Examkin

Rational and Radical Expressions and Equations

35 questions

All types35Multiple choice19Numeric7Fill in the blank9
Question 21Question →

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

Fill in the blanks below

The rational expression x+5x2−4x−12\frac{x + 5}{x^2 - 4x - 12} is undefined for two real values of xx. The smaller of these values is and the larger of these values is .
Show answer & explanation

Answer

The rational expression is undefined for x=−2x = -2 and x=6x = 6.
The expression is undefined when the denominator is zero. Solving x2−4x−12=0x^2 - 4x - 12 = 0 yields (x−6)(x+2)=0(x-6)(x+2) = 0, which gives x=−2x = -2 and x=6x = 6. The smaller value is −2-2 and the larger value is 66.

Step-by-Step Solution

1
Identify the condition that makes a rational expression undefined.
The denominator must equal zero: x2−4x−12=0x^2 - 4x - 12 = 0
Division by zero is undefined in the set of real numbers.
2
Factor the quadratic equation x2−4x−12=0x^2 - 4x - 12 = 0.
(x−6)(x+2)=0(x - 6)(x + 2) = 0
We need to find two numbers that multiply to −12-12 and add to −4-4. These numbers are −6-6 and 22.
3
Solve for xx by setting each factor to zero.
x=6x = 6 or x=−2x = -2
By the zero product property, if (x−6)(x+2)=0(x - 6)(x + 2) = 0, then x−6=0x - 6 = 0 or x+2=0x + 2 = 0.
4
Determine the smaller and larger values to place in the blanks.
Smaller value is −2-2 and larger value is 66.
Comparing the two numbers, −2-2 is less than 66.

Key Concept

A rational expression is undefined when its denominator equals zero.
Question 22Question →

When solving the radical equation 3x+10+x+2=2\sqrt{3x+10} + \sqrt{x+2} = 2 for all real values of xx, squaring both sides yields two potential solutions. What is the numerical value of the extraneous solution?

Fill in the blanks below

The numerical value of the extraneous solution is .
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Answer

The extraneous solution is x=2x = 2.
The value x=2x = 2 is the extraneous solution because substituting it into the original equation yields 16+4=6≠2\sqrt{16} + \sqrt{4} = 6 \neq 2, showing it does not satisfy the original equation despite being generated by the algebraic squaring process.

Step-by-Step Solution

1
Isolate one of the radical terms in the equation.
3x+10=2−x+2\sqrt{3x+10} = 2 - \sqrt{x+2}
Isolating one radical simplifies the squaring process.
2
Square both sides of the equation.
3x+10=4−4x+2+(x+2)3x+10 = 4 - 4\sqrt{x+2} + (x+2), which simplifies to 2x+4=−4x+22x+4 = -4\sqrt{x+2}.
Squaring eliminates the isolated radical.
3
Divide both sides by 2 and square both sides again.
x+2=−2x+2⇒(x+2)2=4(x+2)x+2 = -2\sqrt{x+2} \Rightarrow (x+2)^2 = 4(x+2), which simplifies to x2−4=0x^2 - 4 = 0.
This eliminates the remaining radical term.
4
Solve the quadratic equation for xx.
x=2x = 2 or x=−2x = -2
Factoring (x−2)(x+2)=0(x-2)(x+2) = 0 gives the potential solutions.
5
Check both potential solutions in the original equation.
For x=−2x = -2: 4+0=2\sqrt{4} + \sqrt{0} = 2, which is true. For x=2x = 2: 16+4=6≠2\sqrt{16} + \sqrt{4} = 6 \neq 2, which is false.
This identifies which solution is extraneous.

Key Concept

Solving radical equations and verifying solutions to identify extraneous roots
Estimated Time:2m 0s
Question 23Question →

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

Show answer & explanation

Answer: 2x+1x2+x\frac{2x+1}{x^2+x}

Answer

The correct expression is 2x+1x2+x\frac{2x+1}{x^2+x}.
The correct expression is obtained by finding a common denominator of x(x+1)x(x+1), rewriting the numerators as x+1x+1 and xx, and then adding them to get 2x+12x+1 over the expanded denominator x2+xx^2+x.

