Algebra and Functions

215 soru

Soru 81Soru →

A car rental agency charges a fixed daily rate plus a constant fee per mile driven. A customer who rented a car for 3 days and drove 150 miles paid a total of 180.Anothercustomerwhorentedthesamemodelofcarfor5daysanddrove200milespaidatotalof180. Another customer who rented the same model of car for 5 days and drove 200 miles paid a total of 270. What is the fixed daily rate, in dollars, charged by the rental agency?

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Cevap: 30

Cevap

The fixed daily rate charged by the rental agency is 30 dollars.
Formulating the equations 3d+150m=1803d + 150m = 180 and 5d+200m=2705d + 200m = 270 and solving for the daily rate dd yields d=30d = 30.

Adım Adım Çözüm

1
Define variables and formulate the system of linear equations.
Let dd be the fixed daily rate (in dollars) and mm be the cost per mile driven (in dollars). The scenario translates to:
Equation 1: 3d+150m=1803d + 150m = 180
Equation 2: 5d+200m=2705d + 200m = 270
Total charge is the linear combination of daily fixed costs and per-mile variable costs.
2
Simplify Equation 1 to express dd in terms of mm.
Dividing Equation 1 by 3 yields d+50m=60d + 50m = 60, so d=60−50md = 60 - 50m.
Simplifying equations reduces computation error when using substitution.
3
Substitute the expression for dd into Equation 2 to solve for mm.
5(60−50m)+200m=270⇒300−250m+200m=270⇒−50m=−30⇒m=0.605(60 - 50m) + 200m = 270 \Rightarrow 300 - 250m + 200m = 270 \Rightarrow -50m = -30 \Rightarrow m = 0.60
Substitution eliminates variable dd, leaving a linear equation in one variable.
4
Calculate the fixed daily rate dd.
d=60−50(0.60)=60−30=30d = 60 - 50(0.60) = 60 - 30 = 30
Substituting m=0.60m = 0.60 gives the exact daily fixed rate.

Anahtar Kavram

Setting up and solving a system of two linear equations in two variables using elimination or substitution.
Soru 82Soru →

For all positive real numbers xx, the nested radical expression xx\sqrt{x\sqrt{x}} is mathematically equivalent to x34x^{\frac{3}{4}}.

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Cevap: True

Cevap

The statement is true because expressing xx\sqrt{x\sqrt{x}} in fractional exponent form yields x34x^{\frac{3}{4}} for all positive real numbers xx.
The radical expression xx\sqrt{x\sqrt{x}} simplifies to x34x^{\frac{3}{4}} by systematically applying exponent laws: the inner root gives x12x^{\frac{1}{2}}, multiplying by xx gives x32x^{\frac{3}{2}}, and applying the outer root raises x32x^{\frac{3}{2}} to the 12\frac{1}{2} power, producing x34x^{\frac{3}{4}}.

Adım Adım Çözüm

1
Express the inner radical as a fractional exponent.
x=x12\sqrt{x} = x^{\frac{1}{2}}, making the expression under the outer root x⋅x12x \cdot x^{\frac{1}{2}}.
By definition of fractional exponents, amn=amn\sqrt[n]{a^m} = a^{\frac{m}{n}}.
2
Combine the terms inside the outer radical using the product rule for exponents.
x1⋅x12=x1+12=x32x^1 \cdot x^{\frac{1}{2}} = x^{1 + \frac{1}{2}} = x^{\frac{3}{2}}.
When multiplying terms with the same base, add their exponents.
3
Apply the outer square root as an exponent of 12\frac{1}{2} and simplify using the power rule.
x32=(x32)12=x32⋅12=x34\sqrt{x^{\frac{3}{2}}} = (x^{\frac{3}{2}})^{\frac{1}{2}} = x^{\frac{3}{2} \cdot \frac{1}{2}} = x^{\frac{3}{4}}.
Raising a power to another power requires multiplying the exponents.

Anahtar Kavram

Simplification of nested radical expressions using fractional exponent laws
Soru 83Soru →

An electronics manufacturing plant produces two types of circuit modules: Module X and Module Y. Producing one Module X requires 33 minutes on Machine A and 55 minutes on Machine B. Producing one Module Y requires 44 minutes on Machine A and 22 minutes on Machine B. During a certain shift, Machine A was operated for a total of 230230 minutes and Machine B was operated for a total of 220220 minutes. If both machines operated at full capacity with no idle time, how many Module X units were produced?

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Cevap: 30

Cevap

30
Formulating equations from the machine times gives 3x+4y=2303x + 4y = 230 for Machine A and 5x+2y=2205x + 2y = 220 for Machine B. Multiplying the second equation by 22 yields 10x+4y=44010x + 4y = 440. Subtracting 3x+4y=2303x + 4y = 230 from 10x+4y=44010x + 4y = 440 eliminates yy and yields 7x=2107x = 210, which solves to x=30x = 30.

