Equations and Inequalities - DAT Quantitative Reasoning

1. The solutions of $2x^2-7x-4=0$ are
   Answer: $\left\{-\dfrac12,\;4\right\}$ When you encounter a quadratic equation like $2x^2-7x-4=0$, you need to find the values of $x$ that make the equation true. The most reliable approach is using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. For this equation, $a=2$, $b=-7$, and $c=-4$. Substituting these values: $x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-4)}}{2(2)} = \frac{7 \pm \sqrt{49+32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}$ This gives us two solutions: $x = \frac{7+9}{4} = \frac{16}{4} = 4$ and $x = \frac{7-9}{4} = \frac{-2}{4} = -\frac{1}{2}$ Therefore, the solution set is $\left\{-\frac{1}{2}, 4\right\}$, which matches choice A. Choice B gives $\{-2, 4\}$ - this keeps the correct positive solution but incorrectly gives $-2$ instead of $-\frac{1}{2}$. Choice C gives $\left\{\frac{1}{2}, -4\right\}$ - this has sign errors on both solutions. Choice D gives $\{2, -4\}$ - this also has incorrect values for both solutions, possibly from calculation errors in applying the quadratic formula. Always double-check your solutions by substituting them back into the original equation. For quadratic equations, the quadratic formula is foolproof when factoring seems difficult, so master this formula and be careful with your arithmetic.
2. If $\frac{2^x + 2^{-x}}{2^x - 2^{-x}} = \frac{5}{3}$, what is the value of $2^x$?
   Answer: $2^x = 2$ Let $y = 2^x$. Then $2^{-x} = \frac{1}{2^x} = \frac{1}{y}$. The equation becomes: $\frac{y + \frac{1}{y}}{y - \frac{1}{y}} = \frac{5}{3}$. Multiplying numerator and denominator by $y$: $\frac{y^2 + 1}{y^2 - 1} = \frac{5}{3}$. Cross-multiplying: $3(y^2 + 1) = 5(y^2 - 1)$. Expanding: $3y^2 + 3 = 5y^2 - 5$. Rearranging: $3y^2 + 3 - 5y^2 + 5 = 0$, which gives $-2y^2 + 8 = 0$. So $2y^2 = 8$, thus $y^2 = 4$, and $y = \pm 2$. Since $y = 2^x > 0$ for all real $x$, we have $y = 2$. Therefore $2^x = 2$, which means $x = 1$. Let's verify: If $x = 1$, then $2^x = 2$ and $2^{-x} = 2^{-1} = \frac{1}{2}$. The left side becomes: $\frac{2 + \frac{1}{2}}{2 - \frac{1}{2}} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{2} \cdot \frac{2}{3} = \frac{5}{3}$ ✓. Choice B: If $2^x = 3$, then $2^{-x} = \frac{1}{3}$. The fraction becomes $\frac{3 + \frac{1}{3}}{3 - \frac{1}{3}} = \frac{\frac{10}{3}}{\frac{8}{3}} = \frac{10}{8} = \frac{5}{4} \neq \frac{5}{3}$. Choice C: If $2^x = 4$, then $2^{-x} = \frac{1}{4}$. The fraction becomes $\frac{4 + \frac{1}{4}}{4 - \frac{1}{4}} = \frac{\frac{17}{4}}{\frac{15}{4}} = \frac{17}{15} \neq \frac{5}{3}$. Choice D: If $2^x = 5$, then $2^{-x} = \frac{1}{5}$. The fraction becomes $\frac{5 + \frac{1}{5}}{5 - \frac{1}{5}} = \frac{\frac{26}{5}}{\frac{24}{5}} = \frac{26}{24} = \frac{13}{12} \neq \frac{5}{3}$. Choice E: If $2^x = 6$, the calculation would similarly not yield $\frac{5}{3}$.
