Solving Rational and Radical Equations - Algebra
Card 1 of 30
What restriction must be placed on $x$ for $\sqrt{(x+1)(x-5)}$ to be real?
What restriction must be placed on $x$ for $\sqrt{(x+1)(x-5)}$ to be real?
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$(x+1)(x-5) \ge 0$. Product must be non-negative for real square root.
$(x+1)(x-5) \ge 0$. Product must be non-negative for real square root.
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What restriction must be placed on $x$ for $\sqrt{3-2x}$ to be real?
What restriction must be placed on $x$ for $\sqrt{3-2x}$ to be real?
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$3-2x \ge 0$, so $x \le \frac{3}{2}$. Square root requires $3 - 2x \ge 0$.
$3-2x \ge 0$, so $x \le \frac{3}{2}$. Square root requires $3 - 2x \ge 0$.
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What restriction must be placed on $x$ for $\sqrt{x-7}$ to be real?
What restriction must be placed on $x$ for $\sqrt{x-7}$ to be real?
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$x-7 \ge 0$, so $x \ge 7$. Square root requires non-negative radicand.
$x-7 \ge 0$, so $x \ge 7$. Square root requires non-negative radicand.
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What restriction must be placed on $x$ for $ rac{x+2}{x(x-4)}$ to be defined?
What restriction must be placed on $x$ for $ rac{x+2}{x(x-4)}$ to be defined?
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$x \ne 0$ and $x \ne 4$. Each factor in denominator cannot equal zero.
$x \ne 0$ and $x \ne 4$. Each factor in denominator cannot equal zero.
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What restriction must be placed on $x$ for $ rac{5}{2x+1}$ to be defined?
What restriction must be placed on $x$ for $ rac{5}{2x+1}$ to be defined?
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$x \ne -\frac{1}{2}$. Denominator equals zero when $2x + 1 = 0$.
$x \ne -\frac{1}{2}$. Denominator equals zero when $2x + 1 = 0$.
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What restriction must be placed on $x$ for $ rac{1}{x-3}$ to be defined?
What restriction must be placed on $x$ for $ rac{1}{x-3}$ to be defined?
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$x \ne 3$. Denominator equals zero when $x = 3$.
$x \ne 3$. Denominator equals zero when $x = 3$.
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What must you always do after clearing denominators or squaring both sides?
What must you always do after clearing denominators or squaring both sides?
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Check each candidate solution in the original equation. Verifies solutions aren't extraneous.
Check each candidate solution in the original equation. Verifies solutions aren't extraneous.
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What is an extraneous solution in a rational or radical equation?
What is an extraneous solution in a rational or radical equation?
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A value that solves the transformed equation but not the original equation. Created by operations that aren't reversible.
A value that solves the transformed equation but not the original equation. Created by operations that aren't reversible.
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What is the first step before solving a rational equation with denominators?
What is the first step before solving a rational equation with denominators?
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State the domain restrictions: denominators cannot equal $0$. Prevents division by zero in rational expressions.
State the domain restrictions: denominators cannot equal $0$. Prevents division by zero in rational expressions.
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Which value is extraneous when solving $\sqrt{x-2} = x-4$: $x=3$ or $x=5$?
Which value is extraneous when solving $\sqrt{x-2} = x-4$: $x=3$ or $x=5$?
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$x = 3$ is extraneous. Makes right side negative while left side is positive.
$x = 3$ is extraneous. Makes right side negative while left side is positive.
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What is the least common denominator (LCD) of $ rac{1}{x}$ and $ rac{1}{x-2}$?
What is the least common denominator (LCD) of $ rac{1}{x}$ and $ rac{1}{x-2}$?
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$x(x-2)$. Product of all unique denominators.
$x(x-2)$. Product of all unique denominators.
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What is the solution to $\sqrt{9-x} = 2$?
What is the solution to $\sqrt{9-x} = 2$?
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$x = 5$. Square both sides: $9 - x = 4$.
$x = 5$. Square both sides: $9 - x = 4$.
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What is the solution to $\sqrt{x} = 7$?
What is the solution to $\sqrt{x} = 7$?
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$x = 49$. Square both sides: $x = 49$.
$x = 49$. Square both sides: $x = 49$.
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What is the solution to $ rac{3}{x+2} = 1$ with $x \ne -2$?
What is the solution to $ rac{3}{x+2} = 1$ with $x \ne -2$?
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$x = 1$. Cross-multiply: $3 = 1(x+2)$, so $x = 1$.
$x = 1$. Cross-multiply: $3 = 1(x+2)$, so $x = 1$.
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Solve $ rac{1}{x} + \frac{1}{2} = 1$ with $x \ne 0$.
Solve $ rac{1}{x} + \frac{1}{2} = 1$ with $x \ne 0$.
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$x = 2$. Multiply by LCD $2x$: $2 + x = 2x$.
$x = 2$. Multiply by LCD $2x$: $2 + x = 2x$.
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Solve $ rac{2}{x} = \frac{3}{x+1}$ with $x \ne 0,-1$.
Solve $ rac{2}{x} = \frac{3}{x+1}$ with $x \ne 0,-1$.
