Examkin

Algebra

356 soru

Tüm türler356Çoktan seçmeli181Çoklu seçim90Sayısal85
Soru 81Soru →

Which of the following values are solutions to the equation (x+4)2=49(x + 4)^2 = 49? Select all that apply.

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Cevap: −11-11; 33

Cevap

The correct solutions are −11-11 and 33.
Taking the square root of both sides of (x+4)2=49(x + 4)^2 = 49 yields x+4=7x + 4 = 7 or x+4=−7x + 4 = -7. Solving these two linear equations gives x=3x = 3 and x=−11x = -11. Both values satisfy the original quadratic equation.

Adım Adım Çözüm

1
Apply the square root property to both sides of the quadratic equation.
x+4=±49=±7x + 4 = \pm \sqrt{49} = \pm 7
Taking the square root of both sides accounts for both the positive and negative roots of 4949.
2
Solve the first linear equation for the positive case.
x+4=7  ⟹  x=7−4=3x + 4 = 7 \implies x = 7 - 4 = 3
Subtract 44 from both sides when the expression equals +7+7.
3
Solve the second linear equation for the negative case.
x+4=−7  ⟹  x=−7−4=−11x + 4 = -7 \implies x = -7 - 4 = -11
Subtract 44 from both sides when the expression equals −7-7.

Anahtar Kavram

Square Root Property of Quadratic Equations
Soru 82Soru →

What is the sum of all real solutions to the equation x−3=x+3x - 3 = \sqrt{x + 3}?

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Cevap: 6

Cevap

The sum of all real solutions to the equation is 6.
Squaring both sides of x−3=x+3x - 3 = \sqrt{x + 3} gives x2−6x+9=x+3x^2 - 6x + 9 = x + 3, which simplifies to x2−7x+6=0x^2 - 7x + 6 = 0. Factoring yields (x−6)(x−1)=0(x - 6)(x - 1) = 0, giving candidate solutions x=6x = 6 and x=1x = 1. Substituting x=6x = 6 into the original equation gives 3=9=33 = \sqrt{9} = 3, which is valid. Substituting x=1x = 1 gives −2=4=2-2 = \sqrt{4} = 2, which is false, making x=1x = 1 an extraneous solution. Therefore, x=6x = 6 is the single valid solution, and its sum is 6.

Adım Adım Çözüm

1
Eliminate the radical by squaring both sides of the equation.
(x−3)2=(x+3)2  ⟹  x2−6x+9=x+3(x - 3)^2 = (\sqrt{x + 3})^2 \implies x^2 - 6x + 9 = x + 3
Squaring both sides converts the radical equation into a standard quadratic equation.
2
Rearrange the terms into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0 and factor.
x2−7x+6=0  ⟹  (x−6)(x−1)=0x^2 - 7x + 6 = 0 \implies (x - 6)(x - 1) = 0
Subtracting xx and 33 from both sides isolates zero on one side so the quadratic expression can be factored.
3
Solve for potential values of xx.
x=6x = 6 or x=1x = 1
Setting each factor to zero yields the algebraic solutions to the quadratic equation.
4
Check each potential solution in the original radical equation x−3=x+3x - 3 = \sqrt{x + 3}.
For x=6x = 6: 6−3=36 - 3 = 3 and 6+3=3\sqrt{6 + 3} = 3 (Valid). For x=1x = 1: 1−3=−21 - 3 = -2 and 1+3=2\sqrt{1 + 3} = 2 (Invalid, since −2≠2-2 \neq 2).
Squaring an equation can introduce extraneous solutions that do not satisfy the original principal square root relationship.
5
Calculate the sum of all valid real solutions.
The only valid real solution is 66, so the sum is 66.
Extraneous solutions must be discarded before computing the final sum.

Anahtar Kavram

Solving radical equations using quadratic factoring and identifying extraneous solutions
Tahmini Süre:2m 0s
Soru 83Soru →

A train travels from Station A to Station B, a distance of 180 miles, at a constant speed of vv miles per hour. On the return trip from Station B to Station A, the train travels the first half of the distance at a constant speed that is 20%20\% less than vv, and the remaining half of the distance at a constant speed that is 25%25\% greater than vv. If the total time for the return trip is 6 minutes longer than the total time for the trip from Station A to Station B, what is the value of vv?

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Cevap: 45

Cevap

The value of vv is 4545.
The outbound travel time for 180 miles at speed vv is 180v\frac{180}{v} hours. On the return trip, the first 90 miles at speed 0.80v0.80v require 900.80v=112.5v\frac{90}{0.80v} = \frac{112.5}{v} hours, while the second 90 miles at speed 1.25v1.25v require 901.25v=72v\frac{90}{1.25v} = \frac{72}{v} hours. The total return duration is 112.5+72v=184.5v\frac{112.5 + 72}{v} = \frac{184.5}{v} hours. Setting the difference between the return time and outbound time equal to 6 minutes (0.10.1 hours) gives 184.5v−180v=0.1\frac{184.5}{v} - \frac{180}{v} = 0.1, which simplifies to 4.5v=0.1\frac{4.5}{v} = 0.1, yielding v=45v = 45.

Adım Adım Çözüm

1
Write the expression for the outbound trip duration in terms of vv.
Toutbound=180vT_{\text{outbound}} = \frac{180}{v} hours.
Time is equal to total distance divided by constant speed.
2
Calculate the duration for each half of the return trip in terms of vv.
The first 90 miles take 900.80v=112.5v\frac{90}{0.80v} = \frac{112.5}{v} hours, and the second 90 miles take 901.25v=72v\frac{90}{1.25v} = \frac{72}{v} hours, giving a total return duration of 184.5v\frac{184.5}{v} hours.
The return trip consists of two 90-mile segments driven at 0.80v0.80v and 1.25v1.25v respectively.
3
Equate the difference between return and outbound times to 0.1 hours and solve for vv.
184.5v−180v=0.1  ⟹  4.5v=0.1  ⟹  v=45\frac{184.5}{v} - \frac{180}{v} = 0.1 \implies \frac{4.5}{v} = 0.1 \implies v = 45.
The time difference of 6 minutes is equal to 660=0.1\frac{6}{60} = 0.1 hours.

