Quadratic Equations - GMAT Quantitative Reasoning

1. If $2x^{2}-7x-4=0$, what is the smaller of the two real solutions?
   Answer: -\dfrac{1}{2} When you encounter a quadratic equation like $2x^{2}-7x-4=0$, you need to find its roots using either factoring, the quadratic formula, or completing the square. Since this doesn't factor easily, the quadratic formula is your best approach. For any quadratic $ax^2 + bx + c = 0$, the solutions are $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Here, $a=2$, $b=-7$, and $c=-4$. Substituting: $x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-4)}}{2(2)} = \frac{7 \pm \sqrt{49+32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4}$ This gives you two solutions: $x = \frac{7+9}{4} = \frac{16}{4} = 4$ and $x = \frac{7-9}{4} = \frac{-2}{4} = -\frac{1}{2}$ Since the question asks for the smaller solution, the answer is $-\frac{1}{2}$, which is choice (B). Looking at the wrong answers: (A) -4 is too negative and doesn't satisfy the original equation. (C) $\frac{1}{2}$ is the positive version of the smaller root—a sign error trap. (D) 4 is actually the larger of the two correct solutions, not the smaller one. Strategy tip: Always find both roots first, then identify which one the question asks for. The GMAT often includes both roots as answer choices to test whether you're reading carefully. Also, you can quickly verify your solutions by substituting back into the original equation.
2. A rectangular garden has an area of 48 square meters. The length of the garden is 2 meters greater than its width. What is the perimeter of the garden, in meters?
   Answer: 28 This problem tests your ability to set up and solve a system of equations using geometric relationships. When you encounter word problems involving rectangles with unknown dimensions, look for ways to express both dimensions in terms of a single variable. Let's define the width as $w$ meters. Since the length is 2 meters greater than the width, the length is $(w + 2)$ meters. Using the area formula for rectangles: $\text{Area} = \text{length} \times \text{width}$ $48 = (w + 2) \times w$ $48 = w^2 + 2w$ $w^2 + 2w - 48 = 0$ Factoring this quadratic: $(w + 8)(w - 6) = 0$ This gives us $w = -8$ or $w = 6$. Since width cannot be negative, $w = 6$ meters and length = $6 + 2 = 8$ meters. The perimeter is $2(6 + 8) = 28$ meters. Looking at the wrong answers: (A) 20 might result from incorrectly solving the quadratic or making arithmetic errors in the setup. (B) 24 could come from finding the semi-perimeter (length + width) instead of the full perimeter, or from solving $w^2 + 2w = 24$ instead of 48. (D) 32 might result from confusing the area and perimeter formulas or making calculation errors when factoring. Strategy tip: Always define your variable clearly, double-check that your quadratic setup matches the given area, and remember that perimeter requires adding all four sides—not just length plus width.
3. Consider the equation $2^{2t}-5\cdot2^{t}+6=0$. What is the sum of all real solutions for $t$?
   Answer: $1+\log_{2}3$ When you encounter an exponential equation like this, the key insight is recognizing it as a quadratic equation in disguise. The repeated appearance of $2^t$ suggests using substitution to transform this into a more familiar form. Let $u = 2^t$. This transforms the equation $2^{2t} - 5 \cdot 2^t + 6 = 0$ into $u^2 - 5u + 6 = 0$, since $2^{2t} = (2^t)^2 = u^2$. Factoring the quadratic: $u^2 - 5u + 6 = (u-2)(u-3) = 0$. This gives us $u = 2$ or $u = 3$. Substituting back: $2^t = 2$ or $2^t = 3$. From $2^t = 2$, we get $t = 1$. From $2^t = 3$, we get $t = \log_2 3$. The sum of all solutions is $1 + \log_2 3$, which is answer choice C. Let's examine the wrong answers: Answer A ($\log_2 12$) might result from incorrectly multiplying the solutions instead of adding them, since $2 \times 3 = 6$ and $\log_2(2 \times 6) = \log_2 12$. Answer B ($\log_2 9$) could come from adding the $u$-values incorrectly: $\log_2(2+3) = \log_2 5$ is wrong, but $\log_2(3^2) = \log_2 9$ suggests confusion about logarithm properties. Answer D ($2 + \log_2 3$) results from using $t = 2$ instead of $t = 1$ for the first solution. Remember: when you see exponential equations with the same base appearing in different powers, try substitution to create a quadratic. Always substitute back to find the original variable's values.
