Solving Radical Equations - Accuplacer Aaf

1. Which of the following describes the solution(s) to $\sqrt[3]{x^2 - 7} = 3$?
   Answer: $x = \sqrt{34}$ or $x = -\sqrt{34}$ To solve the equation, cube both sides to eliminate the cube root: $(\sqrt[3]{x^2 - 7})^3 = 3^3$. This simplifies to $x^2 - 7 = 27$. Add 7 to both sides to isolate the $x^2$ term: $x^2 = 34$. Finally, take the square root of both sides, remembering to include both the positive and negative roots: $x = \pm\sqrt{34}$. So the solutions are $x = \sqrt{34}$ and $x = -\sqrt{34}$.
2. What is the value of $x$ that satisfies the equation $\sqrt{4x^2 + 5x - 10} = 2x$?
   Answer: 2 First, note that for the radical to be defined and equal to $2x$, we must have $2x \ge 0$, so $x \ge 0$. Square both sides of the equation: $4x^2 + 5x - 10 = (2x)^2$. This simplifies to $4x^2 + 5x - 10 = 4x^2$. Subtract $4x^2$ from both sides, leaving a linear equation: $5x - 10 = 0$. Solve for $x$: $5x = 10 \Rightarrow x = 2$. This solution satisfies the condition $x \ge 0$. Check it in the original equation: $\sqrt{4(2^2) + 5(2) - 10} = 2(2) \Rightarrow \sqrt{16 + 10 - 10} = 4 \Rightarrow \sqrt{16} = 4$, which is true. The solution is $x = 2$.
3. Solve for $x$: $2\sqrt{x + 1} - x = -2$
   Answer: $x = 8$ First, isolate the radical term: $2\sqrt{x + 1} = x - 2$. Square both sides of the equation: $(2\sqrt{x + 1})^2 = (x - 2)^2$, which simplifies to $4(x + 1) = x^2 - 4x + 4$. Distribute on the left side: $4x + 4 = x^2 - 4x + 4$. Set the equation to zero: $0 = x^2 - 8x$. Factor out $x$: $0 = x(x - 8)$. The potential solutions are $x = 0$ and $x = 8$. Check each solution in the original equation. For $x = 0$: $2\sqrt{0 + 1} - 0 = -2 \Rightarrow 2(1) = -2 \Rightarrow 2 = -2$, which is false. For $x = 8$: $2\sqrt{8 + 1} - 8 = -2 \Rightarrow 2\sqrt{9} - 8 = -2 \Rightarrow 2(3) - 8 = -2 \Rightarrow 6 - 8 = -2$, which is true. The only solution is $x = 8$.
4. What is the solution set for the equation $\sqrt{x + 7} = x - 5$?
   Answer: $\{9\}$ To solve the equation, first square both sides to eliminate the radical: $x + 7 = (x - 5)^2$. Expanding the right side gives $x + 7 = x^2 - 10x + 25$. Rearrange the terms to form a quadratic equation: $x^2 - 11x + 18 = 0$. Factoring the quadratic yields $(x - 2)(x - 9) = 0$, which gives potential solutions $x = 2$ and $x = 9$. Each potential solution must be checked in the original equation. For $x = 9$: $\sqrt{9 + 7} = 9 - 5 \Rightarrow \sqrt{16} = 4 \Rightarrow 4 = 4$, which is true. For $x = 2$: $\sqrt{2 + 7} = 2 - 5 \Rightarrow \sqrt{9} = -3 \Rightarrow 3 = -3$, which is false. Thus, $x = 2$ is an extraneous solution. The only valid solution is $x = 9$.