Anahtar Kavram

Distance, Rate, and Time Modeling with Piecewise Speed Changes
Soru 84Soru →

Pipeline A operating alone can fill a storage tank in xx hours, whereas Pipeline B operating alone takes 50%50\% longer than Pipeline A to fill the same tank. Drainage Pipe C operating alone can empty a full tank in 2x2x hours. If all three pipes are opened simultaneously when the tank is empty, the tank becomes completely full in 1212 hours. What is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 1414

Cevap

The value of xx is 1414.
To find xx, calculate the hourly rate of each pipe: Pipeline A fills 1x\frac{1}{x} of the tank per hour, Pipeline B takes 1.5x=32x1.5x = \frac{3}{2}x hours so it fills 23x\frac{2}{3x} per hour, and Pipe C empties 12x\frac{1}{2x} per hour. The combined rate is 1x+23x−12x=76x\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{7}{6x}. Since the tank fills in 1212 hours, the net rate is 112\frac{1}{12}. Setting 76x=112\frac{7}{6x} = \frac{1}{12} gives 6x=846x = 84, so x=14x = 14.

Adım Adım Çözüm

1
Express the individual work rates per hour in terms of xx.
Pipeline A rate = 1x\frac{1}{x}, Pipeline B time = 1.5x=32x  ⟹  1.5x = \frac{3}{2}x \implies Pipeline B rate = 23x\frac{2}{3x}, Pipe C rate = −12x-\frac{1}{2x}.
Work rate is the reciprocal of the total time required to complete the job alone, with drainage represented as a negative rate.
2
Formulate the equation for the combined rate of all three pipes operating together.
Combined Rate = 1x+23x−12x=1x(1+23−12)=1x(6+4−36)=76x\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{1}{x} \left(1 + \frac{2}{3} - \frac{1}{2}\right) = \frac{1}{x} \left(\frac{6 + 4 - 3}{6}\right) = \frac{7}{6x}.
Simultaneous operation means summing the individual rates.
3
Equate the combined rate to the overall rate required to fill the tank in 12 hours and solve for xx.
\frac{7}{6x} = \frac{1}{12} \implies 6x = 84 \implies x = 14.
The tank is filled in 12 hours, so the net rate per hour is 112\frac{1}{12}.

Anahtar Kavram

Combined Work Rates with Inflow and Outflow
Tahmini Süre:2m 0s
Soru 85Soru →
For all non-zero real numbers uu and vv, the custom operation Δ\Delta is defined by uΔv=uv−vuu \Delta v = \frac{u}{v} - \frac{v}{u} The function ff is defined for all non-zero real numbers xx by f(x)=xΔ2f(x) = x \Delta 2. Which of the following statements must be true for all non-zero real numbers aa and bb? Select all such statements.

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Cevap: aΔb=−(bΔa)a \Delta b = -(b \Delta a); f(−a)=−f(a)f(-a) = -f(a)

Cevap

The statements aΔb=−(bΔa)a \Delta b = -(b \Delta a) and f(−a)=−f(a)f(-a) = -f(a) must be true.
The custom operation is anti-commutative because swapping variables negates the difference of the fractions, making the statement aΔb=−(bΔa)a \Delta b = -(b \Delta a) true. Additionally, substituting −a-a into f(x)=x2−2xf(x) = \frac{x}{2} - \frac{2}{x} yields −f(a)-f(a), showing f(−a)=−f(a)f(-a) = -f(a) is also true.

Adım Adım Çözüm

1
Analyze the first identity aΔb=−(bΔa)a \Delta b = -(b \Delta a).
aΔb=ab−baa \Delta b = \frac{a}{b} - \frac{b}{a} and bΔa=ba−ab=−(ab−ba)=−(aΔb)b \Delta a = \frac{b}{a} - \frac{a}{b} = -\left(\frac{a}{b} - \frac{b}{a}\right) = -(a \Delta b).
Factoring −1-1 out of bΔab \Delta a proves anti-commutativity.
2
Evaluate f(x)f(x) explicitly and test f(−a)f(-a).
f(x)=x2−2xf(x) = \frac{x}{2} - \frac{2}{x}. Substituting −a-a gives f(−a)=−a2−2−a=−(a2−2a)=−f(a)f(-a) = \frac{-a}{2} - \frac{2}{-a} = -\left(\frac{a}{2} - \frac{2}{a}\right) = -f(a).
This confirms that ff is an odd function.
3
Compare f(a2)f(a^2) with (f(a))2(f(a))^2.
f(a2)=a22−2a2f(a^2) = \frac{a^2}{2} - \frac{2}{a^2}, but (f(a))2=a24−2+4a2(f(a))^2 = \frac{a^2}{4} - 2 + \frac{4}{a^2}. Testing a=1a = 1 yields f(1)=−1.5  ⟹  (f(1))2=2.25f(1) = -1.5 \implies (f(1))^2 = 2.25, whereas f(12)=−1.5≠2.25f(1^2) = -1.5 \neq 2.25.
Counterexample shows the equality does not hold in general.
4
Compare f(a+b)f(a+b) with f(a)+f(b)f(a) + f(b).
For a=1,b=1a=1, b=1, f(1+1)=f(2)=22−22=0f(1+1) = f(2) = \frac{2}{2} - \frac{2}{2} = 0, but f(1)+f(1)=−1.5+(−1.5)=−3≠0f(1) + f(1) = -1.5 + (-1.5) = -3 \neq 0.
Functions do not in general preserve additivity f(a+b)=f(a)+f(b)f(a+b) = f(a) + f(b) unless they are linear through the origin.

Anahtar Kavram

Properties of algebraic custom operations and odd functions
Soru 86Soru →

If xx is a real number that satisfies the inequality ∣5−2x3∣≤3\left|\frac{5 - 2x}{3}\right| \le 3, what is the minimum possible value of 4−3x4 - 3x?

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Cevap: −17-17

Cevap

−17-17
To find the minimum possible value of 4−3x4 - 3x, we first solve the inequality ∣5−2x3∣≤3\left|\frac{5 - 2x}{3}\right| \le 3. Multiplying by 33 gives ∣5−2x∣≤9|5 - 2x| \le 9. This unfolds into the compound inequality −9≤5−2x≤9-9 \le 5 - 2x \le 9. Subtracting 55 gives −14≤−2x≤4-14 \le -2x \le 4. Dividing all parts by −2-2 requires flipping the inequality signs, resulting in 7≥x≥−27 \ge x \ge -2, or −2≤x≤7-2 \le x \le 7. Because 4−3x4 - 3x decreases as xx increases, the expression reaches its minimum when xx is at its maximum value of 77. Substituting x=7x = 7 yields 4−3(7)=−174 - 3(7) = -17.