4. Which of the following is equivalent to $x^{4}-5x^{2}+4$?
   Answer: $(x-1)(x+1)(x-2)(x+2)$ When you encounter a polynomial like $x^{4}-5x^{2}+4$, you're dealing with a quadratic in disguise. Notice that this expression only has even powers of $x$, which suggests treating $x^2$ as a single variable to simplify the factoring process. Let's substitute $u = x^2$, transforming our expression into $u^2 - 5u + 4$. This quadratic factors as $(u-1)(u-4)$. Substituting back, we get $(x^2-1)(x^2-4)$. Now we can factor further using the difference of squares formula $a^2-b^2 = (a-b)(a+b)$: - $x^2-1 = (x-1)(x+1)$ - $x^2-4 = (x-2)(x+2)$ Therefore, $x^{4}-5x^{2}+4 = (x-1)(x+1)(x-2)(x+2)$, which matches choice D. Let's verify why the other options are incorrect: Choice A: $(x^{2}-4)(x^{2}+1)$ expands to $x^4 - 3x^2 - 4$, which has the wrong middle term and constant. Choice B: $(x-1)(x-1)(x+2)(x+2) = (x-1)^2(x+2)^2$ expands to $x^4 + 2x^3 - 3x^2 - 4x + 4$, introducing unwanted odd-power terms. Choice C: $(x^{2}-2)(x^{2}+2)$ expands to $x^4 - 4$, missing the middle term entirely. Study tip: When factoring polynomials with only even powers, always try substitution first (let $u = x^2$). This transforms a quartic into a manageable quadratic, then remember to factor completely using difference of squares when possible.
5. If $\dfrac{3}{x}+\dfrac{2}{x-1}=1$ and $x\neq0,1$, what is the sum of all real solutions of the equation?
   Answer: 6 When you encounter rational equations like this one, your goal is to eliminate the fractions by finding a common denominator, then solve the resulting quadratic equation. Remember that the sum of roots of a quadratic can often be found using Vieta's formulas. Starting with $\frac{3}{x}+\frac{2}{x-1}=1$, multiply everything by the common denominator $x(x-1)$: $3(x-1) + 2x = x(x-1)$ Expanding: $3x - 3 + 2x = x^2 - x$ Simplifying: $5x - 3 = x^2 - x$ Rearranging: $x^2 - 6x + 3 = 0$ Now you have a quadratic equation. Rather than solving for individual roots using the quadratic formula, you can use Vieta's formulas. For any quadratic $ax^2 + bx + c = 0$, the sum of roots equals $-\frac{b}{a}$. Here, $a = 1$ and $b = -6$, so the sum of roots is $-\frac{(-6)}{1} = 6$. Looking at the wrong answers: Choice (A) gives 3, which might result from incorrectly identifying the constant term as the sum. Choice (B) gives 4, which could come from miscalculating the coefficients when clearing fractions. Choice (C) gives 5, which might arise from confusing the coefficient of $x$ in the intermediate step $5x - 3 = x^2 - x$ with the final answer. The correct answer is (D) 6. Study tip: When you see rational equations, always check if you can use Vieta's formulas to find the sum or product of roots directly, rather than solving for individual solutions. This saves time and reduces calculation errors.