Step-by-Step Solution

1
Find a common denominator for the two rational terms.
The common denominator for 1x\frac{1}{x} and 1x+1\frac{1}{x+1} is x(x+1)x(x+1), which expands to x2+xx^2+x.
To add fractions with different denominators, they must share a common denominator.
2
Rewrite each fraction with the common denominator by multiplying their numerators and denominators by the appropriate factor.
1x=x+1x(x+1)\frac{1}{x} = \frac{x+1}{x(x+1)} and 1x+1=xx(x+1)\frac{1}{x+1} = \frac{x}{x(x+1)}
Multiplying the numerator and denominator of each fraction by the missing factor of the common denominator preserves the value of the fraction.
3
Add the numerators together over the common denominator and simplify.
(x+1)+xx(x+1)=2x+1x2+x\frac{(x+1) + x}{x(x+1)} = \frac{2x+1}{x^2+x}
Combining the numerators and expanding the denominator gives the simplified equivalent expression.

Key Concept

Adding rational expressions by finding a common denominator
Question 24Question →

What is the positive real solution to the equation 3x−2−2x=1\frac{3}{x-2} - \frac{2}{x} = 1?

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Answer: 4

Answer

The positive real solution to the equation is 4.
By multiplying the entire equation by the least common denominator, x(x−2)x(x-2), the rational equation is cleared of fractions, resulting in 3x−2(x−2)=x(x−2)3x - 2(x-2) = x(x-2). Simplifying this yields the quadratic equation x2−3x−4=0x^2 - 3x - 4 = 0. Factoring the quadratic equation gives (x−4)(x+1)=0(x - 4)(x + 1) = 0, which has roots of x=4x = 4 and x=−1x = -1. The positive solution is 4.

Step-by-Step Solution

1
Multiply both sides of the equation by the least common denominator, x(x−2)x(x-2), to eliminate the fractions.
3x−2(x−2)=x(x−2)3x - 2(x-2) = x(x-2)
Eliminating the denominators simplifies the rational equation into a polynomial equation.
2
Distribute and combine like terms to simplify both sides of the equation.
x+4=x2−2xx + 4 = x^2 - 2x
Simplification is necessary before rearranging the equation into a solvable form.
3
Rearrange the terms into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
x2−3x−4=0x^2 - 3x - 4 = 0
Setting the quadratic expression equal to zero allows for factoring and solving.
4
Factor the quadratic equation.
(x−4)(x+1)=0(x - 4)(x + 1) = 0
Finding the factors reveals the potential values for xx.
5
Solve for the roots and select the positive real solution.
x=4x = 4
The equation yields two solutions, x=4x = 4 and x=−1x = -1, and the question explicitly requests the positive solution.

Key Concept

Solving rational equations by converting them to quadratic equations using the least common denominator.
Question 25Question →

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

Show answer & explanation

Answer: 4x2(x2−1)2\frac{4x^2}{(x^2-1)^2}

Answer

The expression is equivalent to the fraction with numerator four times x squared and denominator the square of the difference x squared minus one.
The correct answer is found by first rewriting the terms inside the parentheses with the common denominator (x2−1)(x^2-1), which simplifies the sum to 2xx2−1\frac{2x}{x^2-1}. Squaring the numerator (2x)2(2x)^2 yields 4x24x^2, and squaring the denominator yields (x2−1)2(x^2-1)^2.

Step-by-Step Solution

1
Find a common denominator for the two fractions inside the parentheses.
The common denominator is (x−1)(x+1)=x2−1(x-1)(x+1) = x^2-1. The expression inside the parentheses becomes x+1x2−1+x−1x2−1\frac{x+1}{x^2-1} + \frac{x-1}{x^2-1}.
To add rational expressions, they must have a common denominator.
2
Combine the numerators over the common denominator.
The sum is (x+1)+(x−1)x2−1=2xx2−1\frac{(x+1) + (x-1)}{x^2-1} = \frac{2x}{x^2-1}.
Combine the like terms in the numerator to simplify the expression before squaring.
3
Square the simplified rational expression by squaring the numerator and the denominator separately.
(2xx2−1)2=(2x)2(x2−1)2=4x2(x2−1)2.\left(\frac{2x}{x^2-1}\right)^2 = \frac{(2x)^2}{(x^2-1)^2} = \frac{4x^2}{(x^2-1)^2}.
Apply the power of a quotient property to obtain the final simplified expression.