3. Solve for $x$: $-3(2-x)>9$
   Answer: $x>5$ When you encounter an inequality with parentheses, you need to distribute first, then isolate the variable while carefully handling the inequality direction. Starting with $-3(2-x)>9$, distribute the $-3$ to both terms inside the parentheses: $-3(2) + (-3)(-x) = -6 + 3x$. So the inequality becomes $-6 + 3x > 9$. Next, add 6 to both sides: $3x > 15$. Finally, divide both sides by 3: $x > 5$. Since we divided by a positive number, the inequality direction stays the same. Looking at the wrong answers: Answer B ($x < 5$) represents the most common error - flipping the inequality sign when dividing by a positive number. Remember, you only flip the inequality when multiplying or dividing by a negative number. Answer C ($x > -5$) likely comes from incorrectly distributing the negative sign, perhaps getting $-6 - 3x > 9$, which would lead to $x < -5$, then incorrectly flipping it. Answer D ($x < -5$) results from the same distribution error but without the extra incorrect flip. You can verify the answer by testing a value: if $x = 6$, then $-3(2-6) = -3(-4) = 12$, and indeed $12 > 9$. The key strategy here is to work systematically through algebraic steps and remember that inequality direction only changes when multiplying or dividing by negative numbers - not positive ones.
4. Solve for $x$: $3^{2x-1}=27$.
   Answer: $2$ When you encounter exponential equations like this one, the key strategy is to express both sides using the same base, then equate the exponents. Start by recognizing that $27$ can be written as a power of $3$: $27 = 3^3$. This transforms your equation into $3^{2x-1} = 3^3$. Since the bases are now equal, the exponents must also be equal: $2x - 1 = 3$. Solving for $x$: $2x = 4$, so $x = 2$. Let's verify: $3^{2(2)-1} = 3^3 = 27$ ✓ Now let's examine why the other answers don't work: **B) $x = 3$**: If you substitute this value, you get $3^{2(3)-1} = 3^5 = 243$, which is far too large. This might result from incorrectly setting $2x - 1 = 5$ instead of $3$. **C) $x = \frac{3}{2}$**: Substituting gives $3^{2(\frac{3}{2})-1} = 3^{3-1} = 3^2 = 9$, which is too small. This error could arise from confusing the relationship between the base and exponent. **D) $x = \frac{5}{2}$**: This yields $3^{2(\frac{5}{2})-1} = 3^{5-1} = 3^4 = 81$, again too large. This might come from misidentifying $27$ as $3^4$ or making an arithmetic error. Strategy tip: Always express both sides of exponential equations using the same base when possible. Common powers to memorize: $3^2 = 9$, $3^3 = 27$, $3^4 = 81$. After solving, substitute your answer back into the original equation to verify it works.
5. A dental lab bills $90 base plus $12 per model; if total is $210, solve $90+12x=210$ for $x$.
   Answer: $x=10$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves determining the number of models in a dental lab bill, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, x=10, is obtained by subtracting 90 from both sides to get 12x=120, then dividing by 12. A common mistake is x=9, which results from miscalculating 12*9 +90 as 210 instead of 198. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.
6. Solve $|x+1|\le4$.
   Answer: $-5\le x\le3$ When you encounter absolute value inequalities, remember that the absolute value represents distance on a number line. The inequality $|x+1|\le4$ asks: "For which values of x is the distance from x to -1 equal to or less than 4?" To solve this systematically, use the definition that $|A| \le B$ is equivalent to $-B \le A \le B$ when B is positive. Here, we have $|x+1| \le 4$, so this becomes: $-4 \le x+1 \le 4$ Subtracting 1 from all parts: $-4-1 \le x \le 4-1$ $-5 \le x \le 3$ You can verify this by testing boundary values: when $x = -5$, we get $|-5+1| = |-4| = 4 \le 4$ ✓, and when $x = 3$, we get $|3+1| = |4| = 4 \le 4$ ✓. Choice A gives $-5 \le x \le 3$, which matches our solution perfectly. Choice B represents the solution to $|x+1| \ge 4$, the opposite inequality. Choice C shifts the interval incorrectly—this would be the solution if the original expression were $|x-1| \le 4$. Choice D uses strict inequalities and represents values where $|x+1| > 4$. Remember this key pattern: $|expression| \le number$ always gives you a single interval (AND condition), while $|expression| \ge number$ gives you two separate intervals (OR condition). Don't confuse the direction of the inequality when converting from absolute value form.