Adım Adım Çözüm

1
Define variables for the unknown quantities.
Let xx be the number of Module X units produced, and let yy be the number of Module Y units produced.
Representing unknown quantities with algebraic variables allows for model formulation.
2
Formulate a system of linear equations from the machine time constraints.
Machine A equation: 3x+4y=2303x + 4y = 230
Machine B equation: 5x+2y=2205x + 2y = 220
The total operational time on each machine equals the sum of times spent producing each module type.
3
Multiply the Machine B equation by 22 to enable elimination of yy.
2×(5x+2y)=2×220  ⟹  10x+4y=4402 \times (5x + 2y) = 2 \times 220 \implies 10x + 4y = 440
Matching the coefficient of yy with the first equation (4y4y) allows elimination via subtraction.
4
Subtract the Machine A equation from the modified Machine B equation and solve for xx.
(10x+4y)−(3x+4y)=440−230  ⟹  7x=210  ⟹  x=30(10x + 4y) - (3x + 4y) = 440 - 230 \implies 7x = 210 \implies x = 30
Subtracting eliminates yy directly, leaving a single linear equation in terms of xx.

Anahtar Kavram

Solving a system of two linear equations in two variables using elimination
Tahmini Süre:1m 30s
Soru 84Soru →

If 3(x−2)+4=2x+113(x - 2) + 4 = 2x + 11, what is the value of xx?

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Cevap: 13

Cevap

The value of xx is 13.
Expanding the left side yields 3x−6+4=3x−23x - 6 + 4 = 3x - 2. Equating this to the right side gives 3x−2=2x+113x - 2 = 2x + 11. Subtracting 2x2x from both sides gives x−2=11x - 2 = 11, and adding 22 to both sides yields x=13x = 13.

Adım Adım Çözüm

1
Expand the left side of the equation
3x−6+4=2x+113x - 6 + 4 = 2x + 11
Apply the distributive property 3(x−2)=3x−63(x - 2) = 3x - 6.
2
Combine like constant terms on the left side
3x−2=2x+113x - 2 = 2x + 11
Combine −6-6 and +4+4 to get −2-2.
3
Subtract 2x2x from both sides
x−2=11x - 2 = 11
Move variable terms to one side of the equation.
4
Add 2 to both sides
x=13x = 13
Isolate the variable xx.

Anahtar Kavram

Solving Linear Equations in One Variable
Soru 85Soru →

If ∣2x−3∣=7|2x - 3| = 7, which of the following is a possible value of xx?

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Cevap: 55

Cevap

The value 55 is a valid solution to the given absolute value equation.
The absolute value equation ∣2x−3∣=7|2x - 3| = 7 splits into two linear equations: 2x−3=72x - 3 = 7 and 2x−3=−72x - 3 = -7. Solving the first equation gives 2x=102x = 10, so x=5x = 5. Solving the second gives 2x=−42x = -4, so x=−2x = -2. Among the listed options, 55 is the correct solution.

Adım Adım Çözüm

1
Set up the two cases for the absolute value equation ∣2x−3∣=7|2x - 3| = 7.
Case 1: 2x−3=72x - 3 = 7; Case 2: 2x−3=−72x - 3 = -7.
An absolute value expression ∣u∣=c|u| = c (where c≥0c \geq 0) resolves to u=cu = c or u=−cu = -c.
2
Solve Case 1: 2x−3=72x - 3 = 7.
2x=10  ⟹  x=52x = 10 \implies x = 5.
Add 3 to both sides and divide by 2.
3
Solve Case 2: 2x−3=−72x - 3 = -7.
2x=−4  ⟹  x=−22x = -4 \implies x = -2.
Add 3 to both sides and divide by 2.
4
Compare the solutions (x=5x = 5 and x=−2x = -2) with the given choices.
The value 55 is present among the options.
Identifies the correct choice matching one of the calculated solutions.

Anahtar Kavram

Solving Linear Absolute Value Equations
Tahmini Süre:45s
Soru 86Soru →

For all real numbers aa and bb, the custom operation ⋆\star is defined by a⋆b=(a+b)2−3ba \star b = (a + b)^2 - 3b. What is the value of 4⋆24 \star 2?

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Cevap: 30

Cevap

30
Substituting a=4a = 4 and b=2b = 2 into the definition (a+b)2−3b(a + b)^2 - 3b gives (4+2)2−3(2)=62−6=36−6=30(4 + 2)^2 - 3(2) = 6^2 - 6 = 36 - 6 = 30.

Adım Adım Çözüm

1
Identify the values assigned to the custom operator variables
a=4a = 4 and b=2b = 2
The expression 4⋆24 \star 2 corresponds to evaluating a⋆ba \star b where aa is the first operand and bb is the second operand.
2
Substitute a=4a = 4 and b=2b = 2 into the operator definition (a+b)2−3b(a + b)^2 - 3b
(4+2)2−3(2)(4 + 2)^2 - 3(2)
Direct substitution into the given algebraic definition.
3
Perform the operations in parentheses and exponentiation
62−6=36−66^2 - 6 = 36 - 6
First add inside the parentheses (4+2=64 + 2 = 6), then square the sum (62=366^2 = 36), and multiply 3×2=63 \times 2 = 6.
4
Subtract to find the final value
30
Subtracting 66 from 3636 yields 3030.