6. Which of the following equals $6y^{2}-19y+10$?
   Answer: $(2y-5)(3y-2)$ When you encounter a quadratic expression that needs to be factored, you're looking to reverse the FOIL process—finding two binomials that multiply to give the original expression. To factor $6y^2 - 19y + 10$, start by identifying what the first terms of each binomial must be. Since the leading coefficient is 6, the first terms could be $6y \cdot y$, $3y \cdot 2y$, etc. Next, find factors of the constant term 10: these could be $10 \cdot 1$, $5 \cdot 2$, or their negatives. Let's verify that choice C works: $(2y - 5)(3y - 2)$. Using FOIL: - First: $2y \cdot 3y = 6y^2$ - Outer: $2y \cdot (-2) = -4y$ - Inner: $(-5) \cdot 3y = -15y$ - Last: $(-5) \cdot (-2) = 10$ Combining: $6y^2 - 4y - 15y + 10 = 6y^2 - 19y + 10$ ✓ Choice A gives $(2y + 1)(3y - 10) = 6y^2 - 20y + 3y - 10 = 6y^2 - 17y - 10$—wrong middle term and wrong sign on the constant. Choice B gives $(2y - 2)(3y - 5) = 6y^2 - 10y - 6y + 10 = 6y^2 - 16y + 10$—correct constant but wrong middle coefficient. Choice D gives $(2y - 5)(3y + 2) = 6y^2 + 4y - 15y - 10 = 6y^2 - 11y - 10$—wrong middle term and wrong sign on the constant. Strategy tip: Always check your factoring by expanding back out using FOIL. This catches sign errors and coefficient mistakes that are common traps in factoring problems.
7. For which of the following values of $k$ does the equation $x^{2}-kx+16=0$ have no real solutions?
   Answer: -7 When you encounter a quadratic equation and need to determine when it has no real solutions, you're working with the discriminant. For any quadratic $ax^2 + bx + c = 0$, the discriminant is $b^2 - 4ac$. When the discriminant is negative, the equation has no real solutions. In the equation $x^2 - kx + 16 = 0$, we have $a = 1$, $b = -k$, and $c = 16$. The discriminant becomes $(-k)^2 - 4(1)(16) = k^2 - 64$. For no real solutions, we need $k^2 - 64 < 0$, which means $k^2 < 64$. This occurs when $-8 < k < 8$. Looking at our options: Only choice (B) -7 falls within this range, giving us $(-7)^2 - 64 = 49 - 64 = -15 < 0$. Choice (A) -9 is incorrect because $(-9)^2 = 81 > 64$, so the discriminant equals $81 - 64 = 17 > 0$, meaning real solutions exist. Choice (C) 8 is incorrect because $8^2 = 64$, making the discriminant $64 - 64 = 0$, which gives exactly one real solution (a repeated root). Choice (D) 9 is incorrect because $9^2 = 81 > 64$, so the discriminant equals $81 - 64 = 17 > 0$, producing two distinct real solutions. Remember: For quadratic equations on the GMAT, quickly identify when you need the discriminant. No real solutions means discriminant < 0, one solution means discriminant = 0, and two solutions means discriminant > 0.
8. If $x^2 - 6x + k = 0$ has two distinct real roots, and one root is twice the other root, what is the value of $k$?
   Answer: $k = 8$ Let the roots be r and 2r. By Vieta's formulas: sum of roots = r + 2r = 3r = 6, so r = 2. Product of roots = r(2r) = 2r² = k. Since r = 2, we have k = 2(2)² = 8. We can verify: x² - 6x + 8 = (x-2)(x-4) has roots 2 and 4, where 4 = 2(2). Choice B gives sum = 6 but discriminant = 0 (equal roots). Choice C gives incorrect root relationship. Choice D gives sum ≠ 6.