Adım Adım Çözüm

1
Clear the denominator from the absolute value inequality
∣5−2x∣≤9|5 - 2x| \le 9
Multiplying both sides of the inequality by the positive number 3 preserves the inequality direction.
2
Express the absolute value inequality as a compound inequality
−9≤5−2x≤9-9 \le 5 - 2x \le 9
For any non-negative constant cc, ∣u∣≤c|u| \le c is equivalent to −c≤u≤c-c \le u \le c.
3
Isolate the variable term by subtracting 5 from all parts
−14≤−2x≤4-14 \le -2x \le 4
Subtracting a constant from all parts of a compound inequality maintains the inequality relationships.
4
Divide by -2 and reverse the inequality signs
7≥x≥−2  ⟹  −2≤x≤77 \ge x \ge -2 \implies -2 \le x \le 7
Dividing an inequality by a negative number reverses the direction of the inequality signs.
5
Determine which bound of xx minimizes 4−3x4 - 3x and evaluate
Minimum value =4−3(7)=−17= 4 - 3(7) = -17
The expression 4−3x4 - 3x has a negative coefficient for xx, making it a decreasing function. Therefore, the minimum value of 4−3x4 - 3x occurs when xx takes its maximum possible value (x=7x = 7).

Anahtar Kavram

Linear Inequalities and Absolute Value
Soru 87Soru →

A logistics company packages cargo using three types of containers: small, medium, and large.

- A shipment of 33 small, 22 medium, and 11 large container has a total weight of 130130 kilograms.
- A shipment of 11 small, 44 medium, and 22 large containers has a total weight of 185185 kilograms.
- A shipment of 22 small, 11 medium, and 33 large containers has a total weight of 160160 kilograms.

What is the weight, in kilograms, of one large container?

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Cevap: 35

Cevap

The weight of one large container is 35 kilograms.
Representing the weights of small, medium, and large containers as variables SS, MM, and LL yields the 3x3 system of linear equations:
1) 3S+2M+L=1303S + 2M + L = 130
2) S+4M+2L=185S + 4M + 2L = 185
3) 2S+M+3L=1602S + M + 3L = 160

Solving for SS in equation (2) gives S=185−4M−2LS = 185 - 4M - 2L. Substituting this into equations (1) and (3) reduces the system to:
- 2M+L=852M + L = 85
- 7M+L=2107M + L = 210

Subtracting the first equation from the second yields 5M=1255M = 125, so M=25M = 25. Substituting M=25M = 25 into 2M+L=852M + L = 85 gives 50+L=8550 + L = 85, which simplifies to L=35L = 35.

Adım Adım Çözüm

1
Set up a system of three linear equations based on the shipment descriptions
3S+2M+L=1303S + 2M + L = 130, S+4M+2L=185S + 4M + 2L = 185, and 2S+M+3L=1602S + M + 3L = 160
Translate the physical constraints of the three shipments into mathematical relationships
2
Isolate variable SS in the second equation and substitute into the first and third equations
Two equations in two variables: 2M+L=852M + L = 85 and 7M+L=2107M + L = 210
Reduce the 3x3 system to a 2x2 system to eliminate variable SS
3
Subtract the two simplified equations to solve for MM
5M=125  ⟹  M=255M = 125 \implies M = 25
Eliminate variable LL to obtain the value of MM
4
Substitute M=25M = 25 back into 2M+L=852M + L = 85 to solve for LL
L=35L = 35
Find the requested value for the weight of one large container

Anahtar Kavram

Solving 3x3 Systems of Linear Equations using Substitution and Elimination
Soru 88Soru →

A pharmaceutical laboratory produces a custom solution in a mixing tank using two automated pumps, Pump A and Pump B. Initially, the tank contains 600600 liters of a solution that is 20%20\% active reagent by volume. Pump A adds pure active reagent to the tank at a constant rate of 1515 liters per minute, while Pump B simultaneously adds distilled water (0%0\% active reagent) to the tank at a constant rate of 2525 liters per minute. At the exact instant when the mixture in the tank reaches 35%35\% active reagent by volume, Pump B is turned off while Pump A continues to add pure active reagent at 1515 liters per minute. How many total minutes, from the moment both pumps were initially started, does it take for the solution in the tank to reach 50%50\% active reagent by volume?

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Cevap: 174

Cevap

The total elapsed time is 174 minutes.
The problem requires setting up a two-stage rational rate model. In the first stage, initial reagent volume is 120120 liters (20%20\% of 600600). With Pump A adding 1515 L/min of reagent and Pump B adding 2525 L/min of water, the mixture concentration after t1t_1 minutes is 120+15t1600+40t1=0.35\frac{120 + 15t_1}{600 + 40t_1} = 0.35, yielding t1=90t_1 = 90 minutes. At t1=90t_1 = 90, the total volume is 4,2004,200 liters and the reagent volume is 1,4701,470 liters. In the second stage, only Pump A is active, so concentration after t2t_2 additional minutes is 1470+15t24200+15t2=0.50\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50, yielding t2=84t_2 = 84 minutes. Summing both intervals gives 90+84=17490 + 84 = 174 total minutes.

Adım Adım Çözüm

1
Determine initial quantities of active reagent and total solution.
Initial total volume V0=600V_0 = 600 liters. Initial volume of active reagent A0=0.20×600=120A_0 = 0.20 \times 600 = 120 liters.
Establishing the baseline volume and solute concentration before any pumps are turned on.
2
Model Phase 1 where both pumps operate concurrently for t1t_1 minutes to reach 35%35\% concentration.
Total volume equation: V(t1)=600+(15+25)t1=600+40t1V(t_1) = 600 + (15 + 25)t_1 = 600 + 40t_1. Reagent volume equation: A(t1)=120+15t1A(t_1) = 120 + 15t_1. Set concentration to 0.350.35: 120+15t1600+40t1=0.35  ⟹  t1=90\frac{120 + 15t_1}{600 + 40t_1} = 0.35 \implies t_1 = 90 minutes.
Both pumps add volume, but only Pump A adds active reagent. Solving 120+15t1600+40t1=720\frac{120 + 15t_1}{600 + 40t_1} = \frac{7}{20} gives 20(120+15t1)=7(600+40t1)  ⟹  2400+300t1=4200+280t1  ⟹  20t1=1800  ⟹  t1=9020(120 + 15t_1) = 7(600 + 40t_1) \implies 2400 + 300t_1 = 4200 + 280t_1 \implies 20t_1 = 1800 \implies t_1 = 90 minutes.
3
Calculate the accumulated total volume and reagent volume at the end of Phase 1.
Total volume V1=600+40(90)=4,200V_1 = 600 + 40(90) = 4,200 liters. Active reagent volume A1=120+15(90)=1,470A_1 = 120 + 15(90) = 1,470 liters.
These amounts serve as the starting parameters for Phase 2.
4
Model Phase 2 where only Pump A operates for t2t_2 additional minutes to reach 50%50\% concentration.
In Phase 2, volume increases at 1515 L/min and reagent increases at 1515 L/min. Set concentration to 0.500.50: 1470+15t24200+15t2=0.50  ⟹  t2=84\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50 \implies t_2 = 84 minutes.
Solving 2(1470+15t2)=4200+15t2  ⟹  2940+30t2=4200+15t2  ⟹  15t2=1260  ⟹  t2=842(1470 + 15t_2) = 4200 + 15t_2 \implies 2940 + 30t_2 = 4200 + 15t_2 \implies 15t_2 = 1260 \implies t_2 = 84 minutes.
5
Sum the durations of Phase 1 and Phase 2 to find the total time.
Total time T=t1+t2=90+84=174T = t_1 + t_2 = 90 + 84 = 174 minutes.
The question asks for the total minutes from the initial start of both pumps.

