadratic equation to find all candidate solutions.
x=6x = 6 or x=−34x = -\frac{3}{4}
Factoring (4x+3)(x−6)=0(4x + 3)(x - 6) = 0 gives the roots of the quadratic equation.
5
Substitute candidate solutions back into the original equation to identify extraneous roots.
The only valid solution is x=6x = 6; the solution x=−34x = -\frac{3}{4} is extraneous.
Squaring equations can introduce extraneous solutions, so candidate solutions must be verified in the original equation.

Key Concept

Solving equations with multiple radicals by isolating terms, squaring both sides, and verifying candidate solutions for extraneous roots.
Question 16Question →

What is the sum of all real solutions to the equation x−5x−1=2x+1−8x2−1\frac{x - 5}{x - 1} = \frac{2}{x + 1} - \frac{8}{x^2 - 1}?

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Answer: 5

Answer

The sum of all real solutions to the equation is 55.
Multiplying the equation by the least common denominator, (x−1)(x+1)(x - 1)(x + 1), yields (x−5)(x+1)=2(x−1)−8(x - 5)(x + 1) = 2(x - 1) - 8. Expanding and simplifying leads to the quadratic equation x2−6x+5=0x^2 - 6x + 5 = 0, which factors as (x−1)(x−5)=0(x - 1)(x - 5) = 0. This gives potential solutions of x=1x = 1 and x=5x = 5. Since x=1x = 1 makes the denominators of the original terms zero, it is extraneous and must be discarded. The only valid solution is x=5x = 5, and therefore the sum of all real solutions is 55.

Step-by-Step Solution

1
Find the least common denominator of the rational terms.
The denominators are x−1x - 1, x+1x + 1, and x2−1x^2 - 1. Since x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1), the least common denominator is (x−1)(x+1)(x - 1)(x + 1).
Finding the least common denominator allows us to eliminate the fractions by multiplying both sides.
2
Multiply the entire equation by the least common denominator (x−1)(x+1)(x - 1)(x + 1), assuming x≠1x \neq 1 and x≠−1x \neq -1.
(x−5)(x+1)=2(x−1)−8(x - 5)(x + 1) = 2(x - 1) - 8
This clears all rational expressions, leaving a polynomial equation.
3
Expand both sides of the equation and combine like terms.
x2−4x−5=2x−2−8⇒x2−4x−5=2x−10x^2 - 4x - 5 = 2x - 2 - 8 \Rightarrow x^2 - 4x - 5 = 2x - 10
Expanding allows the simplification of terms on each side of the equation.
4
Rearrange the equation to set it equal to zero.
x2−6x+5=0x^2 - 6x + 5 = 0
Moving all terms to one side forms a standard quadratic equation which can then be solved.
5
Factor the quadratic equation.
(x−1)(x−5)=0(x - 1)(x - 5) = 0, which gives potential solutions x=1x = 1 and x=5x = 5.
Factoring allows us to find the roots of the quadratic equation.
6
Check the potential solutions in the original equation to identify any extraneous solutions.
Substituting x=1x = 1 results in division by zero in the terms x−5x−1\frac{x - 5}{x - 1} and 8x2−1\frac{8}{x^2 - 1}, so x=1x = 1 is extraneous. Substituting x=5x = 5 yields 0=00 = 0, meaning x=5x = 5 is a valid real solution.
Multiplying by variables can introduce extraneous solutions that make the original denominators zero.
7
Calculate the sum of all valid solutions.
Since x=5x = 5 is the only valid solution, the sum of all real solutions is 55.
The question asks for the sum of all real solutions.

Key Concept

Solving rational equations by clearing denominators and checking for extraneous solutions.
Question 17Question →

If 2x+3x−2=2\frac{2}{x} + \frac{3}{x - 2} = 2 and x>1x > 1, what is the value of x−1x - 1?

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Answer: 3

Answer

The correct answer is 3.
By multiplying both sides of the equation by the common denominator x(x−2)x(x - 2), we obtain the quadratic equation 2x2−9x+4=02x^2 - 9x + 4 = 0. Factoring this equation yields the solutions x=0.5x = 0.5 and x=4x = 4. Since the problem specifies that x>1x > 1, we select x=4x = 4. Substituting this value into the expression x−1x - 1 gives the final answer of 3.

Step-by-Step Solution

1
Multiply the equation by the common denominator x(x−2)x(x - 2) to eliminate the denominators.
2(x−2)+3x=2x(x−2)2(x - 2) + 3x = 2x(x - 2)
This clears the fractions and allows us to work with a polynomial equation.
2
Expand and simplify both sides of the equation.
5x−4=2x2−4x5x - 4 = 2x^2 - 4x
Combining like terms simplifies the expression before solving.
3
Rearrange the terms to set the quadratic equation equal to zero.
2x2−9x+4=02x^2 - 9x + 4 = 0
Putting the equation in standard form ax2+bx+c=0ax^2 + bx + c = 0 is necessary for factoring.
4
Factor the quadratic expression.
(2x−1)(x−4)=0(2x - 1)(x - 4) = 0
Factoring allows us to find the roots of the quadratic equation.
5
Solve for xx and apply the condition x>1x > 1.
x=4x = 4
The equation yields two solutions: x=0.5x = 0.5 and x=4x = 4. The condition x>1x > 1 excludes x=0.5x = 0.5, leaving x=4x = 4 as the only valid solution.
6
Calculate the value of the requested expression x−1x - 1.
33
Substituting x=4x = 4 into x−1x - 1 gives 4−1=34 - 1 = 3.

