Solving Radical Equations
To clear a square root, you square both sides — but squaring can introduce false solutions, so checking is mandatory.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
Isolate the radical first, then square both whole sides. Solve the result, and substitute back into the original to discard any extraneous answers.
How to solve one, step by step
Example: solve $\sqrt{x + 3} = 5$.
- Square both sides: $x + 3 = 25$.
- Solve: $x = 22$. Check: $\sqrt{25} = 5$. Valid.
The mistakes that cost points
- Skipping the check. Squaring can create solutions that don't satisfy the original equation.
- Squaring term by term. Square the entire side at once, not each piece separately.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The equation $\sqrt{3x + k} - 2 = 5$ has the solution $x = 6$. What is the value of $k$?
- A$k = 31$
- B$k = 6$
- C$k = 13$
- D$k = -7$
Show solution
Answer: A, $k = 31$. Substitute $x = 6$: $\sqrt{18 + k} - 2 = 5$. Isolate the radical: $\sqrt{18 + k} = 7$. Square both sides: $18 + k = 49$. Solve: $k = 31$. Check: $\sqrt{18 + 31} = \sqrt{49} = 7$, and $7 - 2 = 5$ ✓.
Medium
The equation $\sqrt{kx - 3} = x - 1$ has the solution $x = 2$. What is the value of $k$?
- A$k = 6$
- B$k = -2$
- C$k = 4$
- D$k = 2$
Show solution
Answer: D, $k = 2$. Substitute $x = 2$ into $\sqrt{kx - 3} = x - 1$: $\sqrt{2k - 3} = 2 - 1 = 1$. Square both sides (the entire right side): $2k - 3 = 1^2 = 1$. Solve: $2k = 4$, so $k = 2$. Check: $\sqrt{2(2) - 3} = \sqrt{1} = 1 = 2 - 1$ ✓.
Hard
The equation $\sqrt{x^2 - kx} = x - 3$ has the solution $x = 9$. What is the value of $k$?
- A$k = 9$
- B$k = -5$
- C$k = 0$
- D$k = 5$
Show solution
Answer: D, $k = 5$. Substitute $x = 9$: $\sqrt{81 - 9k} = 9 - 3 = 6$. Square both sides completely: $81 - 9k = 36$. Subtract 81: $-9k = -45$, so $k = 5$. Check: $\sqrt{81 - 9(5)} = \sqrt{36} = 6 = 9 - 3$ ✓. No extraneous solution.
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