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Irrational equations
This test develops methods for equations containing square and cube roots. Tasks require students to determine the domain, raise both sides to a power and check for extraneous solutions introduced during transformation.
Questions: 10Estimated time: 22 minutes
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Question 1out of 10
After squaring an irrational equation, a candidate root is obtained. What should be done?
The left side of √(x + a) = b is nonnegative, so require b ≥ 0 and x + a ≥ 0. Squaring gives x = b² − a. Check the candidate in the original equation because squaring can introduce extraneous roots.
Question 2out of 10
Why is checking necessary when solving an irrational equation?
The left side of √(x + a) = b is nonnegative, so require b ≥ 0 and x + a ≥ 0. Squaring gives x = b² − a. Check the candidate in the original equation because squaring can introduce extraneous roots.
Question 3out of 10
Find the domain of √(x − 3).
A square root is defined over the real numbers only when its radicand is non-negative.
- Write the condition that the entire expression under the square root must be at least zero.
- Solve the resulting inequality by moving the constant term to the other side. Pay attention to parentheses: subtracting a negative number means adding it.
- Check the boundary value and determine on which side the radicand is non-negative. Zero under the square root is allowed, so the boundary value is included in the domain.
Question 4out of 10
Solve ∛(x³ + 2) = ∛2.
The real cube root is one-to-one for all real inputs. Cubing both sides gives x³ + b = v, so x = ∛(v − b).
Question 5out of 10
Solve 4√x − 8 = 0.
From b√x + a = 0, obtain √x = −a/b for b ≠ 0. Require x ≥ 0 and −a/b ≥ 0. Then square: x = (−a/b)². Check the original equation.
Question 6out of 10
Solve the equation: √(x + 25) = x − 5.
First note that the right-hand side of the equation cannot be negative. Then square both sides and solve the resulting quadratic equation. Check every candidate in the original equation: one of them may be an extraneous root introduced by squaring.
Question 7out of 10
Solve √(x + 2) = 4.
The left side of √(x + a) = b is nonnegative, so require b ≥ 0 and x + a ≥ 0. Squaring gives x = b² − a. Check the candidate in the original equation because squaring can introduce extraneous roots.
Question 8out of 10
Solve √(x + 9) = 4.
The left side of √(x + a) = b is nonnegative, so require b ≥ 0 and x + a ≥ 0. Squaring gives x = b² − a. Check the candidate in the original equation because squaring can introduce extraneous roots.
Question 9out of 10
Solve x − 7√x + 10 = 0. Find the sum of all real solutions.
Let t = √x, so t must be non-negative. Solve the quadratic and recover x from each valid t. Add the x values, not the t values.
Question 10out of 10
Solve the equation: √(x + 11) = x − 1.
First note that the right-hand side of the equation cannot be negative. Then square both sides and solve the resulting quadratic equation. Check every candidate in the original equation: one of them may be an extraneous root introduced by squaring.
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Sample questions
- Solve √(x + ___) = ___.
- Solve ___√x + (___) = 0.
- Solve ∛(x³ + ___) = ∛___.
- After squaring an irrational equation, a candidate root is obtained. What should be done?
Irrational equations
In an irrational equation, the unknown occurs under a radical. The radicand of an even root must be non-negative, and an isolated square root cannot equal a negative expression.
The risk of squaring
Isolate one radical and square both sides, repeating if necessary. Raising both sides to an even power is not equivalent without sign restrictions and may create extraneous roots. Every candidate must therefore be substituted directly into the original equation.
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