Anahtar Kavram

Custom Operator Evaluation
Tahmini Süre:45s
Soru 87Soru →

Company A charges a one-time fixed fee of $20\$20 plus $5\$5 per hour to rent a piece of machinery. Company B charges a one-time fixed fee of $35\$35 plus $2\$2 per hour to rent the same machinery. For how many hours of rental will the total cost charged by both companies be equal?

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Cevap: 5

Cevap

The total cost at both companies will be equal for a rental duration of 5 hours.
The cost charged by Company A is modeled by 20+5h20 + 5h and Company B by 35+2h35 + 2h. Setting these equal yields 20+5h=35+2h20 + 5h = 35 + 2h, which simplifies to 3h=153h = 15, giving h=5h = 5. Thus, 5 hours is the correct rental duration.

Adım Adım Çözüm

1
Define the variable and write total cost equations for each company.
Let hh represent the number of hours. Company A's cost is CA=20+5hC_A = 20 + 5h and Company B's cost is CB=35+2hC_B = 35 + 2h.
Linear cost models combine a fixed base cost with a variable hourly component.
2
Set the two cost equations equal to solve for hh.
20+5h=35+2h20 + 5h = 35 + 2h
The total costs are equal when CA=CBC_A = C_B.
3
Isolate the variable term hh on one side of the equation.
5h−2h=35−20⇒3h=15⇒h=55h - 2h = 35 - 20 \Rightarrow 3h = 15 \Rightarrow h = 5
Subtracting 2h2h and 2020 from both sides simplifies the equation to solve directly for hh.

Anahtar Kavram

Setting up and solving single-variable linear equations from real-world cost contexts.
Tahmini Süre:45s
Soru 88Soru →

If the quadratic equation x2−kx+36=0x^2 - kx + 36 = 0 has two distinct positive integer roots, r1r_1 and r2r_2, and r1+2r_1 + 2 and r2+2r_2 + 2 are the roots of the quadratic equation x2−19x+m=0x^2 - 19x + m = 0, what is the value of mm?

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Cevap: 70

Cevap

The value of mm is 70.
By applying Vieta's formulas to both quadratic equations, we find that the sum of the original roots is r1+r2=15r_1 + r_2 = 15 and their product is r1r2=36r_1 r_2 = 36. Factoring x2−15x+36=(x−3)(x−12)=0x^2 - 15x + 36 = (x-3)(x-12) = 0 confirms the original roots are 33 and 1212. The new roots are 3+2=53+2=5 and 12+2=1412+2=14. Their product m=5×14=70m = 5 \times 14 = 70.

Adım Adım Çözüm

1
Apply Vieta's relations to the first quadratic equation
r1+r2=kr_1 + r_2 = k and r1⋅r2=36r_1 \cdot r_2 = 36
For any quadratic equation x2+bx+c=0x^2 + bx + c = 0, the sum of roots is −b-b and the product of roots is cc.
2
Apply Vieta's relations to the second quadratic equation
(r1+2)+(r2+2)=19  ⟹  r1+r2=15(r_1 + 2) + (r_2 + 2) = 19 \implies r_1 + r_2 = 15
The coefficient of xx in x2−19x+m=0x^2 - 19x + m = 0 dictates that the sum of its roots equals 1919.
3
Determine the roots r1r_1 and r2r_2 and check integer constraints
r1=3r_1 = 3 and r2=12r_2 = 12
The integer factors of 3636 that sum to 1515 are 33 and 1212, satisfying all conditions.
4
Compute the constant term mm for the transformed equation
m=(3+2)(12+2)=5×14=70m = (3 + 2)(12 + 2) = 5 \times 14 = 70
The constant term mm equals the product of the transformed roots (r1+2)(r_1 + 2) and (r2+2)(r_2 + 2).

Anahtar Kavram

Vieta's Formulas and Quadratic Root Transformations
Tahmini Süre:2m 0s
Soru 89Soru →

For all non-zero real numbers aa and bb, the custom operator Δ\Delta is defined by aΔb=a2−b2aba \Delta b = \frac{a^2 - b^2}{ab}. A function f(x)f(x) is defined for all x≠0x \neq 0 and x≠−1x \neq -1 by f(x)=xΔ(x+1)f(x) = x \Delta (x + 1). If kk is a positive constant such that f(k)=−712f(k) = -\frac{7}{12}, what is the value of (k+2)Δk(k + 2) \Delta k?

