Square roots and fractions that create fake solutions.
36practice questions
3levels
The key idea
Squaring both sides or clearing denominators can create solutions that don't satisfy the original equation. Always substitute back: a square root can't equal a negative number, and no denominator can be 0.
The trap
Keeping both roots of the squared equation without checking.
Can Desmos do it?
Graphing each side of the original equation shows only the true intersections, which is a good check. But once a constant is unknown, you have to reason about which roots survive.
College Board skill: Nonlinear equations in one variable and systems of equations in two variables · Advanced Math
Try one at each level
One question from each level. The drill has 12 at every level, and moves you up after four right in a row.
Level 1 · Warm-up
What are all the solutions to x−1=x−3?
A5 only
B2 only
C2 and 5
D6 only
Show the answer
Answer: ASquare both sides: x−1=x2−6x+9, so x2−7x+10=0, with roots 5 and 2. Check: at x=2, the right side is −1, which is negative, so 2 is extraneous. The only solution is 5.
Level 2 · Test level
What is the solution to the equation x−5x−x+56=x2−2550?
Show the answer
Answer: -4Multiply both sides by (x−5)(x+5): x(x+5)−6(x−5)=50, which simplifies to x2−x−20=0, or (x−5)(x+4)=0. But x=5 makes a denominator 0, so the only solution is −4.
Level 3 · Beyond the test
In the equation x−13x−x+1k=x2−16, k is a constant. What is the sum of all values of k for which the equation has no solution?
Show the answer
Answer: 12Clear the denominators: 3x(x+1)−k(x−1)=6, or 3x2+(3−k)x+1k−6=0. Substituting x=1 gives 0 for every k, so 1 is always a root, and it is always excluded. The other root is 3k−2. There is no solution when it is also excluded: 3k−2=1 or 3k−2=−1, so k=9 or k=3. The sum is 12.
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