7. Find the set of $x$ that satisfies both $x+5>2$ and $3x-1<8$.
   Answer: $-3<x<3$ When you encounter compound inequalities like this, you're looking for values of $x$ that satisfy both conditions simultaneously. This requires solving each inequality separately, then finding their intersection. Start with the first inequality: $x + 5 > 2$. Subtracting 5 from both sides gives you $x > -3$. For the second inequality: $3x - 1 < 8$. Adding 1 to both sides yields $3x < 9$, and dividing by 3 gives you $x < 3$. Since you need both conditions to be true, you're looking for values where $x > -3$ AND $x < 3$. This means $x$ must be greater than -3 but less than 3, written as $-3 < x < 3$. This is answer choice A. Let's examine why the other options fail. Answer B ($x < -3$ or $x > 3$) represents the complement of the correct answer - it includes all values outside the solution interval. Answer C ($x < 3$) only satisfies the second inequality but ignores the constraint from the first inequality that $x > -3$. Answer D ($x > -3$) only satisfies the first inequality but ignores the upper bound from the second inequality. Remember that "and" in compound inequalities means intersection - you need the overlap of both solution sets. Always solve each inequality independently first, then find where both conditions are satisfied simultaneously. A number line sketch can help visualize the intersection when you're unsure.
8. Bacteria on a mouthguard increases $20\%$ per hour; if $N(t)=500(1.2)^t$, what is $N(2)$?
   Answer: $N(2)=720$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves modeling bacteria increase on a mouthguard, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, N(2)=720, is obtained by substituting t=2 into 500*$(1.2)^2$ and calculating 500*1.44. A common mistake is N(2)=700, which results from adding 20% twice incorrectly instead of compounding. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.
9. Find the sum of all real solutions of $2^{2x}-6(2^x)+8=0$.
   Answer: 3 Let y = $2^x$. Then $y^2$ - 6y + 8 = 0, so y = 2 or y = 4. This gives $2^x$ = 2, so x = 1, and $2^x$ = 4, so x = 2. The sum is 1 + 2 = 3. The tempting error is adding 2 and 4, the y-values, to get 6, but you must solve for x.
10. What is the solution set of $4^{3-x}\geq 16^{2x}$?
    Answer: $x\leq \frac{3}{5}$ Rewrite both sides as powers of 2: $4^{3-x}$=$2^{6-2x}$ and $16^{2x}$=$2^{8x}$. Since 2 is greater than 1, compare exponents: 6-2x >= 8x, so 6 >= 10x, meaning x <= 3/5. A tempting wrong answer is x >= 3/5 from flipping the inequality when dividing by 10, but the divisor is positive so the direction stays.
11. If $5^{x+1}=125^{\;2-x}$, then $x$ equals
    Answer: $\dfrac54$ When you encounter exponential equations with different bases, your goal is to rewrite both sides using the same base so you can set the exponents equal to each other. Notice that both 5 and 125 are powers of 5: $125 = 5^3$. Let's rewrite the equation using base 5 throughout: $5^{x+1} = 125^{2-x}$ $5^{x+1} = (5^3)^{2-x}$ Using the power rule $(a^m)^n = a^{mn}$, the right side becomes: $5^{x+1} = 5^{3(2-x)}$ $5^{x+1} = 5^{6-3x}$ Since the bases are equal, the exponents must be equal: $x + 1 = 6 - 3x$ Solving for $x$: $x + 3x = 6 - 1$ $4x = 5$ $x = \frac{5}{4}$ This confirms that A) $\frac{5}{4}$ is correct. Let's check why the other options fail. If you substitute B) $x = 1$: the left side gives $5^2 = 25$ while the right side gives $125^1 = 125$ - not equal. For C) $x = \frac{3}{2}$: you get $5^{2.5} \approx 55.9$ versus $125^{0.5} \approx 11.2$ - also unequal. For D) $x = 2$: the left side is $5^3 = 125$ but the right side is $125^0 = 1$ - clearly not equal. Strategy tip: Always look for common bases when solving exponential equations. Convert everything to the smallest possible base (here, 5 instead of 125) to simplify your work and avoid calculation errors.