Key Concept

Simplifying rational expressions and applying the power of a quotient rule.
Estimated Time:1m 30s
Question 26Question →

If xx is a real number, what is the sum of all distinct real solutions to the rational equation 3(x−2)2−2x−2=1\frac{3}{(x-2)^2} - \frac{2}{x-2} = 1?

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Answer: 2

Answer

The sum of all distinct real solutions is 2.
The correct answer is obtained by multiplying the equation by the least common denominator, (x−2)2(x-2)^2, which yields the quadratic equation x2−2x−3=0x^2 - 2x - 3 = 0. Factoring this equation gives two solutions: 33 and −1-1. Since neither value makes the original denominator equal to zero, both are valid solutions. Summing these values gives a result of 22.

Step-by-Step Solution

1
Identify the domain constraints of the rational equation.
The denominator x−2x-2 cannot be zero, so the domain constraint is x≠2x \neq 2.
Division by zero is undefined.
2
Eliminate the denominators by multiplying the entire equation by the least common denominator, which is (x−2)2(x-2)^2.
3−2(x−2)=(x−2)23 - 2(x-2) = (x-2)^2
This transforms the rational equation into a polynomial equation.
3
Expand both sides of the equation and combine like terms.
3−2x+4=x2−4x+4  ⟹  7−2x=x2−4x+43 - 2x + 4 = x^2 - 4x + 4 \implies 7 - 2x = x^2 - 4x + 4
Standard algebraic expansion simplifies the equation into a quadratic form.
4
Rearrange the quadratic equation into standard form ax2+bx+c=0ax^2 + bx + c = 0.
x2−2x−3=0x^2 - 2x - 3 = 0
Setting the equation to zero allows us to solve for xx by factoring.
5
Factor the quadratic equation.
(x−3)(x+1)=0  ⟹  x=3(x-3)(x+1) = 0 \implies x = 3 or x=−1x = -1
Factoring finds the values of xx that satisfy the quadratic equation.
6
Verify if the potential solutions satisfy the domain constraint and calculate their sum.
Both 33 and −1-1 are within the domain (x≠2x \neq 2). The sum is 3+(−1)=23 + (-1) = 2.
Extraneous solutions must be discarded before calculating the final sum.

Key Concept

Solving rational equations by clearing denominators and verifying domain constraints
Question 27Question →

For all real values of xx, what is the real solution to the equation x+13−1=x\sqrt{x+13} - 1 = x?

Show answer & explanation

Answer: 3

Answer

The only real solution to the equation is 3.
Isolating the radical term gives x+13=x+1\sqrt{x+13} = x+1. Squaring both sides yields x+13=x2+2x+1x+13 = x^2 + 2x + 1. Rearranging into standard quadratic form gives x2+x−12=0x^2 + x - 12 = 0, which factors as (x+4)(x−3)=0(x+4)(x-3) = 0. This yields candidate solutions x=3x = 3 and x=−4x = -4. Checking these values in the original equation shows that x=3x = 3 is a valid solution because 3+13−1=3\sqrt{3+13} - 1 = 3, whereas x=−4x = -4 is extraneous because −4+13−1=2≠−4\sqrt{-4+13} - 1 = 2 \neq -4.

Step-by-Step Solution

1
Isolate the radical expression on one side of the equation.
x+13=x+1\sqrt{x+13} = x+1
This prepares the equation for squaring both sides to eliminate the radical.
2
Square both sides of the equation.
x+13=(x+1)2  ⟹  x+13=x2+2x+1x+13 = (x+1)^2 \implies x+13 = x^2 + 2x + 1
Squaring both sides eliminates the radical and yields a polynomial equation.
3
Move all terms to one side to set the quadratic equation equal to zero.
x2+x−12=0x^2 + x - 12 = 0
This puts the equation in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0 so it can be solved by factoring.
4
Factor the quadratic equation.
(x+4)(x−3)=0  ⟹  x=3(x+4)(x-3) = 0 \implies x = 3 or x=−4x = -4
Factoring allows us to find the potential roots of the quadratic equation.
5
Check both candidate solutions in the original equation to identify any extraneous solutions.
For x=3x = 3: 3+13−1=4−1=3\sqrt{3+13} - 1 = 4 - 1 = 3 (valid). For x=−4x = -4: −4+13−1=3−1=2≠−4\sqrt{-4+13} - 1 = 3 - 1 = 2 \neq -4 (extraneous).
Squaring both sides of an equation can introduce extraneous roots that must be checked and discarded.

