Solving Radical Equations - Math 2

1. When solving $\sqrt{x + 4} + \sqrt{x - 1} = 5$, which of the following represents the most efficient approach to avoid extraneous solutions?
   Answer: Isolate $\sqrt{x + 4}$, square both sides, simplify, then square again if necessary Option B is correct because isolating one radical before squaring minimizes the complexity of subsequent algebra and reduces the likelihood of computational errors that can mask extraneous solutions. If we square $\sqrt{x + 4} + \sqrt{x - 1} = 5$ immediately, we get $x + 4 + 2\sqrt{(x+4)(x-1)} + x - 1 = 25$, leading to $2\sqrt{x^2 + 3x - 4} = 22 - 2x$, which requires another squaring step anyway. By isolating first: $\sqrt{x + 4} = 5 - \sqrt{x - 1}$, then squaring gives $x + 4 = 25 - 10\sqrt{x - 1} + x - 1$, which simplifies to $10\sqrt{x - 1} = 20$, so $\sqrt{x - 1} = 2$ and $x = 5$. Option A creates unnecessary algebraic complexity. Option C doesn't apply since the expressions under the radicals aren't perfect squares of $x$. Option D, while potentially accurate, doesn't demonstrate algebraic problem-solving skills.
2. The equation $\sqrt{x + 7} - \sqrt{x - 2} = 1$ has a solution $x = a$. What is the value of $\sqrt{a + 7} + \sqrt{a - 2}$?
   Answer: 9 Let $u = \sqrt{x + 7}$ and $v = \sqrt{x - 2}$. Then $u - v = 1$ and we want $u + v$. Note that $u^2 - v^2 = (x + 7) - (x - 2) = 9$. Since $u^2 - v^2 = (u + v)(u - v) = (u + v)(1) = u + v$, we have $u + v = 9$. Therefore, $\sqrt{a + 7} + \sqrt{a - 2} = 9$. To verify this approach works, we can solve for $a$: From $u - v = 1$ and $u + v = 9$, we get $u = 5$ and $v = 4$. So $\sqrt{x + 7} = 5$ gives $x + 7 = 25$, thus $x = 18$. Check: $\sqrt{18 - 2} = \sqrt{16} = 4$ ✓. The answer is 9.
3. The equation $x = \sqrt{x + 12}$ has how many valid solutions?
   Answer: 1 Square both sides: $x^2 = x + 12$, so $x^2 - x - 12 = 0$. Factor: $(x - 4)(x + 3) = 0$, giving $x = 4$ or $x = -3$. For the original equation to be valid, we need $x \geq 0$ (since $\sqrt{x + 12} = x$ and square roots are non-negative). Check $x = 4$: $\sqrt{16} = 4$ ✓. Check $x = -3$: $\sqrt{9} = 3$ but $x = -3$, so $3 \neq -3$. Therefore, $x = -3$ is extraneous. Only $x = 4$ is valid, so there is exactly 1 solution.
4. If $\sqrt{x^2 - 5x + 6} = x - 3$, then $x$ could equal:
   Answer: 3 only Square both sides: $x^2 - 5x + 6 = (x - 3)^2 = x^2 - 6x + 9$. Simplifying: $-5x + 6 = -6x + 9$, so $x = 3$. We must check this solution and verify the domain restrictions. For the equation to be valid, we need $x - 3 \geq 0$ (since the square root equals $x - 3$), so $x \geq 3$. Also, $x^2 - 5x + 6 \geq 0$. Factoring: $(x - 2)(x - 3) \geq 0$, which holds when $x \leq 2$ or $x \geq 3$. Combining constraints: $x \geq 3$. Since $x = 3$ satisfies this, let's verify: $\sqrt{9 - 15 + 6} = \sqrt{0} = 0$ and $x - 3 = 0$ ✓. The only valid solution is $x = 3$.
5. Which of the following is an extraneous solution to the equation $\sqrt{x + 6} = x$?
   Answer: $x = -2$ Square both sides: $x + 6 = x^2$, so $x^2 - x - 6 = 0$. Factor: $(x - 3)(x + 2) = 0$, giving $x = 3$ or $x = -2$. Check $x = 3$: $\sqrt{9} = 3$ ✓. Check $x = -2$: $\sqrt{4} = 2$ but $x = -2$, so $2 \neq -2$. Therefore, $x = -2$ is extraneous. The issue is that when we squared both sides, we lost the condition that $\sqrt{x + 6} \geq 0$, but $x$ can be negative. For the original equation $\sqrt{x + 6} = x$ to hold, we need $x \geq 0$. Since $x = -2 < 0$, it's extraneous.
6. A student attempts to solve $\sqrt{2x - 3} = x - 3$ and obtains the solutions $x = 4$ and $x = 6$. What should the student conclude after checking these solutions?
   Answer: Only $x = 6$ is valid; $x = 4$ is an extraneous solution First, we need $2x - 3 \geq 0$ (so $x \geq \frac{3}{2}$) and $x - 3 \geq 0$ (so $x \geq 3$) since we're equating a square root to $x - 3$. The domain is $x \geq 3$. Squaring both sides: $2x - 3 = (x - 3)^2 = x^2 - 6x + 9$. Rearranging: $x^2 - 8x + 12 = 0$, which factors as $(x - 4)(x - 6) = 0$. This gives $x = 4$ and $x = 6$. Checking $x = 4$: $\sqrt{2(4) - 3} = \sqrt{5} \approx 2.24$, but $x - 3 = 4 - 3 = 1$. Since $\sqrt{5} \neq 1$, $x = 4$ is extraneous. Checking $x = 6$: $\sqrt{2(6) - 3} = \sqrt{9} = 3$ and $x - 3 = 6 - 3 = 3$. Since $3 = 3$, $x = 6$ is valid.