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Cevap: 1615\frac{16}{15}

Cevap

The value of (k+2)Δk(k + 2) \Delta k is 1615\frac{16}{15}.
Applying the custom operator to f(x)f(x) yields f(x)=x2−(x+1)2x(x+1)=−2x−1x2+xf(x) = \frac{x^2 - (x+1)^2}{x(x+1)} = \frac{-2x - 1}{x^2 + x}. Setting f(k)=−712f(k) = -\frac{7}{12} gives 2k+1k2+k=712\frac{2k + 1}{k^2 + k} = \frac{7}{12}, which simplifies to 7k2−17k−12=07k^2 - 17k - 12 = 0. Factoring (7k+4)(k−3)=0(7k + 4)(k - 3) = 0 and selecting the positive root gives k=3k = 3. Evaluating (3+2)Δ3=5Δ3=52−325⋅3=1615(3+2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \cdot 3} = \frac{16}{15}.

Adım Adım Çözüm

1
Express f(x)f(x) using the definition of the custom operator Δ\Delta.
f(x)=xΔ(x+1)=x2−(x+1)2x(x+1)=x2−(x2+2x+1)x(x+1)=−2x−1x2+xf(x) = x \Delta (x + 1) = \frac{x^2 - (x + 1)^2}{x(x + 1)} = \frac{x^2 - (x^2 + 2x + 1)}{x(x + 1)} = \frac{-2x - 1}{x^2 + x}
Substitute a=xa = x and b=x+1b = x + 1 into aΔb=a2−b2aba \Delta b = \frac{a^2 - b^2}{ab}.
2
Set f(k)=−712f(k) = -\frac{7}{12} and solve for the positive constant kk.
−2k−1k2+k=−712  ⟹  2k+1k2+k=712  ⟹  12(2k+1)=7(k2+k)  ⟹  7k2−17k−12=0\frac{-2k - 1}{k^2 + k} = -\frac{7}{12} \implies \frac{2k + 1}{k^2 + k} = \frac{7}{12} \implies 12(2k + 1) = 7(k^2 + k) \implies 7k^2 - 17k - 12 = 0
Cross-multiply and rearrange into a standard quadratic equation.
3
Factor the quadratic equation to find the positive root kk.
(7k+4)(k−3)=0  ⟹  k=−47(7k + 4)(k - 3) = 0 \implies k = -\frac{4}{7} or k=3k = 3. Since k>0k > 0, k=3k = 3.
The question specifies that kk is a positive constant.
4
Evaluate (k+2)Δk(k + 2) \Delta k for k=3k = 3.
(3+2)Δ3=5Δ3=52−325×3=25−915=1615(3 + 2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \times 3} = \frac{25 - 9}{15} = \frac{16}{15}
Substitute a=5a = 5 and b=3b = 3 into the custom operator definition.

Anahtar Kavram

Function definition substitution, non-commutative custom operators, and quadratic solving.
Tahmini Süre:2m 30s
Soru 90Soru →

For all real numbers xx and yy, the custom operator Δ\Delta is defined by xΔy=x2−2yx \Delta y = x^2 - 2y. If kk is a real number such that (3Δk)Δ4=17(3 \Delta k) \Delta 4 = 17, what is the sum of all possible values of kk?

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Cevap: 9

Cevap

The sum of all possible values of kk is 9.
Applying the custom operation rule xΔy=x2−2yx \Delta y = x^2 - 2y sequentially, the inner evaluation gives 3Δk=9−2k3 \Delta k = 9 - 2k. Applying the operation again to (9−2k)Δ4(9 - 2k) \Delta 4 gives (9−2k)2−8=17(9 - 2k)^2 - 8 = 17, which simplifies to (9−2k)2=25(9 - 2k)^2 = 25. Taking the square root gives two valid solutions for 9−2k9 - 2k: 55 and −5-5. Solving 9−2k=59 - 2k = 5 yields k=2k = 2, and solving 9−2k=−59 - 2k = -5 yields k=7k = 7. The sum of these two values is 2+7=92 + 7 = 9.

Adım Adım Çözüm

1
Evaluate the inner custom operation 3Δk3 \Delta k.
3Δk=32−2k=9−2k3 \Delta k = 3^2 - 2k = 9 - 2k
Apply the definition xΔy=x2−2yx \Delta y = x^2 - 2y with x=3x = 3 and y=ky = k.
2
Substitute the inner result into the outer expression (9−2k)Δ4=17(9 - 2k) \Delta 4 = 17.
(9−2k)2−2(4)=17  ⟹  (9−2k)2−8=17(9 - 2k)^2 - 2(4) = 17 \implies (9 - 2k)^2 - 8 = 17
Apply the definition xΔy=x2−2yx \Delta y = x^2 - 2y with x=9−2kx = 9 - 2k and y=4y = 4.
3
Isolate the squared term and solve for 9−2k9 - 2k.
(9−2k)2=25  ⟹  9−2k=5(9 - 2k)^2 = 25 \implies 9 - 2k = 5 or 9−2k=−59 - 2k = -5
Adding 8 to both sides gives 25; taking the square root requires considering both positive and negative roots.
4
Solve each linear equation for kk and calculate their sum.
Case 1: 9−2k=5  ⟹  2k=4  ⟹  k=29 - 2k = 5 \implies 2k = 4 \implies k = 2.
Case 2: 9−2k=−5  ⟹  2k=14  ⟹  k=79 - 2k = -5 \implies 2k = 14 \implies k = 7.
Sum = 2+7=92 + 7 = 9.
Solving both equations yields all possible values for kk.