9. If $(x-3)(x+2)=5$, what is the sum of all real solutions for $x$?
   Answer: 1 When you encounter a quadratic equation in expanded form like this, you need to convert it to standard form and solve systematically. This tests your ability to manipulate quadratic expressions and apply solution formulas. First, expand the left side: $(x-3)(x+2) = x^2 + 2x - 3x - 6 = x^2 - x - 6$. Setting this equal to 5 gives you $x^2 - x - 6 = 5$, which rearranges to the standard form $x^2 - x - 11 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = -1$, and $c = -11$: $x = \frac{1 \pm \sqrt{1 + 44}}{2} = \frac{1 \pm \sqrt{45}}{2} = \frac{1 \pm 3\sqrt{5}}{2}$ The two solutions are $x_1 = \frac{1 + 3\sqrt{5}}{2}$ and $x_2 = \frac{1 - 3\sqrt{5}}{2}$. Their sum is $\frac{1 + 3\sqrt{5}}{2} + \frac{1 - 3\sqrt{5}}{2} = \frac{2}{2} = 1$. Choice (A) -1 likely comes from incorrectly calculating the coefficient in the quadratic formula. Choice (B) 0 might result from assuming the solutions are opposites, which only happens when there's no linear term. Choice (D) 5 could come from mistakenly thinking the constant term on the right side equals the sum of solutions. Strategy tip: For any quadratic $ax^2 + bx + c = 0$, the sum of solutions always equals $-\frac{b}{a}$. Here, that's $-\frac{(-1)}{1} = 1$, giving you a quick way to verify your answer without fully solving.
10. If $f(x) = x^2 + 2x - 3$ and $g(x) = x^2 - 4x + 3$, for how many values of $x$ does $f(x) = g(x)$?
    Answer: Exactly one value of $x$ Set f(x) = g(x): x² + 2x - 3 = x² - 4x + 3. Subtracting x² from both sides: 2x - 3 = -4x + 3. Adding 4x to both sides: 6x - 3 = 3. Adding 3: 6x = 6, so x = 1. There is exactly one solution. We can verify: f(1) = 1 + 2 - 3 = 0 and g(1) = 1 - 4 + 3 = 0. Choice A assumes no solution exists. Choice C treats this as a quadratic equation. Choice D assumes the functions are identical.
11. The expression $x^2 + bx + c$ can be factored as $(x + p)(x + q)$ where $p$ and $q$ are integers. If $b = -5$ and $c > 0$, how many possible values of $c$ are there?
    Answer: Two possible values of $c$ Expanding (x + p)(x + q) = x² + (p + q)x + pq. So p + q = b = -5 and pq = c > 0. Since c > 0, p and q have the same sign. Since p + q = -5 < 0, both are negative. Integer pairs (p,q) with p + q = -5: (-1,-4), (-2,-3), (-3,-2), (-4,-1). These give c values: (-1)(-4) = 4, (-2)(-3) = 6, (-3)(-2) = 6, (-4)(-1) = 4. So c can be 4 or 6 (two values). Choice B counts duplicate pairs as different. Choice C might count all four pairs. Choice D adds an incorrect value.
12. For what values of $m$ does the equation $mx^2 - (m+2)x + 2 = 0$ have exactly one solution?
    Answer: $m = 0$ or $m = 2$ For exactly one solution, either: (1) the quadratic has discriminant = 0, or (2) m = 0 making it linear. Case 1: For m ≠ 0, discriminant = (m+2)² - 4m(2) = m² + 4m + 4 - 8m = m² - 4m + 4 = (m-2)² = 0, so m = 2. Case 2: If m = 0, equation becomes -2x + 2 = 0, giving x = 1 (one solution). Therefore m = 0 or m = 2. Choice A misses the discriminant case. Choice B misses the linear case. Choice D incorrectly includes m = 1.
13. The height of a ball (in meters) tossed upward is given by $h(t)=-5t^{2}+20t+15$, where $t$ is the time in seconds after the ball is thrown. After how many seconds does the ball reach its maximum height?