Key Concept

Solving rational equations by clearing denominators and solving the resulting quadratic equation while considering domain constraints.

Alternative Method

Instead of factoring, the quadratic formula can be used to solve 2x2−9x+4=02x^2 - 9x + 4 = 0, where x=9±(−9)2−4(2)(4)2(2)=9±54x = \frac{9 \pm \sqrt{(-9)^2 - 4(2)(4)}}{2(2)} = \frac{9 \pm 5}{4}, giving x=0.5x = 0.5 and x=4x = 4. Applying x>1x > 1 leaves x=4x = 4, so x−1=3x - 1 = 3.
Estimated Time:1m 30s
Question 18Question →

Consider the equation:

x+3x−1+4x=4x2−x\frac{x + 3}{x - 1} + \frac{4}{x} = \frac{4}{x^2 - x}

What is the value of x+3x + 3 for the real value of xx that satisfies this equation?

Show answer & explanation

Answer: -5

Answer

-5
The correct answer is −5-5. Multiplying the rational equation by the common denominator x(x−1)x(x - 1) gives the simplified equation x(x+3)+4(x−1)=4x(x + 3) + 4(x - 1) = 4. Expanding and rearranging the terms leads to the quadratic equation x2+7x−8=0x^2 + 7x - 8 = 0. Factoring this quadratic yields (x+8)(x−1)=0(x + 8)(x - 1) = 0, giving the potential solutions x=−8x = -8 and x=1x = 1. Substituting these values back into the original equation shows that x=1x = 1 causes division by zero, making it extraneous. Thus, the only valid solution is x=−8x = -8. Substituting this valid solution into the expression x+3x + 3 yields −8+3=−5-8 + 3 = -5.

Step-by-Step Solution

1
Find the common denominator of the rational terms.
The denominators are x−1x - 1, xx, and x2−xx^2 - x. The common denominator is x(x−1)=x2−xx(x - 1) = x^2 - x.
Finding a common denominator allows us to eliminate the fractions by multiplying both sides.
2
Multiply the entire equation by the common denominator x(x−1)x(x - 1) to clear the denominators.
x(x+3)+4(x−1)=4x(x + 3) + 4(x - 1) = 4
This simplifies the rational equation into a polynomial equation.
3
Expand and simplify the resulting equation.
x2+3x+4x−4=4x^2 + 3x + 4x - 4 = 4, which simplifies to x2+7x−4=4x^2 + 7x - 4 = 4.
Expanding the terms allows us to collect like terms and identify the type of polynomial equation.
4
Write the quadratic equation in standard form and factor it.
x2+7x−8=0x^2 + 7x - 8 = 0, which factors as (x+8)(x−1)=0(x + 8)(x - 1) = 0.
Setting the quadratic equation to zero allows us to find its roots by factoring.
5
Find the potential solutions and check for extraneous values in the original equation.
The potential solutions are x=−8x = -8 and x=1x = 1. The value x=1x = 1 is extraneous because it makes the denominators x−1x - 1 and x2−xx^2 - x equal to zero. The only valid solution is x=−8x = -8.
We must check if the roots make any denominator in the original equation zero, as division by zero is undefined.
6
Evaluate the expression x+3x + 3 using the valid solution.
−8+3=−5-8 + 3 = -5
The question asks for the value of the expression x+3x + 3, not the value of xx itself.

Key Concept

Solving rational equations and checking for extraneous solutions
Question 19Question →

What is the sum of all real solutions to the equation 3x+7−103x+7=3\sqrt{3x + 7} - \frac{10}{\sqrt{3x + 7}} = 3?

Show answer & explanation

Answer: 6

Answer

The only real solution is 6, so the sum of all real solutions is 6.
The correct answer is 6. By substituting u=3x+7u = \sqrt{3x + 7}, the equation becomes u−10/u=3u - 10/u = 3, which simplifies to the quadratic u2−3u−10=0u^2 - 3u - 10 = 0. Factoring gives (u−5)(u+2)=0(u - 5)(u + 2) = 0. Since the principal square root must be non-negative, uu must be positive, so we reject u=−2u = -2 and keep u=5u = 5. Solving 3x+7=5\sqrt{3x + 7} = 5 by squaring both sides yields 3x+7=253x + 7 = 25, which simplifies to x=6x = 6. The extraneous solution x=−1x = -1 (which comes from u=−2u = -2) must be discarded because substituting it back into the original equation results in −3-3 instead of 33. Thus, the only real solution is 6.

