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Radical and Rational Equations

43 questions

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Question 41Question →
An equation is shown below.
x+2x−1−4x=4x2−x\frac{x+2}{x-1} - \frac{4}{x} = \frac{4}{x^2 - x}

What is the real solution to the equation above?

Show answer & explanation

Answer: 2

Answer

The only real solution to the equation is 22.
To solve the rational equation, we multiply all terms by the common denominator x(x−1)x(x-1), assuming x≠0x \neq 0 and x≠1x \neq 1. This results in the equation x(x+2)−4(x−1)=4x(x+2) - 4(x-1) = 4. Expanding and simplifying this gives x2+2x−4x+4=4x^2 + 2x - 4x + 4 = 4, which simplifies to x2−2x=0x^2 - 2x = 0. Factoring the quadratic expression yields x(x−2)=0x(x-2) = 0, which gives the potential solutions x=0x = 0 and x=2x = 2. However, substituting x=0x = 0 back into the original equation causes a division by zero. Therefore, x=0x = 0 is an extraneous solution, and the only valid real solution is 22.

Step-by-Step Solution

1
Find the common denominator for the terms in the rational equation.
The common denominator is x(x−1)=x2−xx(x-1) = x^2 - x, which requires x≠0x \neq 0 and x≠1x \neq 1.
Multiplying by the common denominator allows us to eliminate the fractions.
2
Multiply every term in the equation by the common denominator x(x−1)x(x-1) and simplify.
x(x+2)−4(x−1)=4x(x+2) - 4(x-1) = 4
This clears the denominators and converts the rational equation into a polynomial equation.
3
Expand and simplify the resulting equation to standard quadratic form.
x2+2x−4x+4=4x^2 + 2x - 4x + 4 = 4, which simplifies to x2−2x=0x^2 - 2x = 0.
Grouping like terms is necessary to solve the quadratic equation.
4
Factor the quadratic equation to determine the potential solutions.
x(x−2)=0x(x-2) = 0, giving potential solutions of x=0x = 0 and x=2x = 2.
Factoring allows us to find the roots of the quadratic expression.
5
Check the potential solutions in the original equation to identify any extraneous solutions.
x=0x = 0 is extraneous because it leads to division by zero. Thus, the only real solution is x=2x = 2.
Solutions that make any denominator in the original equation equal to zero must be excluded.

Key Concept

Solving rational equations and identifying extraneous solutions
Question 42Question →

For a real number xx, the equation 3x+10=x+2\sqrt{3x + 10} = x + 2 is given. What is the value of the expression x−1x - 1?

Show answer & explanation

Answer: 1

Answer

The correct answer is the value 1, which is obtained by evaluating the expression for the only valid solution to the equation.
To solve the given radical equation, square both sides to obtain a quadratic equation, which simplifies to x2+x−6=0x^2 + x - 6 = 0. Factoring this equation yields potential solutions of x=2x = 2 and x=−3x = -3. Substituting x=2x = 2 back into the original equation produces a true statement, confirming it as a valid solution. Substituting x=−3x = -3 results in an inequality, identifying it as an extraneous solution. Evaluating the expression for the valid solution yields 2−1=12 - 1 = 1.

Step-by-Step Solution

1
Eliminate the radical by squaring both sides of the equation.
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 removes the square root, allowing the equation to be solved algebraically.
2
Move all terms to one side to set the quadratic equation to zero.
x2+x−6=0x^2 + x - 6 = 0.
A quadratic equation must be in the form ax2+bx+c=0ax^2 + bx + c = 0 to solve it by factoring.
3
Factor the quadratic equation.
(x+3)(x−2)=0(x + 3)(x - 2) = 0, giving potential solutions of x=−3x = -3 and x=2x = 2.
Factoring allows us to find the roots that satisfy the quadratic relationship.
4
Check both potential solutions in the original equation to identify any extraneous solutions.
For x=2x = 2, 3(2)+10=2+2⇒4=4\sqrt{3(2) + 10} = 2 + 2 \Rightarrow 4 = 4 (valid). For x=−3x = -3, 3(−3)+10=−3+2⇒1=−1\sqrt{3(-3) + 10} = -3 + 2 \Rightarrow 1 = -1 (invalid). Thus, x=−3x = -3 is extraneous and x=2x = 2 is the only valid solution.
Squaring both sides of an equation can introduce extraneous roots that do not satisfy the original radical equation.
5
Evaluate the requested expression using the valid solution.
x−1=2−1=1x - 1 = 2 - 1 = 1.
The question asks for the value of the expression rather than the value of the variable itself.

Key Concept

Radical and Rational Equations
Question 43Question →
An equation is shown below.
xx−3−2x+1=8x2−2x−3\frac{x}{x-3} - \frac{2}{x+1} = \frac{8}{x^2-2x-3}
What is the real solution to the equation above?
Show answer & explanation

Answer: 2

Answer

The only real solution is 2.
To solve the rational equation, multiply both sides by the least common denominator, which is (x−3)(x+1)=x2−2x−3(x-3)(x+1) = x^2-2x-3. This yields x(x+1)−2(x−3)=8x(x+1) - 2(x-3) = 8. Simplifying this equation gives x2−x−2=0x^2 - x - 2 = 0. Factoring the quadratic expression gives (x−2)(x+1)=0(x-2)(x+1) = 0, which yields potential solutions of x=2x = 2 and x=−1x = -1. However, substituting x=−1x = -1 into the original equation results in division by zero, making it an extraneous solution. Substituting x=2x = 2 into the original equation is valid, so the only real solution is 2.

Step-by-Step Solution

1
Factor the quadratic denominator on the right side of the equation.
x2−2x−3=(x−3)(x+1)x^2 - 2x - 3 = (x - 3)(x + 1)
This helps identify the least common denominator of the rational terms.
2
Multiply all terms of the equation by the least common denominator, (x−3)(x+1)(x - 3)(x + 1), to eliminate the denominators.
x(x+1)−2(x−3)=8x(x + 1) - 2(x - 3) = 8
Multiplying by the LCD clears the rational expressions, converting the equation into a polynomial equation, under the restriction that x≠3x \neq 3 and x≠−1x \neq -1.
3
Expand and simplify the resulting equation.
x2+x−2x+6=8⇒x2−x−2=0x^2 + x - 2x + 6 = 8 \Rightarrow x^2 - x - 2 = 0
This puts the equation into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
4
Factor the quadratic equation.
(x−2)(x+1)=0(x - 2)(x + 1) = 0
Factoring allows us to find the potential solutions by setting each factor equal to zero.
5
Find the roots of the equation.
x=2 or x=−1x = 2 \text{ or } x = -1
These are the values of xx that satisfy the factored 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 2x+1\frac{2}{x+1} and 8x2−2x−3\frac{8}{x^2-2x-3}, so x=−1x = -1 is extraneous. Substituting x=2x = 2 yields a valid statement: −83=−83-\frac{8}{3} = -\frac{8}{3}.
Solutions that make any denominator in the original rational equation equal to zero are extraneous and must be excluded.

Key Concept

Solving rational equations and identifying extraneous solutions.
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