Radicals & Absolute Values - PSAT Math

Solve for $x$ : $\sqrt{2x+3}=x-1,$ where $x$ is a real number.
Answer: $x=2+\sqrt6$ only When solving radical equations, you need to isolate the radical, square both sides to eliminate it, then check your solutions in the original equation since squaring can introduce extraneous solutions. Starting with $\sqrt{2x+3}=x-1$ , the radical is already isolated. Square both sides: $2x+3=(x-1)^2$ . Expanding the right side gives $2x+3=x^2-2x+1$ . Rearranging into standard form: $0=x^2-4x-2$ . Using the quadratic formula with $a=1$ , $b=-4$ , and $c=-2$ : $x=\frac{4\pm\sqrt{16+8}}{2}=\frac{4\pm\sqrt{24}}{2}=\frac{4\pm2\sqrt{6}}{2}=2\pm\sqrt{6}$ . Now you must check both solutions in the original equation. For $x=2+\sqrt{6}$ : The left side gives $\sqrt{2(2+\sqrt{6})+3}=\sqrt{7+2\sqrt{6}}=1+\sqrt{6}$ . The right side gives $(2+\sqrt{6})-1=1+\sqrt{6}$ . Since both sides equal $1+\sqrt{6}$ , this solution works. For $x=2-\sqrt{6}$ : The right side gives $(2-\sqrt{6})-1=1-\sqrt{6}$ , which is negative (since $\sqrt{6}>1$ ). However, $\sqrt{2x+3}$ must be non-negative, so this solution is extraneous. Choice A gives only the extraneous solution. Choice C includes both solutions without recognizing that one is extraneous. Choice D incorrectly concludes no solution exists. Choice B correctly identifies the single valid solution. Key strategy: Always check solutions in radical equations by substituting back into the original equation, and remember that square roots cannot equal negative values.
Simplify $\sqrt{\dfrac{49}{64}}$ and give the exact value.
Answer: $\dfrac{7}{8}$ The question requires simplifying √(49/64) to its exact value. Start by recognizing that the square root of a fraction is the square root of the numerator over the square root of the denominator, so √(49/64) = √49 / √64. Simplify √49 = 7 and √64 = 8, yielding 7/8. This is already in simplest form as both are integers with no common factors. A key error might be squaring instead of taking roots or miscalculating √64 as 4 instead of 8. Always verify by squaring back: $(7/8)^2$ = 49/64, which matches. For radical simplifications, break down into perfect squares to ensure accuracy.
The graph of an absolute value function passes through the points $(0,4)$ and $(4,4)$ and has its vertex on the x-axis. What is the vertex?
Answer: $(2,0)$ The question seeks the vertex of an absolute value graph passing through (0,4) and (4,4) with vertex on the x-axis. The points are symmetric, so the vertex x-coordinate is the midpoint: (0 + 4)/2 = 2. Since vertex is on x-axis, y = 0, so vertex is (2,0). The function could be y = 2|x - 2|, as at x=0, 2| -2| = 4, and x=4, same. A key error is assuming the vertex is at one of the points instead of the midpoint. Verify by checking symmetry and points. For absolute value graphs, use symmetry and key points to locate the vertex.
Solve the equation $|2x+3|=|x-1|$ . Be careful to consider all sign cases.
Answer: $x=-4$ or $-\dfrac{2}{3}$ The question requires solving |2x+3| = |x-1| considering all sign cases. Square both sides: $(2x+3)^2$ = $(x-1)^2$ , $4x^2$ +12x+9 = $x^2$ -2x+1, $3x^2$ +14x +8=0. Solutions: x = [-14 ± √(196-96)]/6 = [-14 ±10]/6, so x=-4 or x=-2/3. Check: for x=-4, | -8+3|=5, | -4-1|=5; for x=-2/3, | -4/3+3|=5/3, | -2/3-1|=5/3. Both work. A common error is missing a solution by not squaring properly. Always verify in original after solving.
Simplify $\dfrac{3\sqrt{2}}{\sqrt{18}}$ in simplest form.
Answer: $1$ The question requires simplifying ( $\dfrac{3\sqrt{2}}{\sqrt{18}}$ ) to its simplest form. First, simplify the denominator: ( $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ ), so the expression becomes ( $\dfrac{3\sqrt{2}}{3\sqrt{2}} = 1$ ). You can also rationalize or combine, but simplification shows it cancels to 1. A key error is not factoring out the perfect square in ( $\sqrt{18}$ ), leading to unsimplified forms like ( $\sqrt{2}$ ) or ( $\dfrac{1}{\sqrt{2}}$ ). Another mistake might be rewriting it incorrectly as choice A without simplifying. For radicals in fractions, simplify each part separately and cancel common factors before choosing an answer.