Step-by-Step Solution

1
Substitute u=3x+7u = \sqrt{3x + 7} into the equation.
u−10u=3u - \frac{10}{u} = 3, where u>0u > 0.
This substitution simplifies the radical equation into a rational equation of a single variable.
2
Multiply the entire equation by uu and solve the resulting quadratic equation.
u2−3u−10=0  ⟹  (u−5)(u+2)=0u^2 - 3u - 10 = 0 \implies (u - 5)(u + 2) = 0, which yields u=5u = 5 or u=−2u = -2.
Clearing the denominator transforms the rational expression into a standard quadratic equation.
3
Apply the constraint that u=3x+7u = \sqrt{3x + 7} must be non-negative.
u=5u = 5 is the only valid root; u=−2u = -2 has no real solution and must be discarded.
The output of a principal square root is non-negative by definition.
4
Substitute back 3x+7=5\sqrt{3x + 7} = 5 and solve for xx.
3x+7=25  ⟹  3x=18  ⟹  x=63x + 7 = 25 \implies 3x = 18 \implies x = 6.
Squaring both sides eliminates the radical and allows us to isolate the variable xx.
5
Verify if solving 3x+7=−2\sqrt{3x+7} = -2 leads to an extraneous solution.
Squaring both sides of 3x+7=−2\sqrt{3x+7} = -2 gives 3x+7=4  ⟹  x=−13x+7 = 4 \implies x = -1. Substituting x=−1x = -1 into the original equation yields 4−104=2−5=−3≠3\sqrt{4} - \frac{10}{\sqrt{4}} = 2 - 5 = -3 \neq 3. Therefore, x=−1x = -1 is extraneous.
Verifying solutions is necessary to identify and discard values that do not satisfy the original equation.

Key Concept

Solving equations using substitution and identifying extraneous solutions
Estimated Time:2m 30s
Question 20Question →

If xx is a real solution to the equation 2x+7−x=2\sqrt{2x + 7} - x = 2, what is the value of x+4x + 4?

Show answer & explanation

Answer: 5

Answer

5
The correct answer is 5. To solve the equation 2x+7−x=2\sqrt{2x + 7} - x = 2, we first isolate the radical to get 2x+7=x+2\sqrt{2x + 7} = x + 2. Squaring both sides yields 2x+7=x2+4x+42x + 7 = x^2 + 4x + 4. Rearranging this equation into standard quadratic form gives x2+2x−3=0x^2 + 2x - 3 = 0, which factors as (x+3)(x−1)=0(x + 3)(x - 1) = 0. This gives two potential solutions: x=1x = 1 and x=−3x = -3. Checking these in the original equation shows that x=1x = 1 is valid, whereas x=−3x = -3 is extraneous because 2(−3)+7−(−3)=1+3=4≠2\sqrt{2(-3)+7} - (-3) = 1 + 3 = 4 \neq 2. Therefore, the only real solution is x=1x = 1. Substituting this into the expression x+4x + 4 gives 1+4=51 + 4 = 5.

Step-by-Step Solution

1
Isolate the radical expression on one side of the equation.
2x+7=x+2\sqrt{2x + 7} = x + 2
This sets up the equation to eliminate the radical by squaring both sides.
2
Square both sides of the equation to eliminate the radical.
2x+7=(x+2)22x + 7 = (x + 2)^2
Squaring a square root removes the radical, allowing the equation to be solved algebraically.
3
Expand the squared binomial on the right side.
2x+7=x2+4x+42x + 7 = x^2 + 4x + 4
Expanding (x+2)2(x+2)^2 yields a quadratic term, a linear term, and a constant term.
4
Rearrange the equation to set it equal to zero.
x2+2x−3=0x^2 + 2x - 3 = 0
Subtracting 2x2x and 77 from both sides simplifies the equation into standard quadratic form.
5
Factor the quadratic equation.
(x+3)(x−1)=0(x + 3)(x - 1) = 0
Factoring allows us to find the potential solutions for xx.
6
Identify potential solutions and check for extraneous solutions by substituting them back into the original equation.
For x=1x = 1, 2(1)+7−1=3−1=2\sqrt{2(1) + 7} - 1 = 3 - 1 = 2, which is true. For x=−3x = -3, 2(−3)+7−(−3)=1+3=4\sqrt{2(-3) + 7} - (-3) = 1 + 3 = 4, which does not equal 22.
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original radical equation. Thus, x=−3x = -3 is extraneous, leaving x=1x = 1 as the only valid solution.
7
Calculate the value of the expression x+4x + 4 using the valid solution.
1+4=51 + 4 = 5
The question asks for the value of x+4x + 4, so we substitute the valid solution x=1x = 1 into this expression.

Key Concept

Solving radical equations and checking for extraneous solutions.
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Radical and Rational Equations Practice Questions — SAT | Examkin