Anahtar Kavram

Custom Operator Evaluation and Quadratic Equation Root Extraction
Tahmini Süre:2m 0s
Soru 91Soru →

For all real numbers pp and qq, the operation ∇\nabla is defined by p∇q=2p+3q−pqp \nabla q = 2p + 3q - pq. What is the value of 3∇43 \nabla 4?

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Cevap: 6

Cevap

The value of 3∇43 \nabla 4 is 6.
Substituting p=3p = 3 and q=4q = 4 directly into the expression 2p+3q−pq2p + 3q - pq gives 2(3)+3(4)−(3)(4)=6+12−12=62(3) + 3(4) - (3)(4) = 6 + 12 - 12 = 6.

Adım Adım Çözüm

1
Identify the assigned values for the operator inputs.
p=3p = 3 and q=4q = 4.
The expression 3∇43 \nabla 4 matches p∇qp \nabla q with pp as the first argument and qq as the second.
2
Substitute p=3p = 3 and q=4q = 4 into the custom operation formula 2p+3q−pq2p + 3q - pq.
2(3)+3(4)−(3)(4)2(3) + 3(4) - (3)(4)
Direct evaluation of the defined function rule.
3
Perform the multiplication and algebraic addition/subtraction.
6+12−12=66 + 12 - 12 = 6
Simplifying numerical operations gives the final result.

Anahtar Kavram

Custom Operator Evaluation
Soru 92Soru →

Let SS be the set of all real solutions to the polynomial equation (x2−5x+6)(x2−7x+12)=(x−3)2(x−2)(x^2 - 5x + 6)(x^2 - 7x + 12) = (x - 3)^2(x - 2). What is the sum of all elements in set SS?

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Cevap: 10

Cevap

The sum of all distinct real elements in set SS is 10.
The correct answer is obtained by completely factoring both quadratic expressions on the left side into (x−2)(x−3)2(x−4)(x - 2)(x - 3)^2(x - 4). Moving all terms to one side gives (x−2)(x−3)2(x−4)−(x−3)2(x−2)=0(x - 2)(x - 3)^2(x - 4) - (x - 3)^2(x - 2) = 0. Factoring out (x−2)(x−3)2(x - 2)(x - 3)^2 yields (x−2)(x−3)2(x−5)=0(x - 2)(x - 3)^2(x - 5) = 0. The distinct solutions are 2, 3, and 5. Their sum is 10.

Adım Adım Çözüm

1
Factor the quadratic expressions on the left side of the equation.
x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3) and x2−7x+12=(x−3)(x−4)x^2 - 7x + 12 = (x - 3)(x - 4). Substituting these into the left side yields (x−2)(x−3)2(x−4)(x - 2)(x - 3)^2(x - 4).
Factoring simplifies higher-degree polynomial expressions into linear factors.
2
Set up the equation with all terms on one side to avoid dividing by variable expressions.
(x−2)(x−3)2(x−4)−(x−3)2(x−2)=0(x - 2)(x - 3)^2(x - 4) - (x - 3)^2(x - 2) = 0.
Dividing by variable expressions can cause the loss of valid roots.
3
Factor out the common polynomial factor (x−2)(x−3)2(x - 2)(x - 3)^2.
(x−2)(x−3)2[(x−4)−1]=0(x - 2)(x - 3)^2 [(x - 4) - 1] = 0, which simplifies to (x−2)(x−3)2(x−5)=0(x - 2)(x - 3)^2(x - 5) = 0.
Factoring allows finding all roots by setting each factor equal to zero.
4
Identify the distinct real solutions forming set SS and compute their sum.
The roots are x=2x = 2, x=3x = 3, and x=5x = 5. Thus, S={2,3,5}S = \{2, 3, 5\}. The sum of elements in SS is 2+3+5=102 + 3 + 5 = 10.
A set contains only unique elements, so repeated roots are counted only once.

Anahtar Kavram

Factoring polynomial equations completely without dividing by variable expressions
Tahmini Süre:2m 0s
Soru 93Soru →

What is the sum of all real solutions to the equation x(x−3)=10x(x - 3) = 10?

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Cevap: 3

Cevap

The sum of all real solutions to the equation is 3.
Expanding the equation x(x−3)=10x(x - 3) = 10 yields x2−3x−10=0x^2 - 3x - 10 = 0. Factoring this expression gives (x−5)(x+2)=0(x - 5)(x + 2) = 0, yielding solutions x=5x = 5 and x=−2x = -2. The sum of these two real solutions is 5+(−2)=35 + (-2) = 3. Alternatively, applying Vieta's formula directly to x2−3x−10=0x^2 - 3x - 10 = 0 gives the sum of roots as −ba=−−31=3-\frac{b}{a} = -\frac{-3}{1} = 3.