Simplify $\sqrt{98}+\sqrt{8}$ in simplest radical form.
Answer: $9\sqrt{2}$ The question requires simplifying √98 + √8 in simplest radical form. Factor: √98 = √(49*2) = 7√2, √8 = √(4*2) = 2√2. Add: 7√2 + 2√2 = 9√2. This is simplest as like terms are combined. A key error is adding under one radical, like √(98+8), which is wrong. Check by approximating: 7√2 ≈9.899, 2√2≈2.828, sum≈12.727; 9√2≈12.727, matches. Always simplify each radical before combining.
Simplify $\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)$ .
Answer: $3$ The question requires simplifying $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$ . Recognize this as a difference of squares: $(a - b)(a + b) = a^2 - b^2$ , where $a = \sqrt{5}$ and $b = \sqrt{2}$ . Thus, $(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$ . The result is an integer. A key error is expanding incorrectly instead of using the formula. Another mistake might be adding the terms rather than multiplying. When simplifying conjugate products, apply the difference of squares to simplify efficiently and check by expansion if needed.
Solve the inequality $|2x-1|<7$ . Give the solution as an interval.
Answer: $(-3,4)$ The question asks to solve the inequality |2x - 1| < 7 and give the solution as an interval. Rewrite as -7 < 2x - 1 < 7, then add 1: -6 < 2x < 8, and divide by 2: -3 < x < 4. The solution interval is (-3, 4). Verify boundary points: at x = -3, | -6 - 1 | = 7 not less than 7; at x = 4, |8 - 1| = 7 not less. A common error is mishandling the inequality signs or forgetting to consider both cases. Another mistake might be including the endpoints for a strict inequality. When solving absolute value inequalities, always isolate the absolute value and handle the two cases carefully.
A point on a number line is at position $x$ . Its distance from $-3$ is 5 units. Which equation represents this situation, and what are the solutions? Interpreting absolute value as distance helps avoid sign errors.
Answer: $|x+3|=5;\ x=2,-8$ This problem asks us to translate a distance statement into an absolute value equation. If a point at position $x$ is 5 units away from -3, then the distance between $x$ and -3 equals 5. The distance formula on a number line is $|x - (-3)| = |x + 3| = 5$ . To solve this, we consider two cases: $x + 3 = 5$ , giving $x = 2$ , and $x + 3 = -5$ , giving $x = -8$ . We can verify: the distance from 2 to -3 is $|2-(-3)| = |5| = 5$ ✓, and the distance from -8 to -3 is $|-8-(-3)| = |-5| = 5$ ✓. A common error is writing the equation as $|x - 3| = 5$ , which would find points 5 units from positive 3 instead. When setting up distance problems, remember that distance from $x$ to $a$ is $|x - a|$ .
Solve the equation $|x|=|x-6|$ . Interpret absolute value as distance on the number line.
Answer: $x=3$ The question asks to solve |x| = |x - 6| interpreting as distances on the number line. This means distance from x to 0 equals distance to 6, so x is midpoint: x = 3. Check: |3| = 3, |3-6| = 3, equal. Algebraically, consider cases: if x ≥ 6, x = x-6 impossible; if 0 ≤ x < 6, x = 6 - x, 2x=6, x=3; if x < 0, -x = 6 - x impossible. Only x=3. A key error is assuming symmetric solutions like ±3. Always check all intervals defined by critical points.
Simplify $\sqrt{3}\left(\sqrt{12}-\sqrt{27}\right)$ in simplest radical form.
Answer: $-3$ The question requires simplifying ( $\sqrt{3}(\sqrt{12} - \sqrt{27})$ ) in simplest radical form. First, simplify inside: ( $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ ), ( $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ ), so it's ( $\sqrt{3}(2\sqrt{3} - 3\sqrt{3}) = \sqrt{3}(-\sqrt{3}) = -3$ ). Distribute ( $\sqrt{3}$ ) to get the same result. A common error is not simplifying the radicals first, leading to something like ( $\sqrt{39}$ ) by incorrect combination. Another mistake might be computing 6 - 9 without radicals, but that's the unsimplified numerical equivalent. Always simplify each radical before multiplying and check for negative signs in subtraction.
Solve the equation $\sqrt{2x-1}+1=5$ . Be careful not to divide the radical incorrectly; isolate the radical first, then square both sides, and confirm the solution satisfies the original equation.