Key Concept

Solving radical equations and checking for extraneous solutions
Estimated Time:1m 30s
Question 28Question →

The rational expression x2−42x2+5x−3\frac{x^2 - 4}{2x^2 + 5x - 3} is undefined for two real values of xx. If the smaller value is aa and the larger value is bb, what are the values of aa and bb?

Fill in the blanks below

a=a =
b=b =
Show answer & explanation

Answer

The smaller value is −3-3 and the larger value is 1/21/2 (or 0.50.5).
A rational expression is undefined when the denominator is equal to zero. To find these values, set the denominator 2x2+5x−32x^2 + 5x - 3 equal to 00 and solve for xx. Factoring the quadratic yields (2x−1)(x+3)=0(2x - 1)(x + 3) = 0. Setting each factor to zero gives x=1/2x = 1/2 and x=−3x = -3. Since −3-3 is less than 1/21/2, the smaller value is −3-3 and the larger value is 1/21/2 (or 0.50.5).

Step-by-Step Solution

1
Identify the condition that makes a rational expression undefined.
The rational expression is undefined when its denominator is equal to zero: 2x2+5x−3=02x^2 + 5x - 3 = 0.
Division by zero is undefined in the set of real numbers.
2
Factor the quadratic expression in the denominator.
2x2+5x−3=(2x−1)(x+3)=02x^2 + 5x - 3 = (2x - 1)(x + 3) = 0.
Factoring allows us to find the roots of the quadratic equation using the zero product property.
3
Solve for the roots of the factored equation.
2x−1=0⇒x=1/22x - 1 = 0 \Rightarrow x = 1/2, and x+3=0⇒x=−3x + 3 = 0 \Rightarrow x = -3.
Setting each linear factor to zero determines the values of xx that make the denominator zero.
4
Assign the values to the variables based on the inequality constraint.
Since −3<1/2-3 < 1/2, the smaller value is a=−3a = -3 and the larger value is b=1/2b = 1/2 (or 0.50.5).
The question specifies that aa is the smaller value and bb is the larger value.

Key Concept

Identifying domain restrictions of rational expressions by finding where the denominator is equal to zero.
Estimated Time:1m 30s
Question 29Question →

Pipe A can fill a water tank in xx hours, where x>0x > 0. Pipe B can fill the same water tank in x+3x + 3 hours. If both pipes are opened at the same time, they can fill the tank together in exactly 2 hours. What is the value of xx?

Show answer & explanation

Answer: 3

Answer

The value of xx is 3.
The correct value is 3. The work rate equation for the two pipes is 1x+1x+3=12\frac{1}{x} + \frac{1}{x+3} = \frac{1}{2}. Combining the fractions on the left gives 2x+3x2+3x=12\frac{2x+3}{x^2+3x} = \frac{1}{2}. Cross-multiplying yields 4x+6=x2+3x4x + 6 = x^2 + 3x, which simplifies to the quadratic equation x2−x−6=0x^2 - x - 6 = 0. Factoring this quadratic equation gives (x−3)(x+2)=0(x-3)(x+2) = 0, which has the solutions x=3x = 3 and x=−2x = -2. Since xx represents a positive number of hours, the only valid solution is 3.

Step-by-Step Solution

1
Set up the work rate equation based on the individual and combined rates.
1x+1x+3=12\frac{1}{x} + \frac{1}{x+3} = \frac{1}{2}
The sum of the individual rates (in tanks per hour) equals the combined rate of the two pipes.
2
Find a common denominator to combine the fractions on the left side of the equation.
(x+3)+xx(x+3)=12  ⟹  2x+3x2+3x=12\frac{(x+3) + x}{x(x+3)} = \frac{1}{2} \implies \frac{2x+3}{x^2+3x} = \frac{1}{2}
To add fractions, they must share a common denominator, which is x(x+3)x(x+3).
3
Cross-multiply to eliminate the denominators and form a quadratic equation.
2(2x+3)=x2+3x  ⟹  4x+6=x2+3x  ⟹  x2−x−6=02(2x+3) = x^2+3x \implies 4x + 6 = x^2 + 3x \implies x^2 - x - 6 = 0
Cross-multiplication converts the rational equation into a polynomial equation.
4
Factor the quadratic equation to find the solutions for xx.
(x−3)(x+2)=0  ⟹  x=3(x-3)(x+2) = 0 \implies x = 3 or x=−2x = -2
Since xx represents time, it must be positive (x>0x > 0), so the only valid solution is x=3x = 3.

