Algebra and Functions

215 questions

Question 101Question →

For all real numbers aa and bb such that a+b≠0a + b \neq 0, the custom operator ⋆\star is defined by a⋆b=a−ba+ba \star b = \frac{a - b}{a + b}. What is the value of (3⋆1)⋆2(3 \star 1) \star 2?

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Answer: −35-\frac{3}{5}

Answer

−35-\frac{3}{5}
Evaluating the expression step-by-step according to parentheses yields 3⋆1=3−13+1=123 \star 1 = \frac{3-1}{3+1} = \frac{1}{2}. Substituting 12\frac{1}{2} as the first input and 22 as the second input gives 1/2−21/2+2=−3/25/2=−35\frac{1/2 - 2}{1/2 + 2} = \frac{-3/2}{5/2} = -\frac{3}{5}.

Step-by-Step Solution

1
Evaluate the inner custom operation inside parentheses: 3⋆13 \star 1.
3⋆1=3−13+1=24=123 \star 1 = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}.
Follow the order of operations by resolving the grouped expression first using the definition a⋆b=a−ba+ba \star b = \frac{a - b}{a + b} with a=3a = 3 and b=1b = 1.
2
Substitute the result 12\frac{1}{2} back into the main expression to compute (12)⋆2(\frac{1}{2}) \star 2.
(12)⋆2=12−212+2(\frac{1}{2}) \star 2 = \frac{\frac{1}{2} - 2}{\frac{1}{2} + 2}.
Apply the definition of the custom operator again, where the left operand is 12\frac{1}{2} and the right operand is 22.
3
Simplify the complex fraction.
\frac{\frac{1}{2} - \frac{4}{2}}{\frac{1}{2} + \frac{4}{2}} = \frac{-\frac{3}{2}}{\frac{5}{2}} = -\frac{3}{5}.
Combine the fractions in the numerator and denominator, then divide.

Key Concept

Custom Operators and Order of Operations
Estimated Time:1m 30s
Question 102Question →

The quadratic equation x2−6x+a=0x^2 - 6x + a = 0 has two distinct real roots α\alpha and γ\gamma, and the quadratic equation y2−by+21=0y^2 - by + 21 = 0 has two distinct real roots β\beta and δ\delta. If α<β<γ<δ\alpha < \beta < \gamma < \delta and the four roots form an arithmetic progression in that order, what is the value of a+ba + b?

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Answer: 15

Answer

The value of a+ba + b is 15.
By representing the four ordered roots as r,r+d,r+2d,r+3dr, r+d, r+2d, r+3d, Vieta's formula for the sum of roots of the first equation yields r+(r+2d)=2(r+d)=6r + (r+2d) = 2(r+d) = 6, which implies r+d=3r+d = 3. Thus, the second root is β=3\beta = 3. Using the product of roots for the second equation, 3δ=213\delta = 21 gives δ=7\delta = 7. The common difference is d=(7−3)/2=2d = (7-3)/2 = 2, which gives the roots 1,3,5,71, 3, 5, 7. Finally, a=1×5=5a = 1 \times 5 = 5 and b=3+7=10b = 3 + 7 = 10, so a+b=15a + b = 15.

Step-by-Step Solution

1
Apply Vieta's formulas to both quadratic equations.
α+γ=6\alpha + \gamma = 6, αγ=a\alpha\gamma = a, β+δ=b\beta + \delta = b, and βδ=21\beta\delta = 21.
Vieta's relations connect the coefficients of a quadratic polynomial to the sum and product of its roots.
2
Set up the arithmetic progression representation for the roots.
α=r\alpha = r, β=r+d\beta = r + d, γ=r+2d\gamma = r + 2d, and δ=r+3d\delta = r + 3d, where d>0d > 0.
The roots form an increasing arithmetic progression in the order α,β,γ,δ\alpha, \beta, \gamma, \delta.
3
Substitute the expressions for α\alpha and γ\gamma into α+γ=6\alpha + \gamma = 6.
r+(r+2d)=2r+2d=2(r+d)=6  ⟹  r+d=3r + (r + 2d) = 2r + 2d = 2(r + d) = 6 \implies r + d = 3.
Combining terms simplifies the sum of the first and third terms of the arithmetic progression.
4
Identify the value of β\beta and solve for δ\delta.
β=r+d=3\beta = r + d = 3, so βδ=21  ⟹  3δ=21  ⟹  δ=7\beta\delta = 21 \implies 3\delta = 21 \implies \delta = 7.
Since β=r+d\beta = r + d, its value is directly determined as 3, allowing δ\delta to be solved from the product relation.
5
Calculate the common difference dd and the first term rr.
δ−β=2d=7−3=4  ⟹  d=2\delta - \beta = 2d = 7 - 3 = 4 \implies d = 2, and r=3−2=1r = 3 - 2 = 1.
The difference between the fourth and second terms of an AP is equal to 2d2d.
6
Find aa, bb, and their sum a+ba + b.
a=αγ=1×5=5a = \alpha\gamma = 1 \times 5 = 5, b=β+δ=3+7=10b = \beta + \delta = 3 + 7 = 10, so a+b=5+10=15a + b = 5 + 10 = 15.
With all four roots determined (1,3,5,71, 3, 5, 7), the missing coefficients are calculated using Vieta's formulas.

Key Concept

Combining Vieta's Formulas with Arithmetic Progressions to Solve Quadratic Systems
Question 103Question →

For any real numbers uu and vv, the binary operator ⋆\star is defined by u⋆v=2u2−3vu \star v = 2u^2 - 3v. A function ff is defined by f(x)=x⋆4f(x) = x \star 4, and a function gg is defined by g(x)=3x+1g(x) = 3x + 1. If mm is a real number such that f(g(m))=38f(g(m)) = 38, what is the product of all possible values of mm?