Anahtar Kavram

Multi-Phase Algebraic Rate and Mixture Modeling
Soru 89Soru →

If x2+x3=10\frac{x}{2} + \frac{x}{3} = 10, what is the value of xx?

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Cevap: 12

Cevap

The value of xx is 12.
The option showing 12 is correct because converting the fractions x2\frac{x}{2} and x3\frac{x}{3} to have a common denominator of 6 produces 3x+2x6=10\frac{3x + 2x}{6} = 10, or 5x6=10\frac{5x}{6} = 10. Multiplying both sides by 6 gives 5x=605x = 60, and dividing by 5 yields x=12x = 12.

Adım Adım Çözüm

1
Find a common denominator for the fractional terms on the left side of the equation.
The least common denominator of 2 and 3 is 6. Rewriting the fractions gives 3x6+2x6=10\frac{3x}{6} + \frac{2x}{6} = 10.
Fractions must have identical denominators before their numerators can be added.
2
Combine the fractions on the left side.
5x6=10\frac{5x}{6} = 10
Adding the numerators 3x+2x3x + 2x yields 5x5x over the shared denominator 6.
3
Clear the denominator by multiplying both sides of the equation by 6.
5x = 60
Multiplying both sides by 6 eliminates the fraction while maintaining equality.
4
Isolate xx by dividing both sides by 5.
x = 12
Dividing 60 by 5 yields the solution for xx.

Anahtar Kavram

Solving linear equations involving fractions by clearing denominators using the least common multiple.
Tahmini Süre:45s
Soru 90Soru →

If xx is a real number that satisfies the equation x−x+2=4x - \sqrt{x + 2} = 4, what is the value of (x+2)32(x + 2)^{\frac{3}{2}}?

Cevabı ve açıklamayı göster

Cevap: 2727

Cevap

27
Isolating the radical gives x−4=x+2x - 4 = \sqrt{x + 2}. Squaring both sides yields (x−4)2=x+2(x - 4)^2 = x + 2, which expands to x2−8x+16=x+2x^2 - 8x + 16 = x + 2, or x2−9x+14=0x^2 - 9x + 14 = 0. Factoring yields (x−7)(x−2)=0(x - 7)(x - 2) = 0, giving solutions x=7x = 7 and x=2x = 2. Testing x=7x = 7 in the original equation gives 7−9=47 - \sqrt{9} = 4, which is valid. Testing x=2x = 2 gives 2−4=0≠42 - \sqrt{4} = 0 \neq 4, which is extraneous. Substituting the valid root x=7x = 7 into (x+2)32(x + 2)^{\frac{3}{2}} gives (7+2)32=932=(9)3=33=27(7 + 2)^{\frac{3}{2}} = 9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27.

Adım Adım Çözüm

1
Isolate the radical term in the given equation.
x−4=x+2x - 4 = \sqrt{x + 2}
Isolating the radical allows both sides to be squared cleanly.
2
Square both sides to eliminate the square root and form a quadratic equation.
(x−4)2=x+2  ⟹  x2−8x+16=x+2  ⟹  x2−9x+14=0(x - 4)^2 = x + 2 \implies x^2 - 8x + 16 = x + 2 \implies x^2 - 9x + 14 = 0
Squaring eliminates the radical and allows standard quadratic solving techniques.
3
Factor the quadratic equation to find candidate solutions.
(x−7)(x−2)=0  ⟹  x=7 or x=2(x - 7)(x - 2) = 0 \implies x = 7 \text{ or } x = 2
Factoring provides potential real roots.
4
Check candidate solutions in the original equation to eliminate extraneous roots.
For x=7x = 7: 7−7+2=7−3=47 - \sqrt{7 + 2} = 7 - 3 = 4 (valid).
For x=2x = 2: 2−2+2=2−2=0≠42 - \sqrt{2 + 2} = 2 - 2 = 0 \neq 4 (extraneous).
Squaring both sides can introduce false solutions that fail the original radical equation.
5
Evaluate the target expression (x+2)32(x + 2)^{\frac{3}{2}} using x=7x = 7.
(7+2)32=932=(912)3=33=27(7 + 2)^{\frac{3}{2}} = 9^{\frac{3}{2}} = (9^{\frac{1}{2}})^3 = 3^3 = 27
Applying exponent rules (amn=(an)ma^{\frac{m}{n}} = (\sqrt[n]{a})^m) gives the exact required value.

Anahtar Kavram

Solving radical equations requires isolating the radical, squaring both sides, checking for extraneous solutions introduced by squaring, and evaluating fractional exponents via roots and integer powers.
Tahmini Süre:2m 0s
Soru 91Soru →

If x>0x > 0 and x12+x−12=3x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3, what is the value of x2+x−2x^2 + x^{-2}?

Cevabı ve açıklamayı göster

Cevap: 47

Cevap

47
Squaring both sides of x12+x−12=3x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3 gives x+2+x−1=9x + 2 + x^{-1} = 9, which simplifies to x+x−1=7x + x^{-1} = 7. Squaring both sides of x+x−1=7x + x^{-1} = 7 gives x2+2+x−2=49x^2 + 2 + x^{-2} = 49, which yields x2+x−2=47x^2 + x^{-2} = 47.