Key Concept

Solving rational equations in the context of work-rate problems
Question 30Question →

For all real numbers xx such that x≠0x \neq 0 and x≠3x \neq 3, what is the real solution to the equation xx−3+2x=3x−3\frac{x}{x-3} + \frac{2}{x} = \frac{3}{x-3}?

Show answer & explanation

Answer: -2

Answer

The correct answer is -2.
Subtracting xx−3\frac{x}{x-3} from both sides yields 2x=3−xx−3\frac{2}{x} = \frac{3-x}{x-3}. Since 3−x=−(x−3)3-x = -(x-3), the right side simplifies to −1-1 for all x≠3x \neq 3. The equation becomes 2x=−1\frac{2}{x} = -1, which gives x=−2x = -2. Since −2-2 does not violate the domain constraints, it is the correct solution.

Step-by-Step Solution

1
Subtract xx−3\frac{x}{x-3} from both sides of the equation.
2x=3−xx−3\frac{2}{x} = \frac{3-x}{x-3}
To group terms with common denominators on the same side.
2
Simplify the fraction on the right side of the equation.
2x=−1\frac{2}{x} = -1
The numerator 3−x3-x is the negative of the denominator x−3x-3, so their quotient is −1-1 for all x≠3x \neq 3.
3
Solve the simplified equation 2x=−1\frac{2}{x} = -1 for xx.
x=−2x = -2
Multiplying both sides by xx yields 2=−x2 = -x, and dividing by −1-1 gives x=−2x = -2.
4
Check the solution against the domain restrictions x≠0x \neq 0 and x≠3x \neq 3.
The solution x=−2x = -2 is valid.
Since −2-2 is neither 00 nor 33, it does not cause any denominator in the original equation to equal zero.

Key Concept

Solving rational equations by isolating terms with common denominators and checking for extraneous solutions.
Question 31Question →

What is the real solution to the equation 3x+1=x−3\sqrt{3x + 1} = x - 3?

Show answer & explanation

Answer: 8

Answer

The only real solution is 8.
The correct answer is the single value 8. Squaring both sides of the equation 3x+1=x−3\sqrt{3x+1} = x-3 leads to the quadratic equation x2−9x+8=0x^2-9x+8=0, which factors as (x−8)(x−1)=0(x-8)(x-1)=0. This yields potential solutions of 8 and 1. Checking 8 in the original equation gives 25=5\sqrt{25} = 5, which is correct. Checking 1 in the original equation gives 4=−2\sqrt{4} = -2, which is incorrect because the principal square root must be non-negative. Therefore, 1 is an extraneous solution, and 8 is the only valid solution.

Step-by-Step Solution

1
Square both sides of the equation to eliminate the radical.
3x+1=(x−3)23x + 1 = (x - 3)^2
To clear the square root and obtain a polynomial equation.
2
Expand the right side and move all terms to the right side to set the quadratic equation to zero.
x2−9x+8=0x^2 - 9x + 8 = 0
Expanding (x−3)2(x - 3)^2 gives x2−6x+9x^2 - 6x + 9. Subtracting 3x3x and 11 from both sides gives the standard form of the quadratic equation.
3
Factor the quadratic equation.
(x−8)(x−1)=0(x - 8)(x - 1) = 0
Finding two numbers that multiply to 8 and add to -9, which are -8 and -1, allows us to find potential solutions x=8x = 8 and x=1x = 1.
4
Check both potential solutions in the original equation to identify extraneous solutions.
Checking x=8x = 8: 3(8)+1=25=5\sqrt{3(8) + 1} = \sqrt{25} = 5 and 8−3=58 - 3 = 5 (True). Checking x=1x = 1: 3(1)+1=4=2\sqrt{3(1) + 1} = \sqrt{4} = 2 and 1−3=−21 - 3 = -2 (False).
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Solving radical equations and identifying extraneous solutions
Estimated Time:1m 30s
Question 32Question →

For what positive integer value of xx is the rational expression 2x−53x2−10x−8\frac{2x - 5}{3x^2 - 10x - 8} undefined?