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Answer: −83-\frac{8}{3}

Answer

The product of all possible values of mm is −83-\frac{8}{3}.
First, evaluate f(x)=x⋆4=2x2−3(4)=2x2−12f(x) = x \star 4 = 2x^2 - 3(4) = 2x^2 - 12. Next, substitute g(m)=3m+1g(m) = 3m + 1 into f(x)f(x) to get f(g(m))=2(3m+1)2−12f(g(m)) = 2(3m + 1)^2 - 12. Setting this expression equal to 3838 gives 2(3m+1)2−12=382(3m + 1)^2 - 12 = 38, which simplifies to (3m+1)2=25(3m + 1)^2 = 25. Taking the square root gives two possible linear equations: 3m+1=53m + 1 = 5 (which gives m=43m = \frac{4}{3}) and 3m+1=−53m + 1 = -5 (which gives m=−2m = -2). Multiplying these two solutions yields (43)×(−2)=−83\left(\frac{4}{3}\right) \times (-2) = -\frac{8}{3}. Thus, the option equal to −83-\frac{8}{3} is correct.

Step-by-Step Solution

1
Evaluate the function f(x)f(x) using the definition of the custom operator ⋆\star.
f(x)=x⋆4=2x2−3(4)=2x2−12f(x) = x \star 4 = 2x^2 - 3(4) = 2x^2 - 12
Substitute u=xu = x and v=4v = 4 into the formula u⋆v=2u2−3vu \star v = 2u^2 - 3v.
2
Express the nested function f(g(m))f(g(m)) in terms of mm.
f(g(m))=2(g(m))2−12=2(3m+1)2−12f(g(m)) = 2(g(m))^2 - 12 = 2(3m + 1)^2 - 12
Substitute g(m)=3m+1g(m) = 3m + 1 into f(x)f(x).
3
Set f(g(m))f(g(m)) equal to 3838 and solve for (3m+1)2(3m + 1)^2.
2(3m+1)2−12=38  ⟹  2(3m+1)2=50  ⟹  (3m+1)2=252(3m + 1)^2 - 12 = 38 \implies 2(3m + 1)^2 = 50 \implies (3m + 1)^2 = 25
Isolate the squared binomial term using basic algebraic manipulation.
4
Take the square root of both sides to find all possible values of mm.
3m+1=5  ⟹  m=433m + 1 = 5 \implies m = \frac{4}{3} or 3m+1=−5  ⟹  m=−23m + 1 = -5 \implies m = -2
A positive real number has both positive and negative square roots.
5
Calculate the product of the two solutions for mm.
(43)×(−2)=−83\left(\frac{4}{3}\right) \times (-2) = -\frac{8}{3}
Multiply the two roots together as requested by the question stem.

Key Concept

Evaluating custom binary operators and nested composite functions, solving quadratic equations, and finding products of roots.
Question 104Question →

If (x+3)2=25(x + 3)^2 = 25 and x>0x > 0, what is the value of x2−1x^2 - 1?

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Answer: 3

Answer

3
Taking the square root of both sides of (x+3)2=25(x + 3)^2 = 25 gives x+3=±5x + 3 = \pm 5. Since x>0x > 0, we choose x+3=5x + 3 = 5, which gives x=2x = 2. Substituting x=2x = 2 into x2−1x^2 - 1 produces 22−1=32^2 - 1 = 3.

Step-by-Step Solution

1
Take the square root of both sides of the given quadratic equation
x+3=5x + 3 = 5 or x+3=−5x + 3 = -5
The equation (x+3)2=25(x + 3)^2 = 25 implies x+3x + 3 can be either positive or negative 5.
2
Solve for xx under the constraint x>0x > 0
x=2x = 2
From x+3=5x + 3 = 5, we obtain x=2x = 2. The second root x=−8x = -8 is discarded because x>0x > 0.
3
Substitute x=2x = 2 into the target expression x2−1x^2 - 1
22−1=32^2 - 1 = 3
Evaluating x2−1x^2 - 1 at x=2x = 2 yields 4−1=34 - 1 = 3.

Key Concept

Solving Quadratic Equations by Taking Square Roots
Estimated Time:1m 0s
Question 105Question →

For all non-zero real numbers xx and yy, the custom operation ⋆\star is defined by x⋆y=xy−yxx \star y = \frac{x}{y} - \frac{y}{x}. If pp and qq are positive numbers such that p+q=5p + q = 5 and p⋆q=56p \star q = \frac{5}{6}, what is the value of (p+1)⋆q(p + 1) \star q?

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Answer: 32\frac{3}{2}

Answer

32\frac{3}{2}
By simplifying p⋆q=p2−q2pq=(p−q)(p+q)pqp \star q = \frac{p^2 - q^2}{pq} = \frac{(p - q)(p + q)}{pq}, we substitute p+q=5p + q = 5 to get 5(p−q)pq=56\frac{5(p - q)}{pq} = \frac{5}{6}, which reduces to 6(p−q)=pq6(p - q) = pq. Substituting q=5−pq = 5 - p yields the quadratic equation p2+7p−30=0p^2 + 7p - 30 = 0. Since p>0p > 0, we find p=3p = 3 and q=2q = 2. Evaluating (3+1)⋆2=4⋆2=42−24=32(3 + 1) \star 2 = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}.

