Radicals & Absolute Values - SAT Math

1. If $2\sqrt{3x}=18$, what is the value of $5x$?
   Answer: $135$ Divide both sides by $2$ to get $\sqrt{3x}=9$. Squaring gives $3x=81$, so $x=27$ and $5x=135$. Answering $405$ multiplies $3x=81$ by $5$ instead of multiplying $x$ by $5$; dropping the $2$ altogether gives $3x=324$, $x=108$, and $5x=540$.
2. A graph of $g(x)=|x+3|-2$ is shown on a coordinate plane. Which statement gives the vertex of the graph? Use the transformation from $|x|$ to identify the horizontal and vertical shifts.
   Answer: $(-3,-2)$ The function $g(x)=|x+3|-2$ is in the form $|x-h|+k$ where we need to identify $h$ and $k$. Rewriting as $|x-(-3)|+(-2)$, we see that $h=-3$ and $k=-2$, so the vertex is at $(-3,-2)$. The graph shifts 3 units left (since we have $x+3 = x-(-3)$) and 2 units down from the basic $|x|$ graph. A common error is misinterpreting the horizontal shift: $|x+3|$ shifts left by 3, not right. Remember that $|x+a|$ shifts left when $a>0$, while $|x-a|$ shifts right when $a>0$.
3. Simplify $\left(\sqrt{3}+\sqrt{5}\right)^2$. A common incorrect path is to square each term and forget the middle term from $(a+b)^2=a^2+2ab+b^2$.
   Answer: $8+2\sqrt{15}$ 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.
4. $\sqrt[4]{p^{3}}=t^{\frac{1}{8}}$ In the given equation, $p$ and $t$ are constants greater than $1$. If $t=p^{2n-4}$, where $n$ is a constant, what is the value of $n$?
   Answer: $5$ The left side is $p^{\frac{3}{4}}$. Substituting $t=p^{2n-4}$ gives $p^{\frac{2n-4}{8}}$. Setting the exponents equal, $\dfrac{3}{4}=\dfrac{2n-4}{8}$, so $2n-4=6$. Then $2n=10$ and $n=5$. Answering $10$ reports $2n$ rather than $n$.
5. Solve the equation $\sqrt{2x-1}=3$. Which value of $x$ is the solution?
   Answer: $x=5$ To solve $\sqrt{2x-1}=3$, we square both sides to eliminate the radical. Squaring gives us $2x-1 = 9$, which simplifies to $2x = 10$, so $x = 5$. We should verify this solution works in the original equation: $\sqrt{2(5)-1} = \sqrt{10-1} = \sqrt{9} = 3$ ✓. Also, we need to check that the expression under the radical is non-negative: $2(5)-1 = 9 > 0$ ✓. When solving radical equations, always square both sides carefully and verify your solution, as squaring can sometimes introduce extraneous solutions.
6. The coordinate plane shows the graph of $y=-|x-1|+4$. Which statement is true about the graph? Use the negative sign to determine whether the V-shape opens up or down.
   Answer: Vertex at $(1,4)$ 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.
7. $\sqrt{(x-3)^{2}}=\sqrt{3x+31}$ What is the smallest solution to the given equation?
   Answer: $-2$ Squaring both sides gives $(x-3)^{2}=3x+31$, so $x^{2}-6x+9=3x+31$ and $x^{2}-9x-22=0$. Factoring gives $(x-11)(x+2)=0$, so $x=11$ or $x=-2$. Both satisfy the original equation: at $x=-2$ each side equals $\sqrt{25}=5$, and at $x=11$ each side equals $\sqrt{64}=8$. Note that $\sqrt{(x-3)^{2}}=|x-3|$, so the left side is never negative and squaring introduces no extraneous root here. The smaller solution is $-2$.
8. Solve the equation $|x+4|=|2x-1|$. Because both sides are absolute values, consider the different cases created by the expressions inside the absolute values being positive or negative. What are all solutions for $x$?