Adım Adım Çözüm

1
Square both sides of the given equation x12+x−12=3x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3.
(x12+x−12)2=32  ⟹  x+2(x12)(x−12)+x−1=9(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^2 = 3^2 \implies x + 2(x^{\frac{1}{2}})(x^{-\frac{1}{2}}) + x^{-1} = 9
Applying the binomial expansion identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2.
2
Simplify the middle term and solve for x+x−1x + x^{-1}.
x+2(1)+x−1=9  ⟹  x+x−1=7x + 2(1) + x^{-1} = 9 \implies x + x^{-1} = 7
Since x12⋅x−12=x0=1x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = x^0 = 1, subtracting 2 from both sides isolates x+x−1x + x^{-1}.
3
Square both sides of x+x−1=7x + x^{-1} = 7.
(x+x−1)2=72  ⟹  x2+2(x)(x−1)+x−2=49(x + x^{-1})^2 = 7^2 \implies x^2 + 2(x)(x^{-1}) + x^{-2} = 49
Squaring x+x−1x + x^{-1} generates the terms x2x^2 and x−2x^{-2}.
4
Simplify the middle term and solve for x2+x−2x^2 + x^{-2}.
x2+2+x−2=49  ⟹  x2+x−2=47x^2 + 2 + x^{-2} = 49 \implies x^2 + x^{-2} = 47
Subtracting 2 from both sides isolates the desired expression x2+x−2x^2 + x^{-2}.

Anahtar Kavram

Algebraic Exponents and Binomial Expansion
Soru 92Soru →

If xx is an integer that satisfies both ∣2x−5∣≤9|2x - 5| \le 9 and 3−x<53 - x < 5, how many possible values of xx exist?

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Cevap: 9

Cevap

There are 9 possible integer values for x.
Solving ∣2x−5∣≤9|2x - 5| \le 9 gives −2≤x≤7-2 \le x \le 7. Solving 3−x<53 - x < 5 gives x>−2x > -2. Taking the intersection yields −2<x≤7-2 < x \le 7. The integers in this interval are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, amounting to 9 values in total.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣2x−5∣≤9|2x - 5| \le 9
−2≤x≤7-2 \le x \le 7
Unfold ∣2x−5∣≤9|2x - 5| \le 9 as −9≤2x−5≤9-9 \le 2x - 5 \le 9, add 5 to obtain −4≤2x≤14-4 \le 2x \le 14, and divide by 2.
2
Solve the linear inequality 3−x<53 - x < 5
x>−2x > -2
Subtract 3 to get −x<2-x < 2, then divide by −1-1 and reverse the inequality symbol.
3
Determine the intersection of both inequalities
−2<x≤7-2 < x \le 7
Combine −2≤x≤7-2 \le x \le 7 and x>−2x > -2 on the real number line.
4
Count the integer values within the intersection −2<x≤7-2 < x \le 7
9 integer values
The valid integers are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, giving a total count of 7−(−1)+1=97 - (-1) + 1 = 9.

Anahtar Kavram

Linear Inequalities and Absolute Value Bounds
Tahmini Süre:1m 30s
Soru 93Soru →

If xx is a real number that satisfies the equation x+6x−9+x−6x−9=10\sqrt{x + 6\sqrt{x - 9}} + \sqrt{x - 6\sqrt{x - 9}} = 10, what is the value of xx?

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Cevap: 34

Cevap

34
Using the substitution u=x−9≥0u = \sqrt{x - 9} \ge 0, we have x=u2+9x = u^2 + 9. The expressions under the square roots become x+6x−9=u2+6u+9=(u+3)2x + 6\sqrt{x - 9} = u^2 + 6u + 9 = (u + 3)^2 and x−6x−9=u2−6u+9=(u−3)2x - 6\sqrt{x - 9} = u^2 - 6u + 9 = (u - 3)^2. Taking square roots gives (u+3)2+(u−3)2=(u+3)+∣u−3∣=10\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10. For u≥3u \ge 3, this simplifies to (u+3)+(u−3)=10  ⟹  2u=10  ⟹  u=5(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5. Finally, substituting u=5u = 5 back yields x=52+9=34x = 5^2 + 9 = 34.

Adım Adım Çözüm

1
Define a variable substitution to simplify the nested radical structure.
Let u=x−9u = \sqrt{x - 9} where u≥0u \ge 0. Squaring both sides gives u2=x−9u^2 = x - 9, so x=u2+9x = u^2 + 9.
This substitution allows the expressions inside the outer square roots to be rewritten as polynomials in terms of uu.
2
Rewrite the expressions under each square root as perfect square trinomials.
x+6x−9=(u2+9)+6u=(u+3)2x + 6\sqrt{x - 9} = (u^2 + 9) + 6u = (u + 3)^2 and x−6x−9=(u2+9)−6u=(u−3)2x - 6\sqrt{x - 9} = (u^2 + 9) - 6u = (u - 3)^2.
Expressing terms as perfect squares allows the outer radicals to be simplified.
3
Simplify the square root expressions using absolute values.
(u+3)2+(u−3)2=(u+3)+∣u−3∣=10\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10.
For any real number aa, a2=∣a∣\sqrt{a^2} = |a|. Since u≥0u \ge 0, u+3>0u + 3 > 0, so ∣u+3∣=u+3|u + 3| = u + 3.
4
Solve the absolute value equation across valid domain intervals.
If u≥3u \ge 3, ∣u−3∣=u−3|u - 3| = u - 3, giving (u+3)+(u−3)=10  ⟹  2u=10  ⟹  u=5(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5. If 0≤u<30 \le u < 3, ∣u−3∣=3−u|u - 3| = 3 - u, giving (u+3)+(3−u)=6≠10(u + 3) + (3 - u) = 6 \neq 10 (no solution). Thus, u=5u = 5.
Splitting into cases based on the definition of absolute value isolates the valid root.
5
Substitute u=5u = 5 back into the expression for xx.
x=52+9=25+9=34x = 5^2 + 9 = 25 + 9 = 34.
Converting from uu back to xx provides the solution to the original equation.

Anahtar Kavram

Simplifying nested radicals by completing the square under the radical sign and applying the identity a2=∣a∣\sqrt{a^2} = |a|.
Soru 94Soru →

A company allocated a total budget of BB dollars for a project. In the first phase of the project, 25\frac{2}{5} of the total budget plus $3,000\$3,000 was spent. In the second phase, 13\frac{1}{3} of the remaining budget after the first phase was spent. If the unspent amount after both phases is $14,000\$14,000, what was the total initial budget BB?