Answer: $x=\dfrac{17}{2}$ To solve $\sqrt{2x-1} + 1 = 5$ , we first isolate the radical by subtracting 1 from both sides: $\sqrt{2x-1} = 4$ . Now we square both sides to eliminate the square root: $(\sqrt{2x-1})^2 = 4^2$ , which gives us $2x - 1 = 16$ . Solving for $x$ : $2x = 17$ , so $x = \frac{17}{2}$ . Let's verify by substituting back: $\sqrt{2(\frac{17}{2})-1} + 1 = \sqrt{17-1} + 1 = \sqrt{16} + 1 = 4 + 1 = 5$ ✓. A common error is trying to distribute operations across the radical before isolating it, such as incorrectly thinking $\sqrt{2x-1} + 1 = \sqrt{2x} + \sqrt{0}$ . Always isolate the radical term before squaring both sides.
Which expression is equivalent to $\sqrt{45}-\sqrt{20}$ in simplest radical form?
Answer: $\sqrt{5}$ The question seeks the simplest radical equivalent to √45 - √20. Factor: √45 = √(9*5) = 3√5, √20 = √(4*5) = 2√5. Subtract: 3√5 - 2√5 = √5. This is simplified. A common error is subtracting under one radical or misfactoring. Verify: ≈3*2.236 - 2*2.236 ≈6.708 - 4.472 = 2.236, and √5≈2.236, matches. When subtracting radicals, simplify to like terms first.
How many real solutions satisfy $|x-4|=\sqrt{x+5}\,?$
Answer: 2 When you encounter an equation involving both absolute value and a square root, you need to consider the different cases created by the absolute value while respecting the domain restrictions of the square root. First, note that $\sqrt{x+5}$ requires $x \geq -5$ . Since we're setting this equal to $|x-4|$ , which is always non-negative, this constraint makes sense. The absolute value $|x-4|$ creates two cases: - Case 1: When $x \geq 4$ , we have $|x-4| = x-4$ - Case 2: When $x < 4$ , we have $|x-4| = -(x-4) = 4-x$ For Case 1 ( $x \geq 4$ ): Solve $x-4 = \sqrt{x+5}$ . Squaring both sides: $(x-4)^2 = x+5$ , which gives $x^2-8x+16 = x+5$ , so $x^2-9x+11 = 0$ . Using the quadratic formula: $x = \frac{9 \pm \sqrt{81-44}}{2} = \frac{9 \pm \sqrt{37}}{2}$ . Only $x = \frac{9 + \sqrt{37}}{2} \approx 7.54$ satisfies $x \geq 4$ . For Case 2 ( $x < 4$ ): Solve $4-x = \sqrt{x+5}$ . Squaring: $(4-x)^2 = x+5$ , giving $16-8x+x^2 = x+5$ , so $x^2-9x+11 = 0$ . This yields the same solutions, but only $x = \frac{9 - \sqrt{37}}{2} \approx 1.46$ satisfies $x < 4$ . Both solutions check out in the original equation, giving us exactly 2 real solutions. Answer choice (A) suggests no solutions, missing both valid cases. Choice (B) finds only one solution, likely solving just one case. Choice (D) incorrectly assumes more solutions exist. Strategy tip: With absolute value equations, always check your solutions in the original equation since squaring can introduce extraneous solutions.
Which expression is equivalent to $\dfrac{3}{2-\sqrt{5}}$ after rationalizing the denominator?
Answer: $\dfrac{3(2+\sqrt{5})}{-1}$ The question requires finding the equivalent expression to 3/(2 - √5) after rationalizing the denominator. Multiply numerator and denominator by the conjugate 2 + √5: 3(2 + √5) / ((2 - √5)(2 + √5)) = 3(2 + √5) / (4 - 5) = 3(2 + √5) / (-1). This simplifies to -3(2 + √5), matching the form in choice B. Verify numerically: original ≈ 3 / (-0.236) ≈ -12.71, and -3(4.236) ≈ -12.71. A key error is using the wrong conjugate or mishandling the negative denominator. Another mistake could be not distributing the negative sign correctly. When rationalizing, always check your work by verifying numerical equivalence.
Which expression is equivalent to $\dfrac{\sqrt{5}+\sqrt{20}}{\sqrt{5}}$ in simplest form?
Answer: $3$ The question asks which expression is equivalent to ( $\dfrac{\sqrt{5} + \sqrt{20}}{\sqrt{5}}$ ) in simplest form. Split the fraction: ( $= \dfrac{\sqrt{5}}{\sqrt{5}} + \dfrac{\sqrt{20}}{\sqrt{5}} = 1 + \sqrt{\dfrac{20}{5}} = 1 + \sqrt{4} = 1 + 2 = 3$ ). Alternatively, simplify ( $\sqrt{20} = 2\sqrt{5}$ ), so ( $\dfrac{\sqrt{5} + 2\sqrt{5}}{\sqrt{5}} = \dfrac{3\sqrt{5}}{\sqrt{5}} = 3$ ). A common error is not simplifying fully, like leaving it as 1 + \sqrt{4}, which equals 3 but isn't the simplest form. Another mistake might be thinking it's \sqrt{25} = 5 by incorrect addition. When simplifying radical expressions in fractions, break them down and combine like terms before evaluating.