Step-by-Step Solution

1
Express the custom operation using algebraic fractions and factor the numerator
p⋆q=pq−qp=p2−q2pq=(p−q)(p+q)pqp \star q = \frac{p}{q} - \frac{q}{p} = \frac{p^2 - q^2}{pq} = \frac{(p - q)(p + q)}{pq}
Finding a common denominator converts the expression into a form where known identities can be substituted.
2
Substitute p+q=5p + q = 5 and p⋆q=56p \star q = \frac{5}{6} into the simplified expression
5(p−q)pq=56  ⟹  p−qpq=16  ⟹  6(p−q)=pq\frac{5(p - q)}{pq} = \frac{5}{6} \implies \frac{p - q}{pq} = \frac{1}{6} \implies 6(p - q) = pq
Simplifying by dividing both sides by 5 establishes a direct relation between the difference and product of p and q.
3
Substitute q=5−pq = 5 - p to form and solve a quadratic equation for pp
6(2p−5)=p(5−p)  ⟹  12p−30=5p−p2  ⟹  p2+7p−30=0  ⟹  (p+10)(p−3)=06(2p - 5) = p(5 - p) \implies 12p - 30 = 5p - p^2 \implies p^2 + 7p - 30 = 0 \implies (p + 10)(p - 3) = 0
Since p and q are positive numbers, p must be positive, giving p = 3 and q = 2.
4
Evaluate (p+1)⋆q(p + 1) \star q with p=3p = 3 and q=2q = 2
(3+1)⋆2=4⋆2=42−24=2−12=32(3 + 1) \star 2 = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = 2 - \frac{1}{2} = \frac{3}{2}
Applying the custom operator definition to inputs 4 and 2 yields the final value.

Key Concept

Custom Operator Evaluation and Quadratic System Solving
Question 106Question →
What is the sum of all distinct real solutions to the equation x2−8x+15x−3=(x−5)2\frac{x^2 - 8x + 15}{x - 3} = (x - 5)^2?
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Answer: 11

Answer

11
Factoring the numerator x2−8x+15x^2 - 8x + 15 gives (x−3)(x−5)(x - 3)(x - 5). For all x≠3x \neq 3, the original equation simplifies to x−5=(x−5)2x - 5 = (x - 5)^2. Rearranging into standard factored form (x−5)(x−6)=0(x - 5)(x - 6) = 0 yields two distinct real solutions: x=5x = 5 and x=6x = 6. Neither value violates the restriction x≠3x \neq 3. Adding these values together yields 5+6=115 + 6 = 11.

Step-by-Step Solution

1
Determine the domain restriction for the rational expression
The expression x2−8x+15x−3\frac{x^2 - 8x + 15}{x - 3} requires that the denominator x−3≠0x - 3 \neq 0, so x≠3x \neq 3.
Division by zero is undefined in real numbers.
2
Factor the numerator and simplify the left side of the equation
Since x2−8x+15=(x−3)(x−5)x^2 - 8x + 15 = (x - 3)(x - 5), for x≠3x \neq 3, (x−3)(x−5)x−3=x−5\frac{(x - 3)(x - 5)}{x - 3} = x - 5.
Canceling the common non-zero factor (x−3)(x - 3) simplifies the rational equation.
3
Solve the simplified quadratic equation for xx
x−5=(x−5)2  ⟹  (x−5)2−(x−5)=0  ⟹  (x−5)(x−5−1)=0  ⟹  (x−5)(x−6)=0x - 5 = (x - 5)^2 \implies (x - 5)^2 - (x - 5) = 0 \implies (x - 5)(x - 5 - 1) = 0 \implies (x - 5)(x - 6) = 0. Thus, x=5x = 5 or x=6x = 6.
Factoring out the common factor (x−5)(x - 5) preserves all valid solutions.
4
Verify solutions against the domain restriction and calculate the sum
Both x=5x = 5 and x=6x = 6 satisfy x≠3x \neq 3. The sum of all distinct real solutions is 5+6=115 + 6 = 11.
Combining the valid distinct roots gives the required total.

Key Concept

Polynomial factoring, domain restrictions on rational expressions, and avoiding root loss when solving quadratic equations.
Estimated Time:2m 0s
Question 107Question →

A rectangular garden has an area of 4848 square meters. If the length of the garden is 22 meters greater than its width, what is the perimeter of the garden, in meters?

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Answer: 28

Answer

The perimeter of the garden is 28 meters.
By setting up the area equation w(w+2)=48w(w + 2) = 48, we obtain the quadratic equation w2+2w−48=0w^2 + 2w - 48 = 0. Factoring gives (w+8)(w−6)=0(w + 8)(w - 6) = 0. Since width must be positive, w=6w = 6 meters. The length is 6+2=86 + 2 = 8 meters. Thus, the perimeter is 2(6+8)=282(6 + 8) = 28 meters.

Step-by-Step Solution

1
Define variables for width and length.
Width = ww, Length = w+2w + 2.
The length is given as 2 meters greater than the width.
2
Formulate and rearrange the quadratic equation for area.
w2+2w−48=0w^2 + 2w - 48 = 0
Area is length multiplied by width, set equal to 48.
3
Factor the quadratic polynomial.
(w+8)(w−6)=0(w + 8)(w - 6) = 0
Find two numbers that multiply to -48 and add up to +2.
4
Determine the valid physical dimension.
w=6w = 6 meters and length l=8l = 8 meters.
A physical dimension cannot be negative, so w=−8w = -8 is discarded.
5
Compute the perimeter.
Perimeter = 2(6+8)=282(6 + 8) = 28 meters.
Perimeter of a rectangle is twice the sum of its length and width.

Key Concept

Solving quadratic equations by factoring in word problem contexts.
Question 108Question →

For any real number xx, the function hh is defined by h(x)=3x2−5x+4h(x) = 3x^2 - 5x + 4. What is the value of h(3)h(3)?