12. The equation $\dfrac12x^{2}-3x+4=0$ has solutions
    Answer: $\{2,4\}$ When you encounter a quadratic equation, you need to find the values of x that make the equation true. For $\frac{1}{2}x^2 - 3x + 4 = 0$, you can solve this using the quadratic formula or by testing the given options through substitution. Let's verify by substituting the values from option A into the original equation. For x = 2: $\frac{1}{2}(2)^2 - 3(2) + 4 = \frac{1}{2}(4) - 6 + 4 = 2 - 6 + 4 = 0$ ✓ For x = 4: $\frac{1}{2}(4)^2 - 3(4) + 4 = \frac{1}{2}(16) - 12 + 4 = 8 - 12 + 4 = 0$ ✓ Both values satisfy the equation, confirming that A is correct. Option B gives negative values that, when substituted, don't satisfy the equation. For x = -2: $\frac{1}{2}(-2)^2 - 3(-2) + 4 = 2 + 6 + 4 = 12 \neq 0$. Option C fails because x = 1 gives $\frac{1}{2}(1) - 3(1) + 4 = 2.5 \neq 0$. Option D is incorrect since x = -1 yields $\frac{1}{2}(1) + 3 + 4 = 7.5 \neq 0$. For quadratic equations on standardized tests, substitution is often faster than using the quadratic formula, especially when answer choices are provided. Always check both solutions when given a pair, as partial credit traps are common where only one value from a pair might work.
13. A disinfectant-resistant strain grows by factor $1.1$ per hour; if $N(t)=900(1.1)^t$, what is $N(3)$?
    Answer: $N(3)=1{,}197.9$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves predicting growth of a disinfectant-resistant bacteria strain, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, N(3)=1,197.9, is obtained by substituting t=3 into 900*$(1.1)^3$ and calculating 900*1.331. A common mistake is N(3)=1,089.0, which results from using $(1.1)^2$ instead of ^3. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.
14. A dentist mixes $x$ mL of fluoride concentrate at $0.40/mL plus $5 bottle cost; if total is $21, solve $0.40x+5=21$ for $x$.
    Answer: $x=40$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves mixing fluoride concentrate with a bottle cost, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, x=40, is obtained by subtracting 5 from both sides to get 0.40x=16, then dividing by 0.40. A common mistake is x=38, which results from rounding errors or miscalculating 0.40*38 +5 as 21 instead of 20.2. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.
15. Bacteria on a dental impression doubles every hour; if $N(t)=150\cdot2^t$, what does it predict for $t=4$ hours?
    Answer: $N(4)=2{,}400$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves predicting bacteria growth on a dental impression over time, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, N(4)=2,400, is obtained by substituting t=4 into the exponential equation and calculating 150*$2^4$=150*16. A common mistake is N(4)=1,200, which results from using $2^{4/2}$ instead of $2^4$. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.
16. For positive $k,kx^2+4x+1=0$ has one real solution. What is $k$?
    Answer: 4 A quadratic has one real solution when its discriminant is zero. Here, discriminant = $4^2$ - 4(k)(1) = 16 - 4k. Set 16 - 4k = 0, so k = 4. The tempting wrong answer is 16, which comes from ignoring the factor of 4 in the discriminant and would give no real solutions.
17. Find all real numbers $x$ satisfying $|2x-7|=3$.
    Answer: $\{2,\;5\}$ When you encounter absolute value equations, remember that the absolute value represents distance from zero on the number line. The equation $|2x-7|=3$ means the expression $2x-7$ is exactly 3 units away from zero, which gives us two possibilities. To solve $|2x-7|=3$, you need to consider both cases where the expression inside equals positive and negative 3: Case 1: $2x-7=3$ Adding 7 to both sides: $2x=10$ Dividing by 2: $x=5$ Case 2: $2x-7=-3$ Adding 7 to both sides: $2x=4$ Dividing by 2: $x=2$ You can verify: when $x=5$, we get $|2(5)-7|=|3|=3$ ✓, and when $x=2$, we get $|2(2)-7|=|-3|=3$ ✓. Looking at the answer choices: Choice A gives $\{2,5\}$, which matches our solutions perfectly. Choice B $\{1,3\}$ would give us $|2(1)-7|=5$ and $|2(3)-7|=1$, neither of which equals 3. Choice C $\{4,7\}$ yields $|2(4)-7|=1$ and $|2(7)-7|=7$, again missing our target of 3. Choice D $\{5\}$ only includes one of the two correct solutions. Remember that absolute value equations typically have two solutions unless the expression inside the absolute value bars equals zero at the boundary. Always set up both the positive and negative cases, and don't forget to check your work by substituting back into the original equation.