   Answer: $x=-1$ or $5$ 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.
9. Which expression is equivalent to $\dfrac{1}{\sqrt{2}+\sqrt{3}}$ after rationalizing the denominator?
   Answer: $\sqrt{3}-\sqrt{2}$ To rationalize $\frac{1}{\sqrt{2}+\sqrt{3}}$, we multiply by the conjugate of the denominator. The conjugate of $\sqrt{2}+\sqrt{3}$ is $\sqrt{2}-\sqrt{3}$. Multiplying: $\frac{1}{\sqrt{2}+\sqrt{3}} \cdot \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} = \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}$. The denominator becomes $(\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$, giving us $\frac{\sqrt{2}-\sqrt{3}}{-1} = \sqrt{3}-\sqrt{2}$. A common error is using the wrong conjugate or making sign errors when expanding the denominator. Remember that dividing by -1 changes all signs in the numerator.
10. $\sqrt[6]{p^{5}}=t^{\frac{5}{12}}$ In the given equation, $p$ and $t$ are constants greater than $1$. If $t=p^{3n-7}$, where $n$ is a constant, what is the value of $n$?
    Answer: $3$ The left side is $p^{\frac{5}{6}}$. Substituting $t=p^{3n-7}$ gives $p^{\frac{5(3n-7)}{12}}$. Setting the exponents equal, $\dfrac{5}{6}=\dfrac{5(3n-7)}{12}$. Dividing by $5$ and multiplying by $12$ gives $3n-7=2$, so $3n=9$ and $n=3$.
11. Rationalize the denominator and simplify: $\dfrac{5}{\sqrt{3}}$. Your final answer should have no radical in the denominator and should be reduced as much as possible.
    Answer: $\dfrac{5\sqrt{3}}{3}$ To rationalize the denominator of $\frac{5}{\sqrt{3}}$, we multiply both numerator and denominator by $\sqrt{3}$. This gives us $\frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{5\sqrt{3}}{3}$. The key insight is that $\sqrt{3} \cdot \sqrt{3} = 3$, which eliminates the radical from the denominator. A common mistake is to leave the answer as $\frac{5\sqrt{3}}{\sqrt{3}}$ without simplifying the denominator. Always multiply radicals in the denominator to get a rational number when rationalizing.
12. If $5\sqrt{\dfrac{x}{2}}=20$, what is the value of $3x$?
    Answer: $96$ Divide both sides by $5$ to get $\sqrt{\dfrac{x}{2}}=4$. Squaring gives $\dfrac{x}{2}=16$, so $x=32$ and $3x=96$. Answering $16$ stops at $\dfrac{x}{2}$ and answering $48$ triples it instead of tripling $x$.
13. Solve the absolute value equation $|2x-7|=5$.
    Answer: $x=1,6$ The equation $|2x-7| = 5$ requires us to consider two cases since absolute value represents distance from zero. Case 1: $2x-7 = 5$, which gives $2x = 12$, so $x = 6$. Case 2: $2x-7 = -5$, which gives $2x = 2$, so $x = 1$. We can verify: $|2(6)-7| = |12-7| = |5| = 5$ ✓ and $|2(1)-7| = |2-7| = |-5| = 5$ ✓. The key error to avoid is solving only one case—absolute value equations with positive right-hand sides always have two solutions unless the expression inside equals zero at exactly one point.
14. Which expression is equivalent to $\sqrt{12}-\sqrt{27}$ in simplest radical form?
    Answer: $-\sqrt{3}$ To simplify $\sqrt{12}-\sqrt{27}$, we first simplify each radical separately. We have $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$ and $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$. Now we can subtract: $2\sqrt{3} - 3\sqrt{3} = -\sqrt{3}$. A common mistake is trying to combine the radicals before simplifying, such as writing $\sqrt{12-27} = \sqrt{-15}$, which would involve imaginary numbers. Always simplify radicals to their simplest form before combining like terms.