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Answer: 16

Answer

16
To find h(3)h(3), substitute x=3x = 3 into the function h(x)=3x2−5x+4h(x) = 3x^2 - 5x + 4. First, compute 32=93^2 = 9. Multiplying by 3 yields 27. Subtracting 5(3)=155(3) = 15 gives 12, and adding 4 results in the final value of 16.

Step-by-Step Solution

1
Substitute the value x=3x = 3 into the definition of h(x)h(x).
h(3)=3(3)2−5(3)+4h(3) = 3(3)^2 - 5(3) + 4
To evaluate a function at a specific value, replace all instances of the input variable with that value.
2
Evaluate the exponent and carry out the multiplications.
h(3)=3(9)−15+4=27−15+4h(3) = 3(9) - 15 + 4 = 27 - 15 + 4
Follow standard order of operations: compute powers before performing multiplication.
3
Perform the remaining addition and subtraction.
h(3)=16h(3) = 16
Complete the calculation from left to right.

Key Concept

Function Evaluation
Question 109Question →

If x2−9x+18=0x^2 - 9x + 18 = 0, and x1x_1 and x2x_2 are the real roots of the equation such that x1>x2x_1 > x_2, what is the value of 2x1−x22x_1 - x_2?

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Answer: 9

Answer

The value of 2x1−x22x_1 - x_2 is 9.
Factoring the quadratic equation x2−9x+18=0x^2 - 9x + 18 = 0 yields (x−6)(x−3)=0(x - 6)(x - 3) = 0. The roots are x=6x = 6 and x=3x = 3. Given that x1>x2x_1 > x_2, we assign x1=6x_1 = 6 and x2=3x_2 = 3. Evaluating 2x1−x22x_1 - x_2 gives 2(6)−3=92(6) - 3 = 9.

Step-by-Step Solution

1
Factor the quadratic expression
(x−6)(x−3)=0(x - 6)(x - 3) = 0
Find two numbers that multiply to 1818 and add to −9-9, which are −6-6 and −3-3.
2
Solve for the roots of the equation
x=6x = 6 or x=3x = 3
Set each factor equal to zero: x−6=0  ⟹  x=6x - 6 = 0 \implies x = 6 and x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3.
3
Assign root values based on the condition x1>x2x_1 > x_2
x1=6x_1 = 6 and x2=3x_2 = 3
Since 6>36 > 3, the larger root is assigned to x1x_1 and the smaller root to x2x_2.
4
Evaluate the target expression
9
Substitute x1=6x_1 = 6 and x2=3x_2 = 3 into 2x1−x2=2(6)−3=12−3=92x_1 - x_2 = 2(6) - 3 = 12 - 3 = 9.

Key Concept

Solving Quadratic Equations by Factoring
Question 110Question →

For all real numbers xx and yy, the custom operation Δ\Delta is defined by xΔy=(x+1)(y−2)x \Delta y = (x + 1)(y - 2). If the function gg is defined by g(n)=n2−3g(n) = n^2 - 3 for all real numbers nn, what is the value of g(3)Δg(2)g(3) \Delta g(2)?

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Answer: -7

Answer

The value of g(3)Δg(2)g(3) \Delta g(2) is −7-7.
Evaluating g(3)g(3) gives 32−3=63^2 - 3 = 6, and evaluating g(2)g(2) gives 22−3=12^2 - 3 = 1. Substituting these values as x=6x = 6 and y=1y = 1 into the custom operator definition xΔy=(x+1)(y−2)x \Delta y = (x + 1)(y - 2) yields (6+1)(1−2)=7×(−1)=−7(6 + 1)(1 - 2) = 7 \times (-1) = -7.

Step-by-Step Solution

1
Evaluate the inner function g(n)g(n) at n=3n = 3
g(3)=32−3=9−3=6g(3) = 3^2 - 3 = 9 - 3 = 6
Before evaluating the custom operator, the inputs g(3)g(3) and g(2)g(2) must be explicitly calculated.
2
Evaluate the inner function g(n)g(n) at n=2n = 2
g(2)=22−3=4−3=1g(2) = 2^2 - 3 = 4 - 3 = 1
Determine the second operand for the custom operation.
3
Apply the custom operation Δ\Delta to the results from Step 1 and Step 2
6Δ1=(6+1)(1−2)=7×(−1)=−76 \Delta 1 = (6 + 1)(1 - 2) = 7 \times (-1) = -7
Substitute x=6x = 6 and y=1y = 1 into the formula xΔy=(x+1)(y−2)x \Delta y = (x + 1)(y - 2).

Key Concept

Evaluating functions and applying non-commutative custom operators
Question 111Question →

For how many integer values of kk does the quadratic equation (k−2)x2−2(k+1)x+(k+3)=0(k - 2)x^2 - 2(k + 1)x + (k + 3) = 0 have two distinct real roots α\alpha and \beta suchthat such that \frac{1}{\alpha} + \frac{1}{\beta} < 1$?

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Answer: 3

Answer

There are exactly 3 integer values of kk that satisfy all given conditions.
To find the number of valid integer values for kk, we apply the condition for distinct real roots (D>0D > 0), which yields k>−7k > -7 with k≠2k \neq 2. Expressing the sum of reciprocals as α+βαβ=2(k+1)k+3<1\frac{\alpha + \beta}{\alpha\beta} = \frac{2(k+1)}{k+3} < 1, solving k−1k+3<0\frac{k-1}{k+3} < 0 yields −3<k<1-3 < k < 1. The integer values within this range are −2,−1,-2, -1, and 00, giving a total of 3 valid integer values.