18. If $|x - 3| + |x + 2| \geq 7$, which of the following represents the solution set?
    Answer: $(-\infty, -3] \cup [4, \infty)$ We need to consider different cases based on the critical points where the expressions inside the absolute values equal zero: $x = 3$ and $x = -2$. Case 1: $x \leq -2$. Here both $x - 3 < 0$ and $x + 2 \leq 0$, so $|x - 3| = -(x - 3) = 3 - x$ and $|x + 2| = -(x + 2) = -x - 2$. The inequality becomes $(3 - x) + (-x - 2) \geq 7$, which simplifies to $1 - 2x \geq 7$. This gives $-2x \geq 6$, so $x \leq -3$. Since we're in the case $x \leq -2$, the solution for this case is $x \leq -3$. Case 2: $-2 < x < 3$. Here $x - 3 < 0$ and $x + 2 > 0$, so $|x - 3| = 3 - x$ and $|x + 2| = x + 2$. The inequality becomes $(3 - x) + (x + 2) \geq 7$, which simplifies to $5 \geq 7$. This is never true, so there are no solutions in this interval. Case 3: $x \geq 3$. Here both expressions are non-negative, so $|x - 3| = x - 3$ and $|x + 2| = x + 2$. The inequality becomes $(x - 3) + (x + 2) \geq 7$, which simplifies to $2x - 1 \geq 7$. This gives $2x \geq 8$, so $x \geq 4$. Since we're in the case $x \geq 3$, the solution for this case is $x \geq 4$. Combining all cases: $x \leq -3$ or $x \geq 4$, which can be written as $(-\infty, -3] \cup [4, \infty)$. Choice B gives $(-\infty, -2] \cup [3, \infty)$, which would include points like $x = 0$ where $|0-3| + |0+2| = 3 + 2 = 5 < 7$. Choice C gives $(-\infty, -4] \cup [5, \infty)$, which is too restrictive. Choice D gives $(-\infty, -1] \cup [2, \infty)$, which includes $x = 0$ (not a solution). Choice E gives $(-\infty, -5] \cup [6, \infty)$, which is also too restrictive.
19. Bacteria in a dental waterline grows by factor $1.5$ hourly; if $N(t)=200(1.5)^t$, what is $N(3)$?
    Answer: $N(3)=675$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves modeling bacteria growth in a dental waterline, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, N(3)=675, is obtained by substituting t=3 into 200*$(1.5)^3$ and computing 200*3.375. A common mistake is N(3)=600, which results from miscalculating $(1.5)^3$ as 3 instead of 3.375. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.
20. A dental practice budgets $\$2{,}200$–$\$2{,}800$ for software; which is the solution set for $2200\le 150x+850\le 2800$?
    Answer: $9\le x\le 13$ This question tests the ability to solve algebraic equations and inequalities involving linear, quadratic, and exponential expressions, essential for dental quantitative reasoning. Algebra involves manipulating equations and inequalities to find unknown values, using techniques such as substitution, elimination, and the quadratic formula. In this question, the scenario involves budgeting for dental practice software, requiring the application of algebraic principles to solve for the unknown variable. The correct answer, 9≤x≤13, is obtained by subtracting 850 from all parts and dividing by 150. A common mistake is 8≤x≤13, which results from lowering the lower bound incorrectly. To improve, students should practice setting up equations based on real-world scenarios and verify each step of their solution process, using checks like substituting back into the original equation to ensure accuracy.

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