7. How many solutions does the equation $x + 1 = \sqrt{x^2 + 2x + 5}$ have?
   Answer: 0 For the equation to be valid, we need $x + 1 \geq 0$, so $x \geq -1$. Square both sides: $(x + 1)^2 = x^2 + 2x + 5$. Expanding: $x^2 + 2x + 1 = x^2 + 2x + 5$. This simplifies to $1 = 5$, which is impossible. Therefore, there are no solutions. The issue is that while $\sqrt{x^2 + 2x + 5} = \sqrt{(x + 1)^2 + 4} \geq 2$ for all real $x$, we need $x + 1$ to equal this value. But $x + 1 \geq 2$ would require $x \geq 1$. Even then, $(x + 1)^2 = (x + 1)^2 + 4$ leads to $0 = 4$. The equation has no solutions.
8. Consider the equation $x = \sqrt{6x - 5}$. A student claims that since both sides are positive for valid solutions, squaring will not introduce extraneous solutions. How should you evaluate this claim?
   Answer: The claim is incorrect; squaring can still introduce extraneous solutions even when both sides appear positive The student's claim is incorrect. While it's true that we need both sides to be non-negative before squaring, this doesn't guarantee that squaring won't introduce extraneous solutions. Squaring the equation $x = \sqrt{6x - 5}$ gives $x^2 = 6x - 5$, or $x^2 - 6x + 5 = 0$, which factors as $(x - 1)(x - 5) = 0$. This yields $x = 1$ and $x = 5$. However, checking these: For $x = 1$: LHS = 1, RHS = $\sqrt{6(1) - 5} = \sqrt{1} = 1$ ✓. For $x = 5$: LHS = 5, RHS = $\sqrt{6(5) - 5} = \sqrt{25} = 5$ ✓. In this case, both solutions are actually valid, but the key point is that the student's reasoning is flawed. Squaring can introduce extraneous solutions regardless of the apparent signs because the squaring operation can make negative expressions appear to satisfy the squared equation when they don't satisfy the original.
9. For what value of $a$ is $x = 9$ a solution to $\sqrt{x + a} = x - 5$?
   Answer: $a = 7$ When you encounter a radical equation where you need to find a parameter that makes a given value a solution, substitute that value directly into the equation and solve for the unknown parameter. Since $x = 9$ must be a solution to $\sqrt{x + a} = x - 5$, substitute $x = 9$ into both sides of the equation: $\sqrt{9 + a} = 9 - 5$ $\sqrt{9 + a} = 4$ To solve for $a$, square both sides to eliminate the square root: $9 + a = 16$ $a = 7$ Let's verify: when $a = 7$, the equation becomes $\sqrt{9 + 7} = 9 - 5$, which simplifies to $\sqrt{16} = 4$, or $4 = 4$ ✓ Now let's see why the other choices don't work. Choice A gives us $a = 25$: substituting yields $\sqrt{9 + 25} = \sqrt{34} \approx 5.83$, but the right side equals $4$. Choice B gives us $a = 11$: this produces $\sqrt{9 + 11} = \sqrt{20} \approx 4.47 \neq 4$. Choice C gives us $a = 16$: this results in $\sqrt{9 + 16} = \sqrt{25} = 5 \neq 4$. The key strategy here is direct substitution. When a problem tells you that a specific value is a solution to an equation containing a parameter, always substitute that value immediately and solve for the parameter. This approach is much more efficient than trying to solve the general equation first.
10. When solving the equation $\sqrt{x + 7} - \sqrt{x - 5} = 2$, a student obtains potential solutions and must check their validity. Which statement correctly describes the solution process?
    Answer: The equation has exactly one valid solution: $x = 9$ To solve $\sqrt{x + 7} - \sqrt{x - 5} = 2$, we first note the domain requires $x \geq 5$. Isolating one radical: $\sqrt{x + 7} = 2 + \sqrt{x - 5}$. Squaring both sides: $x + 7 = 4 + 4\sqrt{x - 5} + (x - 5)$, which simplifies to $8 = 4\sqrt{x - 5}$ or $\sqrt{x - 5} = 2$. This gives $x - 5 = 4$, so $x = 9$. Checking: $\sqrt{16} - \sqrt{4} = 4 - 2 = 2$ ✓. The answer is A.
11. If $\sqrt[4]{x + 1} = 3$, then $\sqrt{x + 1} =$ ?
    Answer: 9 From $\sqrt[4]{x + 1} = 3$, raise both sides to the 4th power: $x + 1 = 3^4 = 81$, so $x = 80$. Now find $\sqrt{x + 1} = \sqrt{80 + 1} = \sqrt{81} = 9$. Alternatively, note that $\sqrt{x + 1} = (\sqrt[4]{x + 1})^2 = 3^2 = 9$. Both methods give the same answer: 9.

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