SAT Math Quiz: Radicals And Absolute Values

What this quiz covers

This quiz focuses on Radicals And Absolute Values, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $|x+3|=7$ (correct answer)
2. $|x-3|=7$
3. $|x+7|=3$
4. $|x-7|=3$

Explanation: This problem asks for an equation modeling the distance between a point at position x and the point -3, where this distance equals 7 units. The distance between x and -3 is given by |x - (-3)| = |x + 3|. Since this distance equals 7, we have |x + 3| = 7. This correctly models the situation because absolute value represents distance on a number line. A common error is confusing the signs and writing |x - 3| = 7, which would represent the distance from positive 3 instead of -3. When setting up distance problems with absolute value, remember that |x - a| represents the distance from x to a.

Question 2

1. $x=12$
2. $x=\dfrac{13}{2}$
3. $x=13$ (correct answer)
4. $x=\dfrac{26}{2}$

Explanation: This problem asks us to solve the equation √(2x-1) = 5 for x. To solve, we square both sides to eliminate the radical: (√(2x-1))² = 5², which gives us 2x - 1 = 25. Adding 1 to both sides yields 2x = 26, and dividing by 2 gives x = 13. We must verify that this solution makes the expression under the radical nonnegative: 2(13) - 1 = 26 - 1 = 25 ≥ 0, which is true. A common error is forgetting to check that the solution satisfies the domain restriction, or making arithmetic mistakes when solving. When solving radical equations, always verify your answer by substituting back into the original equation.

Question 3

1. $\sqrt{16}$
2. $\sqrt{9}$
3. $3$ (correct answer)
4. $9$

Explanation: We need to simplify $\sqrt{18}/\sqrt{2}$ using the property that $\sqrt{a}/\sqrt{b} = \sqrt{(a/b)}$ for positive a and b. Applying this property: $\sqrt{18}/\sqrt{2} = \sqrt{(18/2)} = \sqrt{9}$. Since $\sqrt{9} = 3$, the simplified form is 3. The common mistake mentioned in the problem is trying to subtract inside the radical (thinking $\sqrt{18}/\sqrt{2} = \sqrt{(18-2)} = \sqrt{16} = 4$), which is incorrect. Another error would be leaving the answer as $\sqrt{9}$ instead of simplifying to 3. When dividing radicals, always use the quotient property before simplifying.

Question 4

1. No real solution
2. $x=4$ only (correct answer)
3. $x=0$ only
4. $x=0$ or $4$

Explanation: We need to solve $\sqrt{x+5} = x-1$ and check which solutions are valid. Squaring both sides gives $x+5 = (x-1)^2 = x^2 - 2x + 1$, which simplifies to $x^2 - 3x - 4 = 0$. Factoring yields $(x-4)(x+1) = 0$, so $x = 4$ or $x = -1$. However, we must check both solutions: for $x = 4$, we get $\sqrt{9} = 3$ and $4-1 = 3$ ✓; for $x = -1$, we get $\sqrt{4} = 2$ but $-1-1 = -2$ ✗. Since square roots are non-negative, only $x = 4$ works. Always check solutions when solving radical equations, as squaring can introduce extraneous solutions.

Question 5

1. $x=4$ only (correct answer)
2. No real solution
3. $x=0$ or $4$
4. $x=0$ only

Explanation: We need to solve $\sqrt{x+5} = x-1$ and check which solutions are valid. Squaring both sides gives $x+5 = (x-1)^2 = x^2 - 2x + 1$, which simplifies to $x^2 - 3x - 4 = 0$. Factoring yields $(x-4)(x+1) = 0$, so $x = 4$ or $x = -1$. However, we must check both solutions: for $x = 4$, we get $\sqrt{9} = 3$ and $4-1 = 3$ ✓; for $x = -1$, we get $\sqrt{4} = 2$ but $-1-1 = -2$ ✗. Since square roots are non-negative, only $x = 4$ works. Always check solutions when solving radical equations, as squaring can introduce extraneous solutions.