Fill in the blanks below

The expression is undefined when xx is equal to .
Show answer & explanation

Answer

The expression is undefined when the positive integer x is equal to 4.
The rational expression is undefined when its denominator is 0. Setting the denominator 3x2−10x−83x^2 - 10x - 8 equal to 0 and factoring gives (3x+2)(x−4)=0(3x + 2)(x - 4) = 0. The roots of this equation are x=−23x = -\frac{2}{3} and x=4x = 4. Since the question restricts the target value to a positive integer, the correct value is 4.

Step-by-Step Solution

1
Set the denominator of the rational expression equal to 0.
3x2−10x−8=03x^2 - 10x - 8 = 0
A rational expression is undefined when its denominator is equal to 0 because division by zero is undefined.
2
Factor the quadratic equation 3x2−10x−8=03x^2 - 10x - 8 = 0.
(3x+2)(x−4)=0(3x + 2)(x - 4) = 0
Factoring the quadratic trinomial helps identify the values of x that satisfy the equation.
3
Solve for the roots by setting each factor to 0.
x=−23x = -\frac{2}{3} or x=4x = 4
According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero.
4
Filter the solutions based on the given constraint.
x=4x = 4
The question specifies that x must be a positive integer. Since −23-\frac{2}{3} is a negative fraction, we reject it and select 4.

Key Concept

Finding values that make a rational expression undefined by setting the denominator to zero and solving the resulting quadratic equation.
Estimated Time:1m 30s
Question 33Question →

If 3x+19−x=3\sqrt{3x + 19} - x = 3, what is the value of xx?

Show answer & explanation

Answer: 2

Answer

The value of xx is 2.
To solve the equation, isolate the radical to obtain 3x+19=x+3\sqrt{3x + 19} = x + 3. Squaring both sides yields 3x+19=x2+6x+93x + 19 = x^2 + 6x + 9. Rearranging terms results in the quadratic equation x2+3x−10=0x^2 + 3x - 10 = 0, which factors into (x+5)(x−2)=0(x + 5)(x - 2) = 0. The potential solutions are x=2x = 2 and x=−5x = -5. Testing these in the original equation shows that 22 is a valid solution because 3(2)+19−2=3\sqrt{3(2) + 19} - 2 = 3. Testing −5-5 results in 3(−5)+19−(−5)=7\sqrt{3(-5) + 19} - (-5) = 7, which does not equal 33. Thus, the only real solution is 22.

Step-by-Step Solution

1
Isolate the radical expression on one side of the equation.
3x+19=x+3\sqrt{3x + 19} = x + 3
Isolating the radical allows us to eliminate it by squaring both sides.
2
Square both sides of the equation.
3x+19=x2+6x+93x + 19 = x^2 + 6x + 9
Squaring a square root removes the radical. Remember to expand (x+3)2(x+3)^2 correctly as x2+6x+9x^2 + 6x + 9.
3
Rearrange the terms into standard quadratic form, ax2+bx+c=0ax^2 + bx + c = 0.
x2+3x−10=0x^2 + 3x - 10 = 0
Moving all terms to one side sets up the quadratic equation for factoring.
4
Factor the quadratic expression.
(x+5)(x−2)=0(x + 5)(x - 2) = 0
Factoring allows us to find the potential values of xx by setting each factor to zero.
5
Solve for the potential values of xx.
x=−5x = -5 or x=2x = 2
Setting x+5=0x + 5 = 0 gives x=−5x = -5, and setting x−2=0x - 2 = 0 gives x=2x = 2.
6
Check both potential solutions in the original equation to identify any extraneous solutions.
For x=2x = 2: 3(2)+19−2=25−2=5−2=3\sqrt{3(2) + 19} - 2 = \sqrt{25} - 2 = 5 - 2 = 3 (valid). For x=−5x = -5: 3(−5)+19−(−5)=4+5=2+5=7≠3\sqrt{3(-5) + 19} - (-5) = \sqrt{4} + 5 = 2 + 5 = 7 \neq 3 (extraneous).
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Key Concept

Solving radical equations and checking for extraneous solutions.
Question 34Question →

What is the value of the larger real solution to the equation 10x−3x−1=2\frac{10}{x} - \frac{3}{x-1} = 2?