5. Solve the equation $x = \sqrt{3x - 2} + 2$.
   Answer: $x = 6$ First, isolate the radical by subtracting 2 from both sides: $x - 2 = \sqrt{3x - 2}$. Square both sides: $(x - 2)^2 = 3x - 2$. Expand the left side: $x^2 - 4x + 4 = 3x - 2$. Move all terms to one side to form a quadratic equation: $x^2 - 7x + 6 = 0$. Factor the quadratic: $(x - 1)(x - 6) = 0$. The potential solutions are $x = 1$ and $x = 6$. Check both in the original equation. For $x = 1$: $1 = \sqrt{3(1) - 2} + 2 \Rightarrow 1 = \sqrt{1} + 2 \Rightarrow 1 = 1 + 2$, which is false. For $x = 6$: $6 = \sqrt{3(6) - 2} + 2 \Rightarrow 6 = \sqrt{16} + 2 \Rightarrow 6 = 4 + 2$, which is true. The only valid solution is $x = 6$.
6. If $\sqrt{x + 4} - \sqrt{x - 1} = 1$, then $x$ equals:
   Answer: 5 Isolate one radical: $\sqrt{x + 4} = 1 + \sqrt{x - 1}$. Square both sides: $x + 4 = 1 + 2\sqrt{x - 1} + (x - 1) = x + 2\sqrt{x - 1}$. Simplifying: $4 = 2\sqrt{x - 1}$, so $\sqrt{x - 1} = 2$. Square again: $x - 1 = 4$, so $x = 5$. Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$ ✓. Also verify domain: $x \geq 1$ for the second radical. Choice B comes from making an error when squaring the binomial. Choice C results from incorrectly isolating the radical. Choice D represents an arithmetic error in the final steps.
7. What is the sum of all real solutions to the equation $\sqrt{3x + 10} = x$?
   Answer: 5 Square both sides of the equation to get $3x + 10 = x^2$. Rearrange this into a standard quadratic form: $x^2 - 3x - 10 = 0$. Factor the quadratic: $(x - 5)(x + 2) = 0$. The potential solutions are $x = 5$ and $x = -2$. Check these in the original equation. For $x = 5$: $\sqrt{3(5) + 10} = 5 \Rightarrow \sqrt{25} = 5 \Rightarrow 5 = 5$, which is correct. For $x = -2$: $\sqrt{3(-2) + 10} = -2 \Rightarrow \sqrt{4} = -2 \Rightarrow 2 = -2$, which is incorrect because the principal square root must be non-negative. Therefore, $x = -2$ is an extraneous solution. The only real solution is 5, so the sum of the solutions is 5.
8. Which of the following equations has no real solutions?
   Answer: $\sqrt{x + 1} = x - 3$ For a radical equation $\sqrt{f(x)} = g(x)$ to have real solutions, we need $f(x) \geq 0$ and $g(x) \geq 0$, and the equation must check out after squaring. For choice A: $\sqrt{x + 1} = x - 3$ requires $x \geq -1$ and $x \geq 3$, so $x \geq 3$. Squaring: $x + 1 = (x - 3)^2 = x^2 - 6x + 9$, so $x^2 - 7x + 8 = 0$. This gives $x = \frac{7 \pm \sqrt{49 - 32}}{2} = \frac{7 \pm \sqrt{17}}{2}$. Both values are less than 3, violating our domain restriction. For choice B: Similar analysis shows this has valid solutions. Choice C and D also yield valid solutions when properly checked. Only choice A has domain restrictions that eliminate all algebraic solutions.
9. What is the solution set for the equation $2 + \sqrt[3]{5x - 2} = 5$?
   Answer: $x = \frac{29}{5}$ First, isolate the radical term by subtracting 2 from both sides: $\sqrt[3]{5x - 2} = 3$. To eliminate the cube root, cube both sides of the equation: $(\sqrt[3]{5x - 2})^3 = 3^3$. This simplifies to $5x - 2 = 27$. Add 2 to both sides: $5x = 29$. Divide by 5 to solve for $x$: $x = \frac{29}{5}$. Odd-indexed roots do not typically introduce extraneous solutions, so checking is mainly for catching calculation errors.