Question 6

1. $x=1,6$ (correct answer)
2. $x=1$
3. $x=6$
4. $x=-1,6$

Explanation: To solve $|2x-7|=5$, we consider two cases based on the definition of absolute value. Case 1: $2x-7=5$ gives $2x=12$, so $x=6$. Case 2: $2x-7=-5$ gives $2x=2$, so $x=1$. Both solutions are valid since substituting back gives $|2(6)-7|=|5|=5$ ✓ and $|2(1)-7|=|-5|=5$ ✓. The key insight is that $|A|=k$ means either $A=k$ or $A=-k$ when $k>0$. Always solve both cases when dealing with absolute value equations.

Question 7

1. $x=0$ (correct answer)
2. $x=4$
3. $x=9$
4. No solution

Explanation: To solve $\sqrt{x+9}-\sqrt{x}=3$, we isolate one radical: $\sqrt{x+9} = 3 + \sqrt{x}$. Squaring both sides gives $x+9 = (3+\sqrt{x})^2 = 9 + 6\sqrt{x} + x$. This simplifies to $x+9 = 9 + 6\sqrt{x} + x$, or $0 = 6\sqrt{x}$, which means $\sqrt{x} = 0$, so $x = 0$. Let's verify: $\sqrt{0+9}-\sqrt{0} = 3-0 = 3$ ✓. The key is recognizing that when squaring $(a+b)^2$, we get $a^2+2ab+b^2$, not just $a^2+b^2$. Always check solutions in radical equations as squaring can introduce extraneous solutions.

Question 8

1. $\dfrac{5\sqrt{3}}{3}$ (correct answer)
2. $\dfrac{5}{3\sqrt{3}}$
3. $\dfrac{15}{\sqrt{3}}$
4. $\dfrac{\sqrt{15}}{3}$

Explanation: We need to rationalize the denominator of $\frac{5}{\sqrt{3}}$. To eliminate the radical from the denominator, we multiply both numerator and denominator by $\sqrt{3}$: $\frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$. The denominator becomes $\sqrt{3} \cdot \sqrt{3} = 3$. A common mistake is multiplying only the denominator by $\sqrt{3}$, which would incorrectly give $\frac{5}{3}$. When rationalizing, always multiply both parts of the fraction by the same expression.

Question 9

1. $|x+2|=7$ (correct answer)
2. $|x-2|=7$
3. $|x+7|=2$
4. $|x-7|=-2$

Explanation: This problem asks us to write an equation representing that point P is 7 units from -2 on a number line. The distance between P (with coordinate x) and -2 is given by $|x - (-2)| = |x + 2|$. Since this distance equals 7, we have $|x + 2| = 7$. This makes sense because solving gives us $x + 2 = 7$ or $x + 2 = -7$, yielding $x = 5$ or $x = -9$, both of which are indeed 7 units from -2. A common error is confusing the signs and writing $|x - 2|$ instead. Remember that distance from point a to point b is $|b - a|$.

Question 10

1. $15$
2. $30$ (correct answer)
3. $\sqrt{68}$
4. $3\sqrt{100}$

Explanation: To simplify $\sqrt{50} \cdot \sqrt{18}$, we use the product rule: $\sqrt{50} \cdot \sqrt{18} = \sqrt{50 \cdot 18} = \sqrt{900}$. Since $900 = 30^2$, we have $\sqrt{900} = 30$. Alternatively, we could simplify each radical first: $\sqrt{50} = 5\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$, giving us $(5\sqrt{2})(3\sqrt{2}) = 15 \cdot 2 = 30$. A common error is adding the numbers under the radicals (getting $\sqrt{68}$) instead of multiplying them. When multiplying radicals, use the rule $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$.