15. $\sqrt[5]{p^{2}}=t^{\frac{4}{15}}$ In the given equation, $p$ and $t$ are constants greater than $1$. If $t=p^{4n}$, where $n$ is a constant, what is the value of $n$?
    Answer: $\dfrac{3}{8}$ The left side is $p^{\frac{2}{5}}$. Substituting $t=p^{4n}$ gives $p^{\frac{16n}{15}}$. Setting the exponents equal, $\dfrac{2}{5}=\dfrac{16n}{15}$. Multiplying both sides by $15$ gives $6=16n$, so $n=\dfrac{3}{8}$. Here $4n=\dfrac{3}{2}$, which is the exponent in $t=p^{4n}$, not the value of $n$.
16. Simplify the expression $\sqrt{12}\,\sqrt{27}$. Write your answer in simplest radical form. A tempting incorrect approach is to add inside the radicals or to stop at $\sqrt{324}$ without simplifying further.
    Answer: $18$ 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.
17. Simplify the expression $\dfrac{3}{\sqrt{5}}$ by rationalizing the denominator. Choose the expression that has no radical in the denominator and is fully simplified.
    Answer: $\dfrac{3\sqrt{5}}{5}$ To rationalize $\frac{3}{\sqrt{5}}$, we multiply both numerator and denominator by $\sqrt{5}$. This gives us $\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{3\sqrt{5}}{5}$. The key is that $\sqrt{5} \cdot \sqrt{5} = 5$, which eliminates the radical from the denominator. A common mistake is to multiply only the denominator by $\sqrt{5}$, forgetting that we must multiply both parts of the fraction by the same value. When rationalizing, always multiply by a form of 1 to maintain equality.
18. Solve the equation $|3x-6|=0$. Some students incorrectly split into two cases even though the absolute value equals zero. What value of $x$ satisfies the equation?
    Answer: $x=2$ To solve $|3x-6|=0$, we recognize that an absolute value equals zero only when the expression inside equals zero. Therefore, we need $3x-6 = 0$, which gives us $3x = 6$, so $x = 2$. We can verify: $|3(2)-6| = |6-6| = |0| = 0$ ✓. A common error is trying to split this into two cases like other absolute value equations, but since the only way $|A| = 0$ is when $A = 0$, there's only one solution. When an absolute value equals zero, simply set the inside expression equal to zero and solve.
19. Solve the inequality $|x-1|<4$. A common incorrect path is to give only one endpoint or to reverse an inequality sign. Which solution set is correct?
    Answer: $-3<x<5$ To solve |x - 1| < 4, we interpret this as the distance from x to 1 being less than 4. This means -4 < x - 1 < 4. Adding 1 to all parts: -4 + 1 < x < 4 + 1, which gives -3 < x < 5. This can also be solved by considering two cases: when x - 1 ≥ 0, we have x - 1 < 4, so x < 5; when x - 1 < 0, we have -(x - 1) < 4, so -x + 1 < 4, giving x > -3. Common errors include using ≤ instead of < or only finding one endpoint. For absolute value inequalities of the form |expression| < a, the solution is always an interval.
20. A point $P$ on a number line is $7$ units from $-2$. Which equation represents this situation using absolute value, and what are the possible values of $P$? Choose the option that matches both the correct equation and both solutions.
    Answer: $|P+2|=7;\ P=5,-9$ The problem states that point P is 7 units from -2, which translates to the distance formula $|P - (-2)| = 7$, or $|P + 2| = 7$. This absolute value equation has two solutions: $P + 2 = 7$ gives $P = 5$, and $P + 2 = -7$ gives $P = -9$. We can verify: the distance from 5 to -2 is $|5-(-2)| = |7| = 7$ ✓, and the distance from -9 to -2 is $|-9-(-2)| = |-7| = 7$ ✓. The key insight is that absolute value represents distance on a number line, and the expression inside should be (variable - reference point). When setting up distance problems, pay careful attention to the signs.

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