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$x = 2$. Cross-multiply: $2(x+1) = 3x$, so $x = 2$.
$x = 2$. Cross-multiply: $2(x+1) = 3x$, so $x = 2$.
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Solve $ rac{1}{x-1} - \frac{1}{x} = 1$ with $x \ne 0,1$.
Solve $ rac{1}{x-1} - \frac{1}{x} = 1$ with $x \ne 0,1$.
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$x = \frac{1}{2}$. Use LCD $x(x-1)$ and solve resulting equation.
$x = \frac{1}{2}$. Use LCD $x(x-1)$ and solve resulting equation.
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Solve $ rac{1}{x} + \frac{1}{2} = 1$ with $x \ne 0$.
Solve $ rac{1}{x} + \frac{1}{2} = 1$ with $x \ne 0$.
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$x = 2$. Multiply by LCD $2x$: $2 + x = 2x$.
$x = 2$. Multiply by LCD $2x$: $2 + x = 2x$.
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What is the solution to $ rac{2x}{x-5} = 3$ with $x \ne 5$?
What is the solution to $ rac{2x}{x-5} = 3$ with $x \ne 5$?
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$x = 15$. Cross-multiply: $2x = 3(x-5)$, so $x = 15$.
$x = 15$. Cross-multiply: $2x = 3(x-5)$, so $x = 15$.
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What is the solution to $ rac{x+1}{x} = 4$ with $x \ne 0$?
What is the solution to $ rac{x+1}{x} = 4$ with $x \ne 0$?
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$x = \frac{1}{3}$. Cross-multiply: $x+1 = 4x$, so $x = \frac{1}{3}$.
$x = \frac{1}{3}$. Cross-multiply: $x+1 = 4x$, so $x = \frac{1}{3}$.
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Solve $ rac{1}{x} + \frac{1}{x} = \frac{1}{3}$ with $x \ne 0$.
Solve $ rac{1}{x} + \frac{1}{x} = \frac{1}{3}$ with $x \ne 0$.
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$x = 6$. Combine: $\frac{2}{x} = \frac{1}{3}$, so $6 = x$.
$x = 6$. Combine: $\frac{2}{x} = \frac{1}{3}$, so $6 = x$.
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Solve $\sqrt{x-1} = \sqrt{5}$.
Solve $\sqrt{x-1} = \sqrt{5}$.
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$x = 6$. Square both sides: $x - 1 = 5$.
$x = 6$. Square both sides: $x - 1 = 5$.
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Which value is extraneous for $\sqrt{x} + 2 = x$: $x=1$ or $x=4$?
Which value is extraneous for $\sqrt{x} + 2 = x$: $x=1$ or $x=4$?
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$x = 1$ is extraneous. Makes left side smaller than right side.
$x = 1$ is extraneous. Makes left side smaller than right side.
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Solve $\sqrt{x+4} = x$ over the real numbers.
Solve $\sqrt{x+4} = x$ over the real numbers.
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$x = 4$. Square both sides, solve quadratic, reject negative solution.
$x = 4$. Square both sides, solve quadratic, reject negative solution.
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Solve $\sqrt{x} + 2 = x$ over the real numbers.
Solve $\sqrt{x} + 2 = x$ over the real numbers.
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$x = 4$. Isolate radical, square both sides, check solutions.
$x = 4$. Isolate radical, square both sides, check solutions.
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Solve $\sqrt{x} + 1 = 4$.
Solve $\sqrt{x} + 1 = 4$.
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$x = 9$. Isolate radical: $\sqrt{x} = 3$.
$x = 9$. Isolate radical: $\sqrt{x} = 3$.
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Which value is extraneous when solving $\sqrt{x+4} = x$: $x=4$ or $x=-1$?
Which value is extraneous when solving $\sqrt{x+4} = x$: $x=4$ or $x=-1$?
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$x = -1$ is extraneous. Makes $\sqrt{x+4}$ negative, which is impossible.
$x = -1$ is extraneous. Makes $\sqrt{x+4}$ negative, which is impossible.
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Solve $\sqrt{x+1} + \sqrt{x+1} = 6$.
Solve $\sqrt{x+1} + \sqrt{x+1} = 6$.
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$x = 8$. Combine like terms: $2\sqrt{x+1} = 6$.
$x = 8$. Combine like terms: $2\sqrt{x+1} = 6$.
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Solve $\sqrt{x+3} = \sqrt{x-1}$ over the real numbers.
Solve $\sqrt{x+3} = \sqrt{x-1}$ over the real numbers.
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No solution. Would require $x + 3 = x - 1$, impossible.
No solution. Would require $x + 3 = x - 1$, impossible.
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Which value is extraneous when solving $\sqrt{x+1} = x$: $\frac{1-\sqrt{5}}{2}$ or $\frac{1+\sqrt{5}}{2}$?
Which value is extraneous when solving $\sqrt{x+1} = x$: $\frac{1-\sqrt{5}}{2}$ or $\frac{1+\sqrt{5}}{2}$?
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$x = \frac{1-\sqrt{5}}{2}$ is extraneous. Negative value fails the original square root equation.
$x = \frac{1-\sqrt{5}}{2}$ is extraneous. Negative value fails the original square root equation.
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