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Cevap: $40,000\$40,000

Cevap

$40,000\$40,000
The correct answer of $40,000\$40,000 is obtained by systematically tracking the remaining budget after each phase. After Phase 1, the remaining amount is B−(25B+3,000)=35B−3,000B - (\frac{2}{5}B + 3,000) = \frac{3}{5}B - 3,000. Spending 13\frac{1}{3} of this balance in Phase 2 leaves 23\frac{2}{3} of it unspent: 23(35B−3,000)=25B−2,000\frac{2}{3}(\frac{3}{5}B - 3,000) = \frac{2}{5}B - 2,000. Setting this equal to the final unspent amount of $14,000\$14,000 gives 25B=16,000\frac{2}{5}B = 16,000, which solves to B=40,000B = 40,000.

Adım Adım Çözüm

1
Express the remaining budget after the first phase in terms of BB.
Amount spent in Phase 1 = 25B+3,000\frac{2}{5}B + 3,000. Remaining after Phase 1 = B−(25B+3,000)=35B−3,000B - \left(\frac{2}{5}B + 3,000\right) = \frac{3}{5}B - 3,000.
Subtracting the first phase expenses from the initial total budget BB determines the balance available for the second phase.
2
Express the unspent budget after the second phase.
Since 13\frac{1}{3} of the remaining budget was spent in Phase 2, 1−13=231 - \frac{1}{3} = \frac{2}{3} of that remaining budget is left. Remaining after Phase 2 = 23(35B−3,000)\frac{2}{3}\left(\frac{3}{5}B - 3,000\right).
Taking 23\frac{2}{3} of the Phase 1 remainder directly gives the final unspent amount.
3
Expand and simplify the algebraic equation setting the unspent amount equal to $14,000\$14,000.
\frac{2}{3}\left(\frac{3}{5}B - 3,000\right) = 14,000 \implies \frac{2}{5}B - 2,000 = 14,000.
Distributing 23\frac{2}{3} across both terms inside the parentheses clears the fraction product.
4
Solve the linear equation for BB.
\frac{2}{5}B = 16,000 \implies B = 16,000 \times \frac{5}{2} = 40,000.
Adding 2,0002,000 to both sides and multiplying by the reciprocal 52\frac{5}{2} yields the total budget BB.

Anahtar Kavram

Formulating and solving multi-step linear equations in one variable with fractional quantities and consecutive remaining balances.
Tahmini Süre:2m 0s
Soru 95Soru →

If xx and yy are positive integers satisfying 3x−3y=7023^x - 3^y = 702 and x+y=3\sqrt{x + y} = 3, what is the value of x2−y2x^2 - y^2?

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Cevap: 2727

Cevap

The correct value of x2−y2x^2 - y^2 is 27.
Squaring x+y=3\sqrt{x + y} = 3 gives x+y=9x + y = 9. Factoring 3x−3y=7023^x - 3^y = 702 gives 3y(3x−y−1)=7023^y(3^{x-y} - 1) = 702. Since 702=33×26702 = 3^3 \times 26 and (3x−y−1)(3^{x-y} - 1) is coprime to 3, we deduce 3y=33  ⟹  y=33^y = 3^3 \implies y = 3, and 3x−3−1=26  ⟹  3x−3=27  ⟹  x=63^{x-3} - 1 = 26 \implies 3^{x-3} = 27 \implies x = 6. Substituting x=6x = 6 and y=3y = 3 into x2−y2x^2 - y^2 gives 36−9=2736 - 9 = 27.

Adım Adım Çözüm

1
Eliminate the radical from the given linear equation.
Squaring both sides of x+y=3\sqrt{x + y} = 3 gives x+y=9x + y = 9.
Squaring both sides removes the square root operator to establish a linear relationship between xx and yy.
2
Factor out the common exponential term 3y3^y from 3x−3y=7023^x - 3^y = 702.
3y(3x−y−1)=7023^y(3^{x-y} - 1) = 702.
Since xx and yy are positive integers and 702>0702 > 0, it must be true that x>yx > y, allowing factoring by exponent rules 3x=3y⋅3x−y3^x = 3^y \cdot 3^{x-y}.
3
Find the prime factorization of 702 and match the power of 3.
702=27×26=33×26702 = 27 \times 26 = 3^3 \times 26, so 3y(3x−y−1)=33×263^y(3^{x-y} - 1) = 3^3 \times 26.
The factor (3x−y−1)(3^{x-y} - 1) is not divisible by 3 because 3x−y3^{x-y} is a multiple of 3 for x>yx > y. Therefore, all powers of 3 in 702 must belong to 3y3^y.
4
Solve for the values of yy and xx.
y=3y = 3 and 3x−3−1=26  ⟹  3x−3=27=33  ⟹  x−3=3  ⟹  x=63^{x-3} - 1 = 26 \implies 3^{x-3} = 27 = 3^3 \implies x - 3 = 3 \implies x = 6.
Equating prime component bases yields y=3y = 3 and x=6x = 6, which satisfies x+y=6+3=9x + y = 6 + 3 = 9.
5
Calculate the target expression x2−y2x^2 - y^2.
x2−y2=62−32=36−9=27x^2 - y^2 = 6^2 - 3^2 = 36 - 9 = 27.
Substituting x=6x = 6 and y=3y = 3 into x2−y2x^2 - y^2 yields 27 (or using (x+y)(x−y)=9×3=27(x+y)(x-y) = 9 \times 3 = 27).

Anahtar Kavram

Factoring exponential expressions using prime factorization, radical simplification, and difference of squares.
Soru 96Soru →

How many integer values of xx satisfy the compound absolute value inequality 1≤∣∣x−4∣−3∣≤51 \le ||x - 4| - 3| \le 5?

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Cevap: 15

Cevap

There are 15 integer values of xx that satisfy the given compound inequality.
Solving 1≤∣∣x−4∣−3∣≤51 \le ||x - 4| - 3| \le 5 requires breaking the nested absolute value into its boundary constraints. The upper bound ∥x−4∣−3∣≤5\|x - 4| - 3| \le 5 restricts xx to [−4,12][-4, 12]. The lower bound ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1 requires either ∣x−4∣≥4|x - 4| \ge 4 (giving x≤0x \le 0 or x≥8x \ge 8) or ∣x−4∣≤2|x - 4| \le 2 (giving 2≤x≤62 \le x \le 6). Taking the intersection produces three distinct inclusive integer intervals: [−4,0][-4, 0], [2,6][2, 6], and [8,12][8, 12]. Each interval contains 5 integers, yielding a total of 15 integer solutions.