10. If $(2x - 1)^{1/2} - x = -2$, what is the value of $x$?
    Answer: 5 Rewrite the equation using a radical: $\sqrt{2x - 1} - x = -2$. Isolate the radical term: $\sqrt{2x - 1} = x - 2$. Square both sides: $2x - 1 = (x - 2)^2$. Expand the right side: $2x - 1 = x^2 - 4x + 4$. Rearrange into a quadratic equation: $x^2 - 6x + 5 = 0$. Factor the quadratic: $(x - 1)(x - 5) = 0$. The potential solutions are $x = 1$ and $x = 5$. Check these solutions. For $x = 1$: $\sqrt{2(1) - 1} - 1 = -2 \Rightarrow \sqrt{1} - 1 = -2 \Rightarrow 1 - 1 = -2 \Rightarrow 0 = -2$, which is false. For $x = 5$: $\sqrt{2(5) - 1} - 5 = -2 \Rightarrow \sqrt{9} - 5 = -2 \Rightarrow 3 - 5 = -2$, which is true. The only solution is $x=5$.
11. The equation $x + \sqrt{x+1} = 5$ has one solution. What is this solution?
    Answer: 3 Isolate the radical term by subtracting $x$ from both sides: $\sqrt{x+1} = 5 - x$. Square both sides: $x+1 = (5 - x)^2$. Expand the right side: $x+1 = 25 - 10x + x^2$. Rearrange into a quadratic equation set to zero: $x^2 - 11x + 24 = 0$. Factor the quadratic: $(x - 3)(x - 8) = 0$. The potential solutions are $x = 3$ and $x = 8$. Check these in the original equation. For $x = 3$: $3 + \sqrt{3+1} = 5 \Rightarrow 3 + \sqrt{4} = 5 \Rightarrow 3 + 2 = 5$, which is true. For $x = 8$: $8 + \sqrt{8+1} = 5 \Rightarrow 8 + \sqrt{9} = 5 \Rightarrow 8 + 3 = 5 \Rightarrow 11 = 5$, which is false. The only valid solution is $x = 3$.
12. Solve for $x$: $\sqrt{x} = \sqrt{x-9} + 1$
    Answer: 25 Square both sides of the equation: $(\sqrt{x})^2 = (\sqrt{x-9} + 1)^2$. This gives $x = (x-9) + 2\sqrt{x-9} + 1$. Simplify the right side: $x = x - 8 + 2\sqrt{x-9}$. Subtract $x$ from both sides and add 8 to both sides to isolate the radical term: $8 = 2\sqrt{x-9}$. Divide by 2: $4 = \sqrt{x-9}$. Square both sides again: $16 = x - 9$. Solve for $x$: $x = 25$. Check the solution in the original equation: $\sqrt{25} = \sqrt{25-9} + 1 \Rightarrow 5 = \sqrt{16} + 1 \Rightarrow 5 = 4 + 1$. The solution is correct.
13. If $\sqrt{x - 3} + \sqrt{x} = 3$, what is the value of $x$?
    Answer: 4 First, isolate one of the radical terms. Let's move $\sqrt{x}$ to the other side: $\sqrt{x - 3} = 3 - \sqrt{x}$. Now, square both sides: $(\sqrt{x - 3})^2 = (3 - \sqrt{x})^2$, which gives $x - 3 = 9 - 6\sqrt{x} + x$. Subtract $x$ from both sides: $-3 = 9 - 6\sqrt{x}$. Isolate the remaining radical term: $-12 = -6\sqrt{x}$, which simplifies to $2 = \sqrt{x}$. Square both sides again to solve for $x$: $4 = x$. Finally, check the solution in the original equation: $\sqrt{4 - 3} + \sqrt{4} = 3 \Rightarrow \sqrt{1} + 2 = 3 \Rightarrow 1 + 2 = 3$. The solution is correct.
