Examkin

Advanced Math

438 questions

All types438Multiple choice220Numeric218
Question 81Question →

The graphs of the equations y−2x=3y - 2x = 3 and y=x2y = x^2 intersect at two points in the xyxy-plane. If (x,y)(x, y) represents an intersection point with a positive xx-coordinate, what is the value of yy?

Show answer & explanation

Answer: 9

Answer

9
Substituting y=x2y = x^2 into the first equation yields x2−2x=3x^2 - 2x = 3. Setting the quadratic equation to zero gives x2−2x−3=0x^2 - 2x - 3 = 0, which factors as (x−3)(x+1)=0(x - 3)(x + 1) = 0. This gives the solutions x=3x = 3 and x=−1x = -1. Because the question specifies a positive xx-coordinate, xx must be 3. Substituting x=3x = 3 back into y=x2y = x^2 yields y=9y = 9. Thus, the correct answer is 9.

Step-by-Step Solution

1
Substitute y=x2y = x^2 into the equation y−2x=3y - 2x = 3.
x2−2x=3x^2 - 2x = 3
This eliminates the variable yy and creates a single quadratic equation in terms of xx to find the xx-coordinates of the intersection points.
2
Rewrite the quadratic equation in standard form.
x2−2x−3=0x^2 - 2x - 3 = 0
Subtracting 3 from both sides sets the quadratic equation to 0, which is necessary for factoring.
3
Factor the quadratic equation to find its solutions.
(x−3)(x+1)=0(x - 3)(x + 1) = 0, which gives x=3x = 3 or x=−1x = -1.
Factoring allows us to find the roots, which represent the xx-coordinates of the intersection points.
4
Apply the constraint that the xx-coordinate must be positive.
x=3x = 3
The question specifies that x>0x > 0, so we discard x=−1x = -1.
5
Substitute x=3x = 3 back into y=x2y = x^2 to find the corresponding value of yy.
y=32=9y = 3^2 = 9
This determines the yy-coordinate of the intersection point with the positive xx-coordinate.

Key Concept

Solving a system of a linear equation and a quadratic equation by substitution.
Question 82Question →

A population of bacteria doubles every 3 hours. If the initial population of the bacteria is 500, what is the population of the bacteria after 9 hours?

Show answer & explanation

Answer: 4000

Answer

The population of the bacteria after 9 hours is 4,000.
The final population is calculated using the formula P(t)=P0×2t/dP(t) = P_0 \times 2^{t/d}, where P0P_0 is the initial population of 500, dd is the doubling period of 3 hours, and tt is the total time of 9 hours. Evaluating this gives P(9)=500×29/3=500×23=500×8=4000P(9) = 500 \times 2^{9/3} = 500 \times 2^3 = 500 \times 8 = 4000.

Step-by-Step Solution

1
Identify the initial population (P0P_0), doubling time (dd), and total time (tt).
P0=500P_0 = 500, d=3d = 3, and t=9t = 9.
These parameters are required to set up the exponential growth model.
2
Calculate the number of doubling periods.
The number of doubling periods is 93=3\frac{9}{3} = 3.
The population doubles once for every 3-hour interval.
3
Calculate the final population using the exponential growth formula.
500×23=500×8=4000500 \times 2^3 = 500 \times 8 = 4000.
Applying the 3 doubling cycles to the initial population of 500 yields the final population.

Key Concept

Exponential Growth Model
Estimated Time:1m 0s
Question 83Question →

In the system of equations below, kk is a positive constant.

x−y=kx - y = k
x2−3xy+y2=5x^2 - 3xy + y^2 = 5

If the system has exactly one real solution (x,y)(x, y), what is the value of kk?

Show answer & explanation

Answer: 2

Answer

The value of kk is 2.
Substituting y=x−ky = x - k into the second equation yields −x2+kx+k2−5=0-x^2 + kx + k^2 - 5 = 0, which can be rewritten in standard form as x2−kx+(5−k2)=0x^2 - kx + (5 - k^2) = 0. For this quadratic equation to have exactly one real solution, its discriminant must equal 0: b2−4ac=(−k)2−4(1)(5−k2)=5k2−20=0b^2 - 4ac = (-k)^2 - 4(1)(5 - k^2) = 5k^2 - 20 = 0. Solving 5k2−20=05k^2 - 20 = 0 gives k2=4k^2 = 4, and since kk must be positive, k=2k = 2.

Step-by-Step Solution

1
Rearrange the first equation to express yy in terms of xx.
y=x−ky = x - k
This allows for substitution into the second equation to eliminate yy.
2
Substitute y=x−ky = x - k into the second equation and expand.
x2−3x(x−k)+(x−k)2=5  ⟹  −x2+kx+k2−5=0x^2 - 3x(x - k) + (x - k)^2 = 5 \implies -x^2 + kx + k^2 - 5 = 0
To create a single quadratic equation in terms of xx.
3
Multiply by −1-1 to write the quadratic equation in standard form ax2+bx