Step-by-Step Solution

1
Ensure the given equation is a valid quadratic equation and determine the condition for two distinct real roots.
For the equation to be quadratic, the coefficient of x2x^2 must be non-zero, so k−2≠0  ⟹  k≠2k - 2 \neq 0 \implies k \neq 2. For distinct real roots, the discriminant D>0D > 0. D=[−2(k+1)]2−4(k−2)(k+3)=4(k2+2k+1)−4(k2+k−6)=4(k+7)D = [-2(k+1)]^2 - 4(k-2)(k+3) = 4(k^2 + 2k + 1) - 4(k^2 + k - 6) = 4(k + 7). Thus, 4(k+7)>0  ⟹  k>−74(k+7) > 0 \implies k > -7.
A quadratic equation has two distinct real roots if and only if its discriminant is strictly positive and the leading coefficient is non-zero.
2
Apply Vieta's formulas to express the sum and product of the roots in terms of kk.
\alpha + \beta = \frac{2(k+1)}{k-2}and and \alpha\beta = \frac{k+3}{k-2}.Notethatfor. Note that for \frac{1}{\alpha} + \frac{1}{\beta}toexist,neitherrootcanbe0,whichrequires to exist, neither root can be 0, which requires \alpha\beta
eq 0 \implies k
eq -3$.
Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.
3
Simplify the condition 1α+1β<1\frac{1}{\alpha} + \frac{1}{\beta} < 1 using Vieta's relations.
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{2(k+1)}{k-2}}{\frac{k+3}{k-2}} = \frac{2(k+1)}{k+3}.Weneed. We need \frac{2k+2}{k+3} < 1$.
Combining the fractions over a common denominator allows direct substitution of the root sum and product.
4
Solve the algebraic inequality 2k+2k+3<1\frac{2k+2}{k+3} < 1.
\frac{2k+2}{k+3} - 1 < 0 \implies \frac{(2k+2) - (k+3)}{k+3} < 0 \implies \frac{k-1}{k+3} < 0 .Thefractionisnegativewhenthenumeratoranddenominatorhaveoppositesigns,whichgivestheinterval. The fraction is negative when the numerator and denominator have opposite signs, which gives the interval -3 < k < 1$.
Subtracting 1 to get 0 on one side prevents sign errors that occur when cross-multiplying variable expressions.
5
Combine all constraints on kk and count the valid integer values.
Constraints on kk: (1) k>−7k > -7, (2) k≠2k \neq 2, (3) k≠−3k \neq -3, and (4) −3<k<1-3 < k < 1. Intersecting these gives −3<k<1-3 < k < 1. The integer values in this open interval are k=−2,−1,0k = -2, -1, 0. Total count = 3.
All structural and domain restrictions must be satisfied simultaneously.

Key Concept

Quadratic Equations and Polynomial Factoring
Estimated Time:3m 0s
Question 112Question →

If x=1x = 1 is a root of the cubic polynomial P(x)=x3+ax2−11x+12P(x) = x^3 + ax^2 - 11x + 12, where aa is a constant, what is the value of the largest real root of P(x)P(x)?

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Answer: 4

Answer

The largest real root of the polynomial P(x)P(x) is 44.
Evaluating P(1)=0P(1) = 0 yields 1+a−11+12=01 + a - 11 + 12 = 0, giving a=−2a = -2. Factoring the polynomial x3−2x2−11x+12x^3 - 2x^2 - 11x + 12 by (x−1)(x - 1) produces the quadratic expression x2−x−12x^2 - x - 12. Factoring this quadratic gives (x−4)(x+3)=0(x - 4)(x + 3) = 0. The roots of the polynomial are −3-3, 11, and 44. Comparing these values, the maximum root is 44.

Step-by-Step Solution

1
Find the constant aa using the given root x=1x = 1.
a=−2a = -2
Since x=1x = 1 is a root, P(1)=13+a(1)2−11(1)+12=0P(1) = 1^3 + a(1)^2 - 11(1) + 12 = 0, which simplifies to a+2=0a + 2 = 0.
2
Factor (x−1)(x - 1) out of P(x)=x3−2x2−11x+12P(x) = x^3 - 2x^2 - 11x + 12.
P(x)=(x−1)(x2−x−12)P(x) = (x - 1)(x^2 - x - 12)
Dividing x3−2x2−11x+12x^3 - 2x^2 - 11x + 12 by (x−1)(x - 1) yields the quotient x2−x−12x^2 - x - 12.
3
Factor the quadratic factor x2−x−12x^2 - x - 12.
(x−4)(x+3)(x - 4)(x + 3)
The quadratic factors into two linear terms whose roots are 44 and −3-3.
4
Determine the largest of all real roots of P(x)P(x).
44
The complete set of real roots is {−3,1,4}\{-3, 1, 4\}, of which 44 is the greatest.

Key Concept

Polynomial Factoring and Remainder/Factor Theorem
Question 113Question →

A specialty coffee shop sells two types of coffee beans: Type A for $8\$8 per pound and Type B for $12\$12 per pound. A customer buys a total of xx pounds of coffee, consisting of aa pounds of Type A and bb pounds of Type B, for a total cost of CC dollars. Which of the following equations correctly model the relationship between these quantities? Select all that apply.