Adım Adım Çözüm

1
Decompose the double inequality into two separate absolute value inequalities: ∣∣x−4∣−3∣≤5||x - 4| - 3| \le 5 and ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1.
Two simultaneous inequalities to solve for xx.
A double inequality a≤∣u∣≤ba \le |u| \le b requires satisfying both ∣u∣≤b|u| \le b and ∣u∣≥a|u| \ge a.
2
Solve the upper bound inequality ∣∣x−4∣−3∣≤5||x - 4| - 3| \le 5.
−5≤∣x−4∣−3≤5  ⟹  −2≤∣x−4∣≤8-5 \le |x - 4| - 3 \le 5 \implies -2 \le |x - 4| \le 8. Since ∣x−4∣≥0≥−2|x - 4| \ge 0 \ge -2 is always true, this simplifies to ∣x−4∣≤8  ⟹  −4≤x≤12|x - 4| \le 8 \implies -4 \le x \le 12.
Absolute value is non-negative, so the lower bound of −2-2 imposes no extra constraint.
3
Solve the lower bound inequality ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1.
This splits into two cases: ∣x−4∣−3≥1|x - 4| - 3 \ge 1 OR ∣x−4∣−3≤−1|x - 4| - 3 \le -1.
Case A: ∣x−4∣≥4  ⟹  x−4≥4|x - 4| \ge 4 \implies x - 4 \ge 4 or x−4≤−4  ⟹  x≥8x - 4 \le -4 \implies x \ge 8 or x≤0x \le 0.
Case B: ∣x−4∣≤2  ⟹  −2≤x−4≤2  ⟹  2≤x≤6|x - 4| \le 2 \implies -2 \le x - 4 \le 2 \implies 2 \le x \le 6.
The absolute value inequality ∣u∣≥1|u| \ge 1 holds when u≥1u \ge 1 or u≤−1u \le -1.
4
Intersect the solution set from the upper bound [−4,12][-4, 12] with the solution set from the lower bound (−∞,0]∪[2,6]∪[8,∞)(-\infty, 0] \cup [2, 6] \cup [8, \infty).
The valid solution set is x∈[−4,0]∪[2,6]∪[8,12]x \in [-4, 0] \cup [2, 6] \cup [8, 12].
Both conditions must hold simultaneously.
5
Count the integer values in each of the three valid intervals.
Interval [−4,0][-4, 0] has 5 integers: {−4,−3,−2,−1,0}\{-4, -3, -2, -1, 0\}.
Interval [2,6][2, 6] has 5 integers: {2,3,4,5,6}\{2, 3, 4, 5, 6\}.
Interval [8,12][8, 12] has 5 integers: {8,9,10,11,12}\{8, 9, 10, 11, 12\}.
Total integer solutions = 5+5+5=155 + 5 + 5 = 15.
The number of integers in an inclusive integer range [a,b][a, b] is b−a+1b - a + 1.

Anahtar Kavram

Linear Inequalities and Absolute Value
Soru 97Soru →

A company's annual budget of $84,000\$84,000 is split among three departments: Research, Marketing, and Operations. The Marketing department receives 23\frac{2}{3} as much funding as the Research department. The Operations department receives $6,000\$6,000 more than half of the combined funding of the Research and Marketing departments. What is the amount, in dollars, allocated to the Research department?

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Cevap: 31,20031,200

Cevap

31,20031,200
Defining Research funding as xx, Marketing funding becomes 23x\frac{2}{3}x, and Operations funding becomes 12(x+23x)+6000=56x+6000\frac{1}{2}(x + \frac{2}{3}x) + 6000 = \frac{5}{6}x + 6000. Summing all three department allocations yields x+23x+56x+6000=84,000x + \frac{2}{3}x + \frac{5}{6}x + 6000 = 84,000. Combining the variable terms gives 52x+6000=84,000\frac{5}{2}x + 6000 = 84,000, which simplifies to 52x=78,000\frac{5}{2}x = 78,000 and yields x=31,200x = 31,200.

Adım Adım Çözüm

1
Define the unknown variable for the target quantity.
Let xx represent the dollar amount allocated to the Research department.
The problem asks specifically for the Research department allocation.
2
Express the allocations of Marketing and Operations in terms of xx.
Marketing =23x= \frac{2}{3}x. Combined Research and Marketing =x+23x=53x= x + \frac{2}{3}x = \frac{5}{3}x. Operations =12(53x)+6,000=56x+6,000= \frac{1}{2}\left(\frac{5}{3}x\right) + 6,000 = \frac{5}{6}x + 6,000.
Translating word problem relationships into algebraic expressions.
3
Set up the single-variable linear equation for the total budget.
x+23x+(56x+6,000)=84,000x + \frac{2}{3}x + \left(\frac{5}{6}x + 6,000\right) = 84,000
The sum of allocations across all three departments must equal the total budget of $84,000\$84,000.
4
Combine like terms using a common denominator.
66x+46x+56x+6,000=84,000  ⟹  156x+6,000=84,000  ⟹  52x+6,000=84,000\frac{6}{6}x + \frac{4}{6}x + \frac{5}{6}x + 6,000 = 84,000 \implies \frac{15}{6}x + 6,000 = 84,000 \implies \frac{5}{2}x + 6,000 = 84,000
Simplifying fractional coefficients by finding the common denominator 6.
5
Isolate xx to solve the linear equation.
52x=78,000  ⟹  5x=156,000  ⟹  x=31,200\frac{5}{2}x = 78,000 \implies 5x = 156,000 \implies x = 31,200
Subtracting 6,0006,000 from both sides and multiplying by 25\frac{2}{5}.

Anahtar Kavram

Linear Equations in One Variable
Soru 98Soru →

If xx is a real number such that ∣4−3x∣≤13|4 - 3x| \leq 13, and yy is an integer such that −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3, what is the least possible integer value of x2−yx^2 - y?