Show answer & explanation

Answer: 2.5

Answer

The larger real solution is 2.52.5.
Multiplying the equation by the least common denominator x(x−1)x(x-1) yields 10(x−1)−3x=2x(x−1)10(x-1) - 3x = 2x(x-1). Simplifying this leads to 2x2−9x+10=02x^2 - 9x + 10 = 0. Factoring the quadratic yields (2x−5)(x−2)=0(2x-5)(x-2) = 0, which gives the solutions x=2.5x = 2.5 and x=2x = 2. The larger of these two solutions is 2.52.5.

Step-by-Step Solution

1
Multiply the entire equation by the least common denominator, which is x(x−1)x(x-1), for x≠0x \neq 0 and x≠1x \neq 1.
10(x−1)−3x=2x(x−1)10(x-1) - 3x = 2x(x-1)
This clears the fractions from the rational equation.
2
Expand both sides of the equation.
10x−10−3x=2x2−2x10x - 10 - 3x = 2x^2 - 2x
Expanding the terms allows us to combine like terms.
3
Combine like terms on the left side and move all terms to one side to set the quadratic equation equal to zero.
2x2−9x+10=02x^2 - 9x + 10 = 0
Setting the quadratic equation to zero is required to solve it by factoring.
4
Factor the quadratic equation by grouping.
(2x−5)(x−2)=0(2x-5)(x-2) = 0
Factoring allows us to find the roots of the quadratic equation.
5
Solve for xx by setting each factor equal to zero.
x=2.5x = 2.5 or x=2x = 2
By the zero product property, at least one of the factors must be zero.
6
Check for extraneous solutions and select the larger real value.
Both 22 and 2.52.5 are valid because they do not make the original denominators zero. The larger value is 2.52.5.
The question specifically asks for the larger of the two real solutions.

Key Concept

Solving rational equations by clearing denominators
Estimated Time:1m 30s
Question 35Question →
Consider the system of equations below:
y=x+37y = \sqrt{x + 37}
y=x−5y = x - 5
If (x,y)(x, y) is a real solution to this system, what is the value of xyxy?
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Answer: 84

Answer

84
The correct answer is 84 because solving the system yields the quadratic equation x2−11x−12=0x^2 - 11x - 12 = 0, which has roots x=12x = 12 and x=−1x = -1. Substituting these back into the original equations shows that only x=12x = 12 is valid (giving y=7y = 7), while x=−1x = -1 is extraneous (since 36=6≠−6\sqrt{36} = 6 \neq -6). The product of the coordinates of the valid solution is 12×7=8412 \times 7 = 84.

Step-by-Step Solution

1
Set the two expressions for yy equal to each other to solve for xx.
x+37=x−5\sqrt{x + 37} = x - 5
Since both equations are solved for yy, their right-hand sides must be equal at the point of intersection.
2
Square both sides of the equation to eliminate the radical.
x+37=(x−5)2⇒x+37=x2−10x+25x + 37 = (x - 5)^2 \Rightarrow x + 37 = x^2 - 10x + 25
Squaring is the inverse operation of the square root. The binomial on the right must be expanded fully using (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2.
3
Rearrange the quadratic equation into standard form.
x2−11x−12=0x^2 - 11x - 12 = 0
Subtracting xx and 3737 from both sides sets the quadratic equation to zero so it can be solved.
4
Factor the quadratic equation to find the algebraic solutions.
(x−12)(x+1)=0⇒x=12 or x=−1(x - 12)(x + 1) = 0 \Rightarrow x = 12 \text{ or } x = -1
Factoring the trinomial allows us to find the roots by setting each linear factor to zero.
5
Substitute each solution back into the original system to check for extraneous solutions.
For x=12x = 12: y=12−5=7y = 12 - 5 = 7 and y=12+37=7y = \sqrt{12 + 37} = 7 (Valid). For x=−1x = -1: y=−1−5=−6y = -1 - 5 = -6, but y=−1+37=6≠−6y = \sqrt{-1 + 37} = 6 \neq -6 (Extraneous).
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original radical relationship because the principal square root must be non-negative.
6
Calculate the product xyxy of the coordinates of the valid solution.
xy=12×7=84xy = 12 \times 7 = 84
The question asks for the product of xx and yy for the real solution (12,7)(12, 7).

Key Concept

Solving systems containing radical equations and verifying for extraneous solutions.
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