Select all that apply

Show answer & explanation

Answer: a=x−ba = x - b; C=12x−4aC = 12x - 4a; b=C−8a12b = \frac{C - 8a}{12}

Answer

The correct formulations are a=x−ba = x - b, C=12x−4aC = 12x - 4a, and b=C−8a12b = \frac{C - 8a}{12}.
The system is defined by two fundamental linear relationships: weight total a+b=xa + b = x and cost total 8a+12b=C8a + 12b = C. Isolating aa from the weight total gives a=x−ba = x - b. Substituting b=x−ab = x - a into the cost equation yields C=8a+12(x−a)=12x−4aC = 8a + 12(x - a) = 12x - 4a. Isolating bb from the cost equation gives 12b=C−8a  ⟹  b=C−8a1212b = C - 8a \implies b = \frac{C - 8a}{12}. All three expressions are mathematically equivalent and correct.

Step-by-Step Solution

1
Set up the basic linear system from the word problem.
Total weight equation: a+b=xa + b = x; Total cost equation: 8a+12b=C8a + 12b = C.
Word problems with total quantity and total monetary value translate into two distinct linear equations.
2
Rearrange the total weight equation to express aa in terms of xx and bb.
a=x−ba = x - b.
Subtracting bb from both sides isolates aa.
3
Substitute b=x−ab = x - a into the total cost equation.
C=8a+12(x−a)=8a+12x−12a=12x−4aC = 8a + 12(x - a) = 8a + 12x - 12a = 12x - 4a.
Expressing cost in terms of a single variable aa and total weight xx simplifies multi-variable models.
4
Isolate bb in the total cost equation.
12b=C−8a  ⟹  b=C−8a1212b = C - 8a \implies b = \frac{C - 8a}{12}.
Subtracting the cost contribution of Type A (8a8a) and dividing by the unit cost of Type B (1212) isolates bb.

Key Concept

Linear Equation Modeling in Word Problems
Question 114Question →

If x2(2x−3)=9(2x−3)x^2(2x - 3) = 9(2x - 3), what is the sum of all distinct real solutions for xx?

Show answer & explanation

Answer: 1.51.5

Answer

The sum of all distinct real solutions for xx is 1.51.5.
To find all solutions without losing roots, set the equation to zero: x2(2x−3)−9(2x−3)=0x^2(2x - 3) - 9(2x - 3) = 0. Factoring out (2x−3)(2x - 3) gives (x2−9)(2x−3)=0(x^2 - 9)(2x - 3) = 0, which fully factors into (x−3)(x+3)(2x−3)=0(x - 3)(x + 3)(2x - 3) = 0. The distinct real roots are 33, −3-3, and 1.51.5. Adding these values together yields 3+(−3)+1.5=1.53 + (-3) + 1.5 = 1.5.

Step-by-Step Solution

1
Move all terms to one side of the equation to avoid losing solutions by dividing by a variable expression.
x2(2x−3)−9(2x−3)=0x^2(2x - 3) - 9(2x - 3) = 0
Dividing by an expression containing variables can eliminate valid roots.
2
Factor out the common term (2x−3)(2x - 3).
(x2−9)(2x−3)=0(x^2 - 9)(2x - 3) = 0
Factoring allows application of the zero-product property.
3
Factor the quadratic term x2−9x^2 - 9 as a difference of squares.
(x−3)(x+3)(2x−3)=0(x - 3)(x + 3)(2x - 3) = 0
a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
4
Set each factor to zero to find all real solutions.
x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3, x+3=0  ⟹  x=−3x + 3 = 0 \implies x = -3, and 2x−3=0  ⟹  x=1.52x - 3 = 0 \implies x = 1.5.
If the product of factors is zero, at least one factor must equal zero.
5
Sum all distinct real solutions.
3+(−3)+1.5=1.53 + (-3) + 1.5 = 1.5
The question asks for the sum of all distinct real solutions.

Key Concept

Factoring Polynomial Equations and Avoiding Variable Cancellation
Estimated Time:1m 30s
Question 115Question →

For all non-zero real numbers aa and bb, the custom binary operator ⋆\star is defined by a⋆b=a2−b2aba \star b = \frac{a^2 - b^2}{ab}. A function ff is defined for all positive numbers xx by f(x)=(x+1)⋆xf(x) = (x + 1) \star x. What is the value of f(3)⋆f(2)f(3) \star f(2)?

Show answer & explanation

Answer: −5170-\frac{51}{70}

Answer

−5170-\frac{51}{70}
First, evaluate the inner function values: f(3)=4⋆3=42−324⋅3=712f(3) = 4 \star 3 = \frac{4^2 - 3^2}{4 \cdot 3} = \frac{7}{12} and f(2)=3⋆2=32−223⋅2=56=1012f(2) = 3 \star 2 = \frac{3^2 - 2^2}{3 \cdot 2} = \frac{5}{6} = \frac{10}{12}. Next, evaluate the outer operation (712)⋆(1012)=(7/12)2−(10/12)2(7/12)(10/12)=49−10070=−5170\left(\frac{7}{12}\right) \star \left(\frac{10}{12}\right) = \frac{(7/12)^2 - (10/12)^2}{(7/12)(10/12)} = \frac{49 - 100}{70} = -\frac{51}{70}. Hence, the correct response is −5170-\frac{51}{70}.