    Answer: 2 When you encounter a quadratic function representing the motion of a projectile, you're looking for the vertex of a parabola. Since the coefficient of $t^2$ is negative (-5), this parabola opens downward, meaning the vertex represents the maximum height. For any quadratic function in the form $f(t) = at^2 + bt + c$, the time at which the maximum (or minimum) occurs is given by $t = -\frac{b}{2a}$. In this height function $h(t) = -5t^2 + 20t + 15$, we have $a = -5$ and $b = 20$. Substituting into the vertex formula: $t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2$ seconds. You can verify this makes sense by checking that the height increases before $t = 2$ and decreases after. At $t = 1$: $h(1) = -5(1) + 20(1) + 15 = 30$ meters. At $t = 2$: $h(2) = -5(4) + 20(2) + 15 = 35$ meters. At $t = 3$: $h(3) = -5(9) + 20(3) + 15 = 30$ meters. Choice (A) 1 second gives a height of 30 meters, which is less than the maximum. Choice (C) 3 seconds also gives 30 meters, the same height as at 1 second due to the parabola's symmetry. Choice (D) 4 seconds would give an even lower height of 15 meters. Remember: for projectile motion problems, always use the vertex formula $t = -\frac{b}{2a}$ to find when maximum or minimum values occur. This saves time compared to completing the square or taking derivatives.
14. If $x^2 - 4x + k = (x - a)^2 + b$ for all values of $x$, and $k = 7$, what is the value of $a^2 + b^2$?
    Answer: $13$ Complete the square: x² - 4x + 7 = (x² - 4x + 4) + 7 - 4 = (x - 2)² + 3. So a = 2 and b = 3. Therefore, a² + b² = 4 + 9 = 13. Choice A equals k only. Choice B equals a² + 1 only. Choice D equals (a + b)².
15. The quadratic $ax^2 + bx + c = 0$ has discriminant $\Delta = 25$. If the quadratic $ax^2 + bx + (c + 6) = 0$ has discriminant $\Delta'$, what is $\Delta - \Delta'$?
    Answer: $24a$ When you encounter questions about discriminants and how they change when coefficients are modified, focus on the discriminant formula: $\Delta = b^2 - 4ac$. For the original quadratic $ax^2 + bx + c = 0$, we know $\Delta = b^2 - 4ac = 25$. For the modified quadratic $ax^2 + bx + (c + 6) = 0$, the discriminant becomes: $\Delta' = b^2 - 4a(c + 6) = b^2 - 4ac - 24a$ Since $b^2 - 4ac = 25$, we can substitute: $\Delta' = 25 - 24a$ Therefore: $\Delta - \Delta' = 25 - (25 - 24a) = 25 - 25 + 24a = 24a$ This confirms answer choice B is correct. Looking at the wrong answers: Choice A gives $-24a$, which results from incorrectly calculating $\Delta' - \Delta$ instead of $\Delta - \Delta'$. Choice C gives $6a$, which might come from mistakenly thinking the discriminant changes by $4a \times 6 = 24a$ but then dividing by 4. Choice D gives $-6a$, which could result from thinking the change is simply $-a$ times the constant term change, missing the factor of 4 in the discriminant formula. Remember this key insight: when the constant term of a quadratic increases by $k$, the discriminant decreases by $4ak$. The discriminant and constant term move in opposite directions, and there's always that factor of $4a$ involved due to the structure of the discriminant formula.
16. The quadratic equation $x^{2}+mx+n=0$ has integer coefficients, and one of its roots is $2+\sqrt{3}$. Which of the following is the value of $n$?
    Answer: 1 When you encounter a quadratic equation with integer coefficients that has an irrational root involving a square root, remember that irrational roots must come in conjugate pairs. Since one root is $2+\sqrt{3}$, the other root must be $2-\sqrt{3}$. For any quadratic $x^2 + mx + n = 0$ with roots $r_1$ and $r_2$, we can use Vieta's formulas: the sum of roots equals $-m$ and the product of roots equals $n$. The sum of our roots is: $(2+\sqrt{3}) + (2-\sqrt{3}) = 4$, so $m = -4$. The product of our roots is: $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$, so $n = 1$. We can verify this by constructing the quadratic: $x^2 - 4x + 1 = 0$. Using the quadratic formula: $x = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$. ✓ Looking at the wrong answers: (B) 3 would give us different roots entirely. (C) 4 comes from incorrectly using the sum instead of the product. (D) 7 has no clear mathematical justification from our given information. The answer is (A) 1. Strategy tip: Whenever you see a quadratic with integer coefficients and one irrational root involving $\sqrt{n}$, immediately identify the conjugate root. Then use Vieta's formulas—the product will always eliminate the radical terms, giving you a clean integer result.