Question 11

1. $2\sqrt{2}$ (correct answer)
2. $4\sqrt{2}$
3. $6\sqrt{2}$
4. $\sqrt{62}$

Explanation: This problem asks us to simplify an expression with radicals by first simplifying each term. We have $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$, $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$, and $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$. Substituting these gives us $6\sqrt{2} - 2(3\sqrt{2}) + 2\sqrt{2} = 6\sqrt{2} - 6\sqrt{2} + 2\sqrt{2} = 2\sqrt{2}$. The key error to avoid is trying to combine the numbers under the radicals before simplifying - you must simplify each radical first, then combine like terms. When working with radicals on the SAT, always factor out perfect squares first.

Question 12

1. $x=4$ only
2. $x=-4$ or $10$
3. $x=4$ or $10$ (correct answer)
4. $x=10$ only

Explanation: This problem asks us to find all points that are 3 units away from 7 on a number line, which translates to solving $|x-7|=3$. The absolute value equation splits into two cases: either $x-7 = 3$ or $x-7 = -3$. From the first case, $x = 7 + 3 = 10$. From the second case, $x = 7 - 3 = 4$. We can verify: $|10-7| = |3| = 3$ ✓ and $|4-7| = |-3| = 3$ ✓. The key insight is that absolute value represents distance, so there are two points equidistant from 7. When solving $|A| = k$ where $k > 0$, always consider both positive and negative cases: $A = k$ and $A = -k$.

Question 13

1. $5(\sqrt{3}+\sqrt{2})$
2. $\dfrac{5(\sqrt{3}+\sqrt{2})}{5}$
3. $\sqrt{3}+\sqrt{2}$
4. $5\sqrt{3}+5\sqrt{2}$ (correct answer)

Explanation: To rationalize $\frac{5}{\sqrt{3}-\sqrt{2}}$, we multiply by the conjugate $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$. The numerator becomes $5(\sqrt{3}+\sqrt{2})$. The denominator becomes $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$ using the difference of squares formula. Therefore, the final answer is $\frac{5(\sqrt{3}+\sqrt{2})}{1} = 5\sqrt{3} + 5\sqrt{2}$. The common error is forgetting to multiply both numerator and denominator by the conjugate, or not recognizing that the denominator simplifies to 1. When rationalizing denominators, always use the conjugate and simplify completely.

Question 14

1. $18$ (correct answer)
2. $9\sqrt{2}$
3. $6\sqrt{6}$
4. $\sqrt{39}$

Explanation: To simplify $\sqrt{12} \cdot \sqrt{27}$, we can use the property that $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. This gives us $\sqrt{12 \cdot 27} = \sqrt{324}$. Since $324 = 18^2$, we have $\sqrt{324} = 18$. Alternatively, we could simplify each radical first: $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$, then multiply: $(2\sqrt{3})(3\sqrt{3}) = 6 \cdot 3 = 18$. Common errors include adding the numbers under the radicals instead of multiplying, or stopping at $\sqrt{324}$ without recognizing it as a perfect square. When multiplying radicals, combine under one radical sign first, then check if the result is a perfect square.

Question 15

1. $25xy$
2. $5x\sqrt{2y}$ (correct answer)
3. $\sqrt{25x^2}\sqrt{2y}$
4. $10x\sqrt{y}$

Explanation: To simplify $\sqrt{50x^2y}$ where $x \geq 0$ and $y \geq 0$, we factor out perfect squares from under the radical. First, we factor 50 as $25 \cdot 2$, so $\sqrt{50x^2y} = \sqrt{25 \cdot 2 \cdot x^2 \cdot y}$. Since 25 and $x^2$ are perfect squares, we can write this as $\sqrt{25} \cdot \sqrt{x^2} \cdot \sqrt{2y} = 5 \cdot x \cdot \sqrt{2y} = 5x\sqrt{2y}$. Note that since $x \geq 0$, we have $\sqrt{x^2} = x$ (not $\pm x$). A common error is leaving the expression as $\sqrt{25x^2}\sqrt{2y}$ without simplifying the perfect squares. Always extract perfect square factors completely when simplifying radicals.