Adım Adım Çözüm

1
Expand the left side of the equation.
x2−3x=10x^2 - 3x = 10
Multiply xx across the binomial (x−3)(x - 3) to convert to standard polynomial form.
2
Rearrange the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−3x−10=0x^2 - 3x - 10 = 0
Subtract 10 from both sides so the equation equals zero.
3
Factor the quadratic expression.
(x−5)(x+2)=0(x - 5)(x + 2) = 0
Find two numbers that multiply to −10-10 and add up to −3-3, which are −5-5 and 22.
4
Solve for the roots of the equation.
x=5x = 5 or x=−2x = -2
Apply the zero-product property: set each factor equal to zero.
5
Calculate the sum of the solutions.
Sum =5+(−2)=3= 5 + (-2) = 3
Add the two real solutions together. Alternatively, by Vieta's formulas, the sum of roots is −ba=−−31=3-\frac{b}{a} = -\frac{-3}{1} = 3.

Anahtar Kavram

Quadratic Equations and Polynomial Factoring
Tahmini Süre:1m 0s
Soru 94Soru →

For all real numbers xx and yy, the custom operation ♢\diamondsuit is defined by x♢y=3x−2y+1x \diamondsuit y = 3x - 2y + 1. What is the value of 5♢45 \diamondsuit 4?

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Cevap: 8

Cevap

The value of 5♢45 \diamondsuit 4 is 8.
Substituting x=5x = 5 and y=4y = 4 into the operation x♢y=3x−2y+1x \diamondsuit y = 3x - 2y + 1 yields 3(5)−2(4)+1=15−8+1=83(5) - 2(4) + 1 = 15 - 8 + 1 = 8.

Adım Adım Çözüm

1
Substitute the operands into the custom operator rule
5 \diamondsuit 4 = 3(5) - 2(4) + 1
The first term 55 corresponds to xx and the second term 44 corresponds to yy in the formula.
2
Simplify the numeric expression
15 - 8 + 1 = 8
Multiply first, then perform subtraction and addition from left to right.

Anahtar Kavram

Function Definitions, Evaluation, and Custom Operators
Soru 95Soru →

For all non-zero real numbers xx, the function ff is defined by f(x)=x2−42xf(x) = \frac{x^2 - 4}{2x}, and the function gg is defined by g(x)=3x−1g(x) = 3x - 1. What is the positive value of aa for which f(g(a))=0f(g(a)) = 0?

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Cevap: 1

Cevap

The positive value of aa for which f(g(a))=0f(g(a)) = 0 is 1.
To solve f(g(a))=0f(g(a)) = 0, first find the values of yy where f(y)=0f(y) = 0. The numerator of f(y)=y2−42yf(y) = \frac{y^2 - 4}{2y} equals zero when y2−4=0y^2 - 4 = 0, yielding y=2y = 2 and y=−2y = -2. Setting g(a)=3a−1g(a) = 3a - 1 equal to these roots gives 3a−1=2  ⟹  a=13a - 1 = 2 \implies a = 1 and 3a−1=−2  ⟹  a=−1/33a - 1 = -2 \implies a = -1/3. Since aa must be positive, the correct value is 1.

Adım Adım Çözüm

1
Determine the values of the argument yy that satisfy f(y)=0f(y) = 0.
y=2y = 2 or y=−2y = -2.
A fraction equals zero when its numerator is zero and its denominator is non-zero. Setting y2−4=0y^2 - 4 = 0 gives y=±2y = \pm 2.
2
Substitute g(a)=3a−1g(a) = 3a - 1 into yy to solve for aa.
Solving 3a−1=23a - 1 = 2 yields a=1a = 1; solving 3a−1=−23a - 1 = -2 yields a=−1/3a = -1/3.
Setting the expression for g(a)g(a) equal to each root of f(y)=0f(y) = 0 identifies all potential values for aa.
3
Select the value of aa matching the positivity condition.
a=1a = 1.
The question explicitly specifies finding the positive value of aa.

Anahtar Kavram

Function composition and evaluation of nested functional equations.
Soru 96Soru →

For all non-zero real numbers aa and bb, the binary operator ⋄\diamond is defined by a⋄b=a−baa \diamond b = a - \frac{b}{a}. If kk is a positive integer such that (k⋄3)⋄3=12(k \diamond 3) \diamond 3 = \frac{1}{2}, what is the value of kk?

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Cevap: 3

Cevap

The positive integer value of kk is 3.
Applying the non-commutative custom operator definition a⋄b=a−baa \diamond b = a - \frac{b}{a} to the nested expression (k⋄3)⋄3=12(k \diamond 3) \diamond 3 = \frac{1}{2} requires working from the inside out. Letting u=k⋄3=k−3ku = k \diamond 3 = k - \frac{3}{k}, the outer equation becomes u⋄3=u−3u=12u \diamond 3 = u - \frac{3}{u} = \frac{1}{2}. Multiplying through by 2u2u gives 2u2−u−6=02u^2 - u - 6 = 0, which factors as (2u+3)(u−2)=0(2u + 3)(u - 2) = 0, yielding u=2u = 2 or u=−32u = -\frac{3}{2}. Substituting u=2u = 2 back into k−3k=2k - \frac{3}{k} = 2 produces k2−2k−3=0k^2 - 2k - 3 = 0, which factors as (k−3)(k+1)=0(k - 3)(k + 1) = 0. Since kk must be a positive integer, k=3k = 3. Testing u=−32u = -\frac{3}{2} gives 2k2+3k−6=02k^2 + 3k - 6 = 0, which has no integer solutions.