Step-by-Step Solution

1
Evaluate f(3)f(3) using the definition of f(x)f(x) and the custom operator ⋆\star.
f(3)=(3+1)⋆3=4⋆3=42−324×3=16−912=712f(3) = (3 + 1) \star 3 = 4 \star 3 = \frac{4^2 - 3^2}{4 \times 3} = \frac{16 - 9}{12} = \frac{7}{12}
By definition, f(x)=(x+1)⋆xf(x) = (x+1) \star x. Setting x=3x=3 yields 4⋆34 \star 3.
2
Evaluate f(2)f(2) using the definition of f(x)f(x) and the custom operator ⋆\star.
f(2)=(2+1)⋆2=3⋆2=32−223×2=9−46=56f(2) = (2 + 1) \star 2 = 3 \star 2 = \frac{3^2 - 2^2}{3 \times 2} = \frac{9 - 4}{6} = \frac{5}{6}
Setting x=2x=2 in f(x)f(x) yields 3⋆23 \star 2.
3
Evaluate f(3)⋆f(2)=(712)⋆(56)f(3) \star f(2) = \left(\frac{7}{12}\right) \star \left(\frac{5}{6}\right).
(712)⋆(1012)=(712)2−(1012)2(712)(1012)=49−10014470144=−5170\left(\frac{7}{12}\right) \star \left(\frac{10}{12}\right) = \frac{\left(\frac{7}{12}\right)^2 - \left(\frac{10}{12}\right)^2}{\left(\frac{7}{12}\right)\left(\frac{10}{12}\right)} = \frac{\frac{49 - 100}{144}}{\frac{70}{144}} = -\frac{51}{70}
Apply the rule u⋆v=u2−v2uvu \star v = \frac{u^2 - v^2}{uv} with u=712u = \frac{7}{12} and v=1012v = \frac{10}{12}.

Key Concept

Custom non-commutative binary operators and function evaluation
Estimated Time:2m 0s
Question 116Question →

Working together at their respective constant rates, Machine X and Machine Y can complete a specialized production order in 1212 hours. If Machine X works alone for 44 hours and Machine Y works alone for 99 hours, they complete 712\frac{7}{12} of the total order. Machine Z operates at a constant rate that is 50%50\% greater than the rate of Machine X. How many hours would it take Machine Y and Machine Z working together to complete the entire order?

Show answer & explanation

Answer: 1010

Answer

1010 hours
The correct answer is 1010 hours. Solving the system of equations yields individual rates of 130\frac{1}{30} order/hour for Machine X and 120\frac{1}{20} order/hour for Machine Y. A 50%50\% rate increase gives Machine Z a rate of 120\frac{1}{20} order/hour. Adding the rates of Machine Y and Machine Z yields a combined rate of 110\frac{1}{10} order/hour, which requires 1010 hours to complete one job.

Step-by-Step Solution

1
Define variables and model the joint rate equation.
Let xx be the rate of Machine X (order/hour) and yy be the rate of Machine Y (order/hour). Their combined rate equation is x+y=112x + y = \frac{1}{12}.
Working together for 1212 hours completes 11 full order.
2
Set up the second equation based on the partial work completed.
4x+9y=7124x + 9y = \frac{7}{12}.
Machine X works for 44 hours and Machine Y works for 99 hours to complete 712\frac{7}{12} of the order.
3
Solve the system of linear equations for xx and yy.
Express 4x+9y4x + 9y as 4(x+y)+5y=7124(x + y) + 5y = \frac{7}{12}. Substituting x+y=112x + y = \frac{1}{12} gives 4(112)+5y=712  ⟹  412+5y=712  ⟹  5y=312=14  ⟹  y=1204\left(\frac{1}{12}\right) + 5y = \frac{7}{12} \implies \frac{4}{12} + 5y = \frac{7}{12} \implies 5y = \frac{3}{12} = \frac{1}{4} \implies y = \frac{1}{20}. Subsequently, x=112−120=5−360=260=130x = \frac{1}{12} - \frac{1}{20} = \frac{5 - 3}{60} = \frac{2}{60} = \frac{1}{30}.
Determines the individual work rates of Machine X and Machine Y.
4
Calculate the work rate of Machine Z.
Rate of Machine Z = 1.5×x=1.5×130=32×130=1201.5 \times x = 1.5 \times \frac{1}{30} = \frac{3}{2} \times \frac{1}{30} = \frac{1}{20}.
Machine Z is 50%50\% faster than Machine X.
5
Calculate the combined rate and total time for Machine Y and Machine Z.
Combined rate = y+z=120+120=220=110y + z = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10}. Total time = 1110=10\frac{1}{\frac{1}{10}} = 10 hours.
The inverse of the combined rate gives the total time required to finish one job.

Key Concept

Work Rates and Simultaneous Linear Modeling
Estimated Time:2m 30s
Question 117Question →

For all real numbers aa and bb, the custom binary operation ⋆\star is defined by a⋆b=a2−2b2+aba \star b = a^2 - 2b^2 + ab. The function gg is defined by g(x)=x⋆3g(x) = x \star 3. What is the positive value of xx for which g(x)=10g(x) = 10?

Show answer & explanation

Answer: 4

Answer

The positive value of xx for which g(x)=10g(x) = 10 is 44.
Evaluating g(x)=x⋆3g(x) = x \star 3 gives x2−2(3)2+x(3)=x2+3x−18x^2 - 2(3)^2 + x(3) = x^2 + 3x - 18. Setting this expression equal to 1010 produces the quadratic equation x2+3x−28=0x^2 + 3x - 28 = 0. Factoring yields (x+7)(x−4)=0(x + 7)(x - 4) = 0, giving solutions x=−7x = -7 and x=4x = 4. Since xx must be positive, the correct value is 44.