Question 16

1. $8+\sqrt{15}$
2. $8+2\sqrt{15}$ (correct answer)
3. $3+5$
4. $15+2\sqrt{8}$

Explanation: To simplify $(\sqrt{3}+\sqrt{5})^2$, we use the formula $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a=\sqrt{3}$ and $b=\sqrt{5}$, so we get $(\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{5}) + (\sqrt{5})^2 = 3 + 2\sqrt{15} + 5 = 8 + 2\sqrt{15}$. The middle term $2\sqrt{3}\sqrt{5} = 2\sqrt{15}$ is often forgotten, leading to the incorrect answer $3+5=8$. When squaring a binomial with radicals, always remember to include the cross term $2ab$. This is a direct application of the FOIL method or the perfect square formula.

Question 17

1. Vertex at $(1,4)$ (correct answer)
2. Vertex at $(-1,4)$
3. Opens upward
4. Vertex at $(4,1)$

Explanation: For the function $y = -|x-1| + 4$, we need to analyze its properties. The standard form $y = a|x-h| + k$ tells us the vertex is at $(h,k) = (1,4)$. The negative coefficient in front of the absolute value ($a = -1$) means the graph opens downward, creating an inverted V-shape with its peak at the vertex. Without the negative sign, the graph would open upward; the negative sign reflects the graph across the horizontal line through the vertex. When working with transformations, remember that the sign in front of the absolute value determines the direction of opening.

Question 18

1. $5$ (correct answer)
2. $\sqrt{25}$
3. $\dfrac{25}{2}$
4. $\sqrt{100}$

Explanation: To simplify $\frac{\sqrt{50}}{\sqrt{2}}$, we use the property that $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ for positive $a$ and $b$. This gives us $\sqrt{\frac{50}{2}} = \sqrt{25}$. Since $\sqrt{25} = 5$, the expression simplifies to 5. We can verify this is correct by checking that $5 = \frac{\sqrt{50}}{\sqrt{2}} = \frac{\sqrt{25 \cdot 2}}{\sqrt{2}} = \frac{5\sqrt{2}}{\sqrt{2}} = 5$ ✓. A common error is trying to simplify by subtracting the numbers under the radicals. When dividing radicals, combine them under a single radical sign first, then simplify.

Question 19

1. $x=5$ only
2. $x=-1$ only
3. $x=-1$ or $5$ (correct answer)
4. $x=-5$ or $1$

Explanation: To solve $|x+4|=|2x-1|$, we consider when the expressions inside the absolute values are equal or opposite. This gives us two cases: Case 1: $x+4 = 2x-1$, which yields $4 = x-1$, so $x = 5$. Case 2: $x+4 = -(2x-1) = -2x+1$, which gives $3x = -3$, so $x = -1$. Let's verify: For $x = 5$: $|5+4| = |9| = 9$ and $|2(5)-1| = |9| = 9$ ✓. For $x = -1$: $|-1+4| = |3| = 3$ and $|2(-1)-1| = |-3| = 3$ ✓. When both sides of an equation have absolute values, consider both when the insides are equal and when they are opposites.

Question 20

1. $y=|x-3|-1$ (correct answer)
2. $y=|x-3|+1$
3. $y=|x+3|-1$
4. $y=-|x-3|-1$

Explanation: We need to find the equation of an absolute value function with vertex at $(3, -1)$ that opens upward. The general form is $y = |x - h| + k$ where $(h, k)$ is the vertex. Substituting our vertex coordinates: $y = |x - 3| + (-1) = |x - 3| - 1$. Since the graph opens upward, we don't need a negative sign in front of the absolute value (that would make it open downward). A common error is writing $|x + 3|$ instead of $|x - 3|$; remember that the standard form uses subtraction. Always double-check by substituting $x = 3$ to verify you get $y = -1$.