The function $f(x)=\sqrt{x^2-8x+12}$ is defined only for real numbers $x$ that satisfy which condition?
Answer: $x\le2\;\text{or}\;x\ge6$ When you encounter a function with a square root, you need to determine its domain—the set of values for which the function is defined. Since square roots of negative numbers aren't real, the expression under the radical must be non-negative. For $f(x)=\sqrt{x^2-8x+12}$ to be defined, we need $x^2-8x+12 \geq 0$ . Let's factor this quadratic expression. We're looking for two numbers that multiply to 12 and add to -8: those numbers are -6 and -2. So $x^2-8x+12 = (x-2)(x-6)$ . Now we need $(x-2)(x-6) \geq 0$ . A product of two factors is non-negative when both factors have the same sign (both positive or both negative). The factors equal zero when $x = 2$ and $x = 6$ , which divides the number line into three intervals. Testing each region: when $x < 2$ , both factors are negative, so their product is positive. When $2 < x < 6$ , one factor is positive and one is negative, so their product is negative. When $x > 6$ , both factors are positive, so their product is positive. Including the boundary points where the expression equals zero, the solution is $x \leq 2$ or $x \geq 6$ . Choice A suggests $x > 2$ and $x < 6$ , which is exactly where the expression is negative. Choice B includes this problematic interval with its endpoints. Choice C correctly identifies the regions but excludes the crucial boundary points where the function is still defined. Strategy tip: When finding domains of square root functions, always factor the expression under the radical and use a sign chart to identify where it's non-negative.
Solve the absolute value equation $|x-1|+2=7$ .
Answer: $x=6$ or $-4$ The question requires solving the absolute value equation |x - 1| + 2 = 7. Isolate the absolute value: |x - 1| = 5. Consider two cases: x - 1 = 5 gives x = 6, and x - 1 = -5 gives x = -4. Both satisfy the original: |6-1| + 2 = 7, |-4-1| + 2 = 7. A common error is forgetting to isolate before splitting cases. Always check solutions in the original equation to confirm. For absolute value equations, isolate first then handle positive and negative scenarios.
Which expression equals $\dfrac{5}{2+\sqrt3}$ when the denominator is rationalized?
Answer: $10-5\sqrt3$ When you encounter a fraction with a radical in the denominator, you need to rationalize it by eliminating the square root from the bottom. This process uses the conjugate method. To rationalize $\dfrac{5}{2+\sqrt{3}}$ , multiply both numerator and denominator by the conjugate of the denominator. The conjugate of $2+\sqrt{3}$ is $2-\sqrt{3}$ : $\dfrac{5}{2+\sqrt{3}} \cdot \dfrac{2-\sqrt{3}}{2-\sqrt{3}} = \dfrac{5(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$ The numerator becomes: $5(2-\sqrt{3}) = 10-5\sqrt{3}$ The denominator uses the difference of squares pattern: $(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4-3 = 1$ Therefore: $\dfrac{10-5\sqrt{3}}{1} = 10-5\sqrt{3}$ , which is answer choice B. Looking at the wrong answers: A) $5+\sqrt{3}$ might result from incorrectly adding terms or confusing the rationalization process. C) $10+5\sqrt{3}$ comes from using the wrong sign—multiplying by $2+\sqrt{3}$ instead of the conjugate $2-\sqrt{3}$ . D) $5-\sqrt{3}$ suggests someone forgot to distribute the 5 properly in the numerator. Study tip: Always use the conjugate when rationalizing denominators containing addition or subtraction with radicals. The conjugate flips the middle sign, and when multiplied together, the radicals eliminate due to the difference of squares pattern: $(a+b)(a-b) = a^2-b^2$ .
What is the solution set of the absolute value equation $|x-4|=9$ ? (Give both solutions.)
Answer: $\{-5,13\}$ The question asks for the solution set of the absolute value equation |x - 4| = 9, requiring both solutions. To solve, consider the two cases: x - 4 = 9 or x - 4 = -9, leading to x = 13 or x = -5. The solution set is {-5, 13}. Verify by substitution: | -5 - 4 | = | -9 | = 9, and |13 - 4| = 9, both correct. A common error is only considering the positive case and missing the negative solution. Another mistake might be incorrect arithmetic when solving the linear equations. When solving absolute value equations, always handle both cases and verify solutions to ensure accuracy.

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