Adım Adım Çözüm

1
Substitute a temporary variable for the inner custom operator expression.
Let u=k⋄3=k−3ku = k \diamond 3 = k - \frac{3}{k}.
Simplifies the nested expression (k⋄3)⋄3=12(k \diamond 3) \diamond 3 = \frac{1}{2} to a single variable operation u⋄3=12u \diamond 3 = \frac{1}{2}.
2
Apply the operator definition to u⋄3u \diamond 3 and solve the resulting quadratic equation for uu.
u−3u=12  ⟹  2u2−u−6=0  ⟹  (2u+3)(u−2)=0u - \frac{3}{u} = \frac{1}{2} \implies 2u^2 - u - 6 = 0 \implies (2u + 3)(u - 2) = 0, so u=2u = 2 or u=−32u = -\frac{3}{2}.
Clearing the denominator uu transforms the custom operator equation into a standard quadratic equation.
3
Solve for positive integer kk using the obtained values of uu.
Setting k−3k=2k - \frac{3}{k} = 2 yields k2−2k−3=0  ⟹  (k−3)(k+1)=0k^2 - 2k - 3 = 0 \implies (k - 3)(k + 1) = 0. Since kk must be a positive integer, k=3k = 3. Setting k−3k=−32k - \frac{3}{k} = -\frac{3}{2} yields 2k2+3k−6=02k^2 + 3k - 6 = 0, which has non-integer discriminant 57\sqrt{57}.
Only u=2u = 2 yields an integer solution for kk, satisfying all problem constraints.

Anahtar Kavram

Custom Operator Evaluation and Nested Functional Equations
Soru 97Soru →

For all non-zero real numbers xx, the function ff satisfies the equation 2f(x)+f(1x)=6x2f(x) + f\left(\frac{1}{x}\right) = 6x. What is the value of f(2)f(2)?

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Cevap: 7

Cevap

The value of f(2)f(2) is 7.
Substituting x=2x = 2 into the functional equation gives 2f(2)+f(12)=122f(2) + f\left(\frac{1}{2}\right) = 12. Substituting x=12x = \frac{1}{2} gives 2f(12)+f(2)=32f\left(\frac{1}{2}\right) + f(2) = 3. Multiplying the first equation by 2 and subtracting the second equation eliminates f(12)f\left(\frac{1}{2}\right), producing 3f(2)=213f(2) = 21, so f(2)=7f(2) = 7.

Adım Adım Çözüm

1
Substitute x=2x = 2 into the functional equation
2f(2)+f(12)=122f(2) + f\left(\frac{1}{2}\right) = 12
This establishes a linear relation between f(2)f(2) and f(12)f\left(\frac{1}{2}\right).
2
Substitute x=12x = \frac{1}{2} into the functional equation
2f(12)+f(2)=32f\left(\frac{1}{2}\right) + f(2) = 3
This establishes a second linear relation involving the reciprocal input.
3
Eliminate f(12)f\left(\frac{1}{2}\right) by solving the system of equations
f(2)=7f(2) = 7
Multiplying the first equation by 2 yields 4f(2)+2f(12)=244f(2) + 2f\left(\frac{1}{2}\right) = 24. Subtracting the second equation gives 3f(2)=213f(2) = 21, which simplifies to f(2)=7f(2) = 7.

Anahtar Kavram

Evaluating functional equations by reciprocal variable substitution to form and solve a linear system.
Tahmini Süre:2m 0s
Soru 98Soru →

If 2 is a root of the quadratic equation x2−kx+24=0x^2 - kx + 24 = 0, where kk is a constant, and the quadratic equation x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0 has exactly one real solution, what is the value of mm?

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Cevap: 64

Cevap

The value of mm is 64.
Substituting x=2x = 2 into x2−kx+24=0x^2 - kx + 24 = 0 gives 4−2k+24=04 - 2k + 24 = 0, which simplifies to 2k=282k = 28 or k=14k = 14. Substituting k=14k = 14 into x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0 produces x2−16x+m=0x^2 - 16x + m = 0. For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 to have exactly one real solution, its discriminant b2−4acb^2 - 4ac must equal 0. Therefore, (−16)2−4(1)(m)=0(-16)^2 - 4(1)(m) = 0, which means 256−4m=0256 - 4m = 0, giving m=64m = 64.