Step-by-Step Solution

1
Apply the definition of the custom binary operator ⋆\star to find g(x)g(x).
g(x)=x⋆3=x2−2(3)2+x(3)=x2−18+3x=x2+3x−18g(x) = x \star 3 = x^2 - 2(3)^2 + x(3) = x^2 - 18 + 3x = x^2 + 3x - 18
Substitute a=xa = x and b=3b = 3 into the given formula a⋆b=a2−2b2+aba \star b = a^2 - 2b^2 + ab.
2
Set the function expression equal to 1010 and rearrange into standard quadratic form.
x2+3x−18=10  ⟹  x2+3x−28=0x^2 + 3x - 18 = 10 \implies x^2 + 3x - 28 = 0
Subtract 1010 from both sides to form a quadratic equation equal to zero.
3
Factor the quadratic equation to find the values of xx.
(x+7)(x−4)=0  ⟹  x=−7 or x=4(x + 7)(x - 4) = 0 \implies x = -7 \text{ or } x = 4
Find two numbers that multiply to −28-28 and add to 33, which are 77 and −4-4.
4
Select the positive value of xx as specified in the stem.
x=4x = 4
The question asks specifically for the positive value of xx.

Key Concept

Evaluating non-commutative custom operators and solving polynomial equations resulting from function evaluations.
Estimated Time:2m 0s
Question 118Question →

A local theater sold a total of 120120 tickets for a weekend performance, consisting only of adult tickets and student tickets. Adult tickets were sold for $15\$15 each, and student tickets were sold for $10\$10 each. If the total revenue from the ticket sales was $1500\$1{}500, how many adult tickets were sold?

Show answer & explanation

Answer: 60

Answer

The correct answer is 60 adult tickets.
By setting up the total ticket equation a+s=120a + s = 120 and total revenue equation 15a+10s=150015a + 10s = 1500, substituting s=120−as = 120 - a gives 15a+10(120−a)=150015a + 10(120 - a) = 1500. Simplifying yields 5a=3005a = 300, so a=60a = 60. Thus, 60 adult tickets were sold.

Step-by-Step Solution

1
Set up the linear equation in one variable.
15a+10(120−a)=150015a + 10(120 - a) = 1500
Express the number of student tickets as (120−a)(120 - a) so the revenue equation relies on a single variable.
2
Expand and combine like terms.
5a+1200=15005a + 1200 = 1500
Distribute 1010 across (120−a)(120 - a) to simplify the equation.
3
Isolate the variable aa.
a=60a = 60
Subtract 12001200 from both sides and divide by 55 to find the number of adult tickets.

Key Concept

Algebraic Modeling of Linear Systems in Two Variables
Estimated Time:1m 0s
Question 119Question →

How many integer values of xx satisfy the inequality ∣x−3∣≤4|x - 3| \leq 4?

Show answer & explanation

Answer: 9

Answer

There are 9 integer values of xx that satisfy the inequality.
Unfolding ∣x−3∣≤4|x - 3| \leq 4 yields −4≤x−3≤4-4 \leq x - 3 \leq 4. Adding 3 across all terms gives −1≤x≤7-1 \leq x \leq 7. The integer values satisfying this range are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, which totals 9 values.

Step-by-Step Solution

1
Convert the absolute value inequality into a compound linear inequality.
−4≤x−3≤4-4 \leq x - 3 \leq 4
The absolute value inequality ∣k∣≤c|k| \leq c (for c≥0c \geq 0) is equivalent to −c≤k≤c-c \leq k \leq c.
2
Isolate xx by adding 3 across all parts of the inequality.
−1≤x≤7-1 \leq x \leq 7
Adding the same constant to all parts of an inequality maintains the inequality relationships.
3
Count the total number of integer values within the inclusive range [−1,7][-1, 7].
9
The number of integers in an inclusive range [a,b][a, b] is calculated as b−a+1b - a + 1, which gives 7−(−1)+1=97 - (-1) + 1 = 9.

Key Concept

Solving absolute value inequalities and counting integer solutions within a bounded interval
Question 120Question →

For all real numbers aa and bb, the custom binary operator ⋆\star is defined by a⋆b=2a−ba \star b = 2a - b. The function ff is defined for all real numbers by f(x)=x2−1f(x) = x^2 - 1. If (x⋆3)⋆f(3)=6(x \star 3) \star f(3) = 6, what is the value of xx?

Show answer & explanation

Answer: 5

Answer

5
First, evaluate the function f(3)=32−1=8f(3) = 3^2 - 1 = 8. Next, evaluate the first custom operation x⋆3=2x−3x \star 3 = 2x - 3. Substituting these into the overall expression yields (2x−3)⋆8=2(2x−3)−8=4x−14(2x - 3) \star 8 = 2(2x - 3) - 8 = 4x - 14. Setting this equal to 6 gives 4x−14=64x - 14 = 6, which simplifies to 4x=204x = 20, so x=5x = 5.

Step-by-Step Solution

1
Evaluate the inner function f(3)f(3)
f(3)=32−1=9−1=8f(3) = 3^2 - 1 = 9 - 1 = 8
Substitute x=3x = 3 into the given definition f(x)=x2−1f(x) = x^2 - 1.
2
Evaluate the inner custom operation x⋆3x \star 3
x⋆3=2x−3x \star 3 = 2x - 3
Apply the custom operator rule a⋆b=2a−ba \star b = 2a - b with a=xa = x and b=3b = 3.
3
Evaluate the outer custom operation (x⋆3)⋆f(3)(x \star 3) \star f(3)
(2x−3)⋆8=2(2x−3)−8=4x−6−8=4x−14(2x - 3) \star 8 = 2(2x - 3) - 8 = 4x - 6 - 8 = 4x - 14
Substitute the results from Step 1 and Step 2 into the custom operator definition.
4
Solve the algebraic equation for xx
4x−14=6  ⟹  4x=20  ⟹  x=54x - 14 = 6 \implies 4x = 20 \implies x = 5
Set the simplified expression equal to 6 as given in the problem stem and isolate xx.

Key Concept

Custom Operator Evaluation and Nested Functions
Estimated Time:1m 30s
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