14. Which value of $x$ satisfies $\sqrt{x^2 - 8} = x - 2$?
    Answer: $x = 3$ only Square both sides: $x^2 - 8 = (x - 2)^2 = x^2 - 4x + 4$. Simplifying: $-8 = -4x + 4$, so $4x = 12$ and $x = 3$. We must check this solution: For $x = 3$: $\sqrt{9 - 8} = \sqrt{1} = 1$ and $3 - 2 = 1$ ✓. We must also check the domain: $x^2 - 8 \geq 0$ requires $x \leq -2\sqrt{2}$ or $x \geq 2\sqrt{2} \approx 2.83$, and $x - 2 \geq 0$ requires $x \geq 2$. Since $x = 3$ satisfies both conditions, it's valid. Choice A assumes $x = 2$ works without checking. Choice C comes from solving incorrectly and not checking domain restrictions. Choice D assumes both solutions from a quadratic would be valid without verification.
15. Find the solution set for $\sqrt{x^2 - 3x + 3} = 1$.
    Answer: $\{1, 2\}$ To solve, square both sides of the equation: $x^2 - 3x + 3 = 1^2$, which simplifies to $x^2 - 3x + 3 = 1$. Set the equation to zero: $x^2 - 3x + 2 = 0$. Factor the quadratic expression: $(x - 1)(x - 2) = 0$. This gives two potential solutions: $x = 1$ and $x = 2$. Checking both solutions in the original equation is necessary. For $x = 1$: $\sqrt{1^2 - 3(1) + 3} = \sqrt{1 - 3 + 3} = \sqrt{1} = 1$. This is true. For $x = 2$: $\sqrt{2^2 - 3(2) + 3} = \sqrt{4 - 6 + 3} = \sqrt{1} = 1$. This is also true. Both solutions are valid.
16. Solve for $y$: $3\sqrt{y} = \sqrt{4y + 5}$.
    Answer: 1 To solve this equation, square both sides: $(3\sqrt{y})^2 = (\sqrt{4y + 5})^2$. This simplifies to $9y = 4y + 5$. Next, solve the resulting linear equation for $y$. Subtract $4y$ from both sides: $5y = 5$. Divide by 5: $y = 1$. Finally, check the solution in the original equation: $3\sqrt{1} = \sqrt{4(1) + 5} \Rightarrow 3(1) = \sqrt{9} \Rightarrow 3 = 3$. The solution is valid.
17. What is the sum of the solutions for the equation $\sqrt{2x + 1} - \sqrt{x} = 1$?
    Answer: 4 First, isolate one radical: $\sqrt{2x + 1} = 1 + \sqrt{x}$. Square both sides: $2x + 1 = (1 + \sqrt{x})^2$, which expands to $2x + 1 = 1 + 2\sqrt{x} + x$. Simplify and isolate the remaining radical term: $x = 2\sqrt{x}$. Square both sides again: $x^2 = (2\sqrt{x})^2$, which gives $x^2 = 4x$. Rearrange into $x^2 - 4x = 0$ and factor: $x(x - 4) = 0$. The potential solutions are $x = 0$ and $x = 4$. Check both solutions in the original equation. For $x = 0$: $\sqrt{2(0)+1} - \sqrt{0} = 1 \Rightarrow 1 - 0 = 1$, which is true. For $x = 4$: $\sqrt{2(4)+1} - \sqrt{4} = 1 \Rightarrow \sqrt{9} - 2 = 1 \Rightarrow 3 - 2 = 1$, which is true. Both solutions are valid. The sum of the solutions is $0 + 4 = 4$.
18. How many solutions does the equation $\sqrt{6 - x} = x + 2$ have?
    Answer: Exactly one solution Square both sides: $6 - x = (x + 2)^2 = x^2 + 4x + 4$. Rearranging: $x^2 + 5x - 2 = 0$. Using the quadratic formula: $x = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2}$. This gives two algebraic solutions, but we must check both. For the equation to be valid, we need $6 - x \geq 0$ (so $x \leq 6$) and $x + 2 \geq 0$ (so $x \geq -2$). The solution $x = \frac{-5 + \sqrt{33}}{2} \approx 0.37$ satisfies these conditions and checks out. The solution $x = \frac{-5 - \sqrt{33}}{2} \approx -5.37$ violates $x \geq -2$, so it's extraneous. Choice A suggests no solutions exist. Choice C assumes both algebraic solutions are valid. Choice D misunderstands the nature of radical equations.

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