Adım Adım Çözüm

1
Substitute the known root x=2x = 2 into the equation x2−kx+24=0x^2 - kx + 24 = 0.
k=14k = 14
Since x=2x = 2 is a root of the quadratic equation, evaluating the expression at x=2x = 2 must equal zero.
2
Substitute k=14k = 14 into the second quadratic equation x2−(k+2)x+m=0x^2 - (k + 2)x + m = 0.
x2−16x+m=0x^2 - 16x + m = 0
This determines the linear coefficient of the second quadratic equation.
3
Set the discriminant of x2−16x+m=0x^2 - 16x + m = 0 to zero.
m=64m = 64
A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has exactly one real solution if and only if its discriminant b2−4acb^2 - 4ac equals zero.

Anahtar Kavram

Solving quadratic equations via root substitution and applying the discriminant condition for repeated roots.
Tahmini Süre:1m 30s
Soru 99Soru →

For all real numbers xx except 00 and 11, the function f(x)f(x) satisfies the relation f(x)+2f(11−x)=9xf(x) + 2f\left(\frac{1}{1-x}\right) = 9x. What is the value of f(2)f(2)?

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Cevap: 6

Cevap

The value of f(2)f(2) is 66.
Evaluating the functional equation f(x)+2f(11−x)=9xf(x) + 2f\left(\frac{1}{1-x}\right) = 9x at the values x=2x = 2, x=−1x = -1, and x=12x = \frac{1}{2} creates a system of three linear equations in terms of f(2)f(2), f(−1)f(-1), and f(12)f\left(\frac{1}{2}\right). Solving this system yields 9f(2)=549f(2) = 54, which simplifies directly to f(2)=6f(2) = 6.

Adım Adım Çözüm

1
Evaluate the functional equation at x=2x = 2
f(2)+2f(11−2)=9(2)  ⟹  f(2)+2f(−1)=18f(2) + 2f\left(\frac{1}{1-2}\right) = 9(2) \implies f(2) + 2f(-1) = 18
Applying the input x=2x = 2 creates an equation connecting f(2)f(2) and f(−1)f(-1).
2
Evaluate the functional equation at x=−1x = -1
f(−1)+2f(11−(−1))=9(−1)  ⟹  f(−1)+2f(12)=−9f(-1) + 2f\left(\frac{1}{1-(-1)}\right) = 9(-1) \implies f(-1) + 2f\left(\frac{1}{2}\right) = -9
Evaluating at the new input x=−1x = -1 generates a second equation connecting f(−1)f(-1) and f(12)f\left(\frac{1}{2}\right).
3
Evaluate the functional equation at x=12x = \frac{1}{2}
f(12)+2f(11−1/2)=9(12)  ⟹  f(12)+2f(2)=4.5f\left(\frac{1}{2}\right) + 2f\left(\frac{1}{1-1/2}\right) = 9\left(\frac{1}{2}\right) \implies f\left(\frac{1}{2}\right) + 2f(2) = 4.5
Evaluating at x=12x = \frac{1}{2} completes the cyclic chain by linking back to f(2)f(2).
4
Solve the system of three linear equations for f(2)f(2)
From equation 3, f(12)=4.5−2f(2)f\left(\frac{1}{2}\right) = 4.5 - 2f(2). Substituting into equation 2 yields f(−1)=4f(2)−18f(-1) = 4f(2) - 18. Substituting into equation 1 gives f(2)+2(4f(2)−18)=18  ⟹  9f(2)=54  ⟹  f(2)=6f(2) + 2(4f(2) - 18) = 18 \implies 9f(2) = 54 \implies f(2) = 6.
Eliminating f(12)f\left(\frac{1}{2}\right) and f(−1)f(-1) isolated f(2)f(2) to find its exact numerical value.

Anahtar Kavram

Cyclic Functional Equations and Substitution Systems
Soru 100Soru →

If 2x2+5x−3=02x^2 + 5x - 3 = 0 and x>0x > 0, what is the value of xx?

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Cevap: 0.5

Cevap

The value of xx is 0.5.
Factoring the quadratic equation 2x2+5x−3=02x^2 + 5x - 3 = 0 yields (2x−1)(x+3)=0(2x - 1)(x + 3) = 0. Setting each linear factor equal to zero gives two possible solutions for xx: x=0.5x = 0.5 and x=−3x = -3. Because the problem stipulates that x>0x > 0, the negative solution is discarded, leaving x=0.5x = 0.5.

Adım Adım Çözüm

1
Factor the quadratic expression
(2x−1)(x+3)=0(2x - 1)(x + 3) = 0
Splitting the middle term 5x5x into 6x−x6x - x allows grouping to factor by grouping.
2
Find the roots of the equation
x=0.5x = 0.5 or x=−3x = -3
By the zero-product property, if the product of two factors is zero, at least one factor must be zero.
3
Apply the positivity constraint x>0x > 0
x=0.5x = 0.5
The root x=−3x = -3 violates the given condition that xx must be strictly greater than zero.

Anahtar Kavram

Solving quadratic equations by factoring and applying domain constraints
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Algebra and Functions Alıştırma Soruları — GMAT — Sayfa 5 | Examkin