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Cevap: −7-7

Cevap

The least possible integer value of x2−yx^2 - y is −7-7.
To find the minimum value of x2−yx^2 - y, we must minimize x2x^2 and maximize yy. The absolute value inequality ∣4−3x∣≤13|4 - 3x| \leq 13 simplifies to −3≤x≤173-3 \leq x \leq \frac{17}{3}, which contains 00, so the minimum of x2x^2 is 00. The inequality −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3 simplifies to −4≤y<8-4 \leq y < 8. Since yy is an integer, its maximum value is 77. Therefore, the minimum possible value of x2−yx^2 - y is 0−7=−70 - 7 = -7.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣4−3x∣≤13|4 - 3x| \leq 13 for xx.
−13≤4−3x≤13  ⟹  −17≤−3x≤9  ⟹  173≥x≥−3  ⟹  −3≤x≤173-13 \leq 4 - 3x \leq 13 \implies -17 \leq -3x \leq 9 \implies \frac{17}{3} \geq x \geq -3 \implies -3 \leq x \leq \frac{17}{3}.
An absolute value inequality ∣u∣≤k|u| \leq k expands to −k≤u≤k-k \leq u \leq k. Dividing by a negative number flips the inequality signs.
2
Determine the minimum possible value of x2x^2.
Minimum x2=0x^2 = 0.
Since xx can take any real value in the interval [−3,173][-3, \frac{17}{3}], which contains 00, the minimum square of any real number in this interval is 00 at x=0x = 0.
3
Solve the double inequality −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3 for yy.
−15<1−2y≤9  ⟹  −16<−2y≤8  ⟹  8>y≥−4  ⟹  −4≤y<8-15 < 1 - 2y \leq 9 \implies -16 < -2y \leq 8 \implies 8 > y \geq -4 \implies -4 \leq y < 8.
Multiplying by 33 preserves inequalities, subtracting 11 preserves inequalities, and dividing by −2-2 reverses all inequality signs.
4
Find the maximum integer value of yy.
Maximum integer y=7y = 7.
The solution set for yy is the half-open interval [−4,8)[-4, 8). Since yy is constrained to be an integer, the largest integer strictly less than 88 is 77.
5
Minimize the expression x2−yx^2 - y.
Minimum (x2−y)=Minimum (x2)−Maximum (y)=0−7=−7\text{Minimum } (x^2 - y) = \text{Minimum } (x^2) - \text{Maximum } (y) = 0 - 7 = -7.
To minimize a difference A−BA - B, one must minimize the minuend AA and maximize the subtrahend BB.

Anahtar Kavram

Linear Inequalities and Absolute Value
Tahmini Süre:2m 0s
Soru 99Soru →

For all real numbers x>0x > 0, which of the following expressions are equivalent to (x3/2⋅x3x1/6)2\left(\frac{x^{3/2} \cdot \sqrt[3]{x}}{x^{1/6}}\right)^2? Select all such expressions.

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Cevap: x103\sqrt[3]{x^{10}}; x3x3x^3 \sqrt[3]{x}; (x53)2\left(\sqrt[3]{x^5}\right)^2

Cevap

The expressions equivalent to the given expression are x103\sqrt[3]{x^{10}}, x3x3x^3 \sqrt[3]{x}, and (x53)2\left(\sqrt[3]{x^5}\right)^2.
Simplifying the given expression by converting all radical forms to fractional exponents yields x10/3x^{10/3}. The three expressions x103\sqrt[3]{x^{10}}, x3x3x^3 \sqrt[3]{x}, and (x53)2\left(\sqrt[3]{x^5}\right)^2 each rewrite to x10/3x^{10/3} when evaluated using standard power laws.

Adım Adım Çözüm

1
Convert radical expressions to fractional exponent form inside the parentheses.
x3=x1/3\sqrt[3]{x} = x^{1/3}, so the numerator becomes x3/2⋅x1/3x^{3/2} \cdot x^{1/3}.
Converting all terms to exponent notation allows applying standard exponent addition and subtraction rules.
2
Simplify the numerator by adding exponents.
x3/2+1/3=x9/6+2/6=x11/6x^{3/2 + 1/3} = x^{9/6 + 2/6} = x^{11/6}.
When multiplying exponential terms with the same base, add their exponents using a common denominator.
3
Divide by the denominator by subtracting exponents.
x11/6x1/6=x11/6−1/6=x10/6=x5/3\frac{x^{11/6}}{x^{1/6}} = x^{11/6 - 1/6} = x^{10/6} = x^{5/3}.
When dividing exponential terms with the same base, subtract the denominator's exponent from the numerator's exponent.
4
Apply the outer exponent of 2.
(x5/3)2=x(5/3)⋅2=x10/3\left(x^{5/3}\right)^2 = x^{(5/3) \cdot 2} = x^{10/3}.
When raising a power to another power, multiply the inner and outer exponents.
5
Compare x10/3x^{10/3} to each option.
x103=x10/3\sqrt[3]{x^{10}} = x^{10/3}, x3x3=x3+1/3=x10/3x^3 \sqrt[3]{x} = x^{3 + 1/3} = x^{10/3}, and (x53)2=(x5/3)2=x10/3\left(\sqrt[3]{x^5}\right)^2 = (x^{5/3})^2 = x^{10/3} are all equivalent.
Matching each candidate expression in fractional exponent form identifies all equivalent choices.

Anahtar Kavram

Simplification of algebraic expressions using rules of fractional exponents and radicals
Tahmini Süre:1m 30s
Soru 100Soru →

If −3∣x−2∣+4>−11-3|x - 2| + 4 > -11, which of the following inequalities represents all possible real values of xx?

Cevabı ve açıklamayı göster

Cevap: −3<x<7-3 < x < 7

Cevap

−3<x<7-3 < x < 7
Subtracting 44 from both sides gives −3∣x−2∣>−15-3|x - 2| > -15. Dividing by −3-3 and reversing the inequality sign results in ∣x−2∣<5|x - 2| < 5. Converting to the compound inequality −5<x−2<5-5 < x - 2 < 5 and adding 22 to each part yields the correct interval −3<x<7-3 < x < 7.

Adım Adım Çözüm

1
Subtract 4 from both sides of the inequality
−3∣x−2∣>−15-3|x - 2| > -15
Isolate the absolute value term on the left side.
2
Divide both sides by −3-3 and reverse the inequality sign
∣x−2∣<5|x - 2| < 5
Dividing an inequality by a negative number reverses the direction of the inequality symbol.
3
Express the absolute value inequality as a compound inequality
−5<x−2<5-5 < x - 2 < 5
An inequality of the form ∣u∣<c|u| < c (where c>0c > 0) is equivalent to −c<u<c-c < u < c.
4
Add 2 to all three parts of the compound inequality
−3<x<7-3 < x < 7
Isolate xx to find the complete range of solution values.

Anahtar Kavram

Linear Inequalities and Absolute Value
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Algebra Alıştırma Soruları — GRE General Test — Sayfa 5 | Examkin