17. The equation $x^4 - 5x^2 + 4 = 0$ can be solved by substitution. How many distinct real solutions does this equation have?
    Answer: Four distinct real solutions Let y = x². Then x⁴ - 5x² + 4 = 0 becomes y² - 5y + 4 = 0. Factoring: (y - 1)(y - 4) = 0, so y = 1 or y = 4. When y = x² = 1, we get x = ±1. When y = x² = 4, we get x = ±2. Therefore, there are four distinct real solutions: x = -2, -1, 1, 2. Choice A ignores that both y values are positive. Choice B counts only positive x values. Choice C miscounts the solutions.
18. If $x^2 - px + q = 0$ has roots $r$ and $s$, and $x^2 - qx + p = 0$ has roots $t$ and $u$, what is the relationship between $r + s + t + u$ and $p + q$?
    Answer: $r + s + t + u = p + q$ When you encounter quadratic equations with given roots, immediately think about Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. For any quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, Vieta's formulas tell us that the sum of the roots equals $-\frac{b}{a}$ and the product equals $\frac{c}{a}$. For the first equation $x^2 - px + q = 0$ with roots $r$ and $s$, we have $a = 1$, $b = -p$, and $c = q$. Therefore: $r + s = -\frac{(-p)}{1} = p$ For the second equation $x^2 - qx + p = 0$ with roots $t$ and $u$, we have $a = 1$, $b = -q$, and $c = p$. Therefore: $t + u = -\frac{(-q)}{1} = q$ Adding these results: $r + s + t + u = p + q$ Looking at the incorrect answers: Choice A suggests $2p + 2q$, which would mean each sum of roots equals $p + q$ rather than the individual coefficients. Choice B is algebraically identical to A, representing the same error of doubling the correct result. Choice C adds an unexplained constant of 2, which has no basis in Vieta's formulas. The key insight is recognizing that when the coefficients of two quadratic equations are swapped ($p$ and $q$ switch positions), the sums of their respective roots will simply be those swapped coefficients. Remember: for monic quadratics (leading coefficient = 1), the sum of roots always equals the negative of the linear coefficient.
19. Which of the following intervals contains all real numbers $x$ satisfying $x^{2}-3x-4\le0$?
    Answer: $-1\le x\le4$ When you encounter a quadratic inequality like $x^2 - 3x - 4 \le 0$, you're looking for the values of $x$ that make the quadratic expression negative or zero. The key is to first find where the quadratic equals zero, then determine which intervals satisfy the inequality. Start by factoring $x^2 - 3x - 4 = 0$. You need two numbers that multiply to $-4$ and add to $-3$. These are $-4$ and $1$, so the factored form is $(x - 4)(x + 1) = 0$. This gives you roots at $x = 4$ and $x = -1$. These roots divide the number line into three intervals: $x < -1$, $-1 < x < 4$, and $x > 4$. Since the parabola opens upward (positive leading coefficient), the quadratic is negative between the roots and positive outside them. Testing a point like $x = 0$: $(0)^2 - 3(0) - 4 = -4 < 0$, confirming the expression is negative between the roots. Since we want $\le 0$, we include the endpoints where the expression equals zero. Therefore, the solution is $-1 \le x \le 4$. Choice A gives $x \le -1$ or $x \ge 4$, which represents where the quadratic is positive or zero—the opposite of what we want. Choice C includes only $x < -1$, missing the interval between the roots. Choice D correctly identifies the interval between roots but excludes the endpoints where the quadratic equals zero, ignoring the "equal to" part of $\le$. Strategy tip: Always check whether your inequality includes "or equal to" and remember to include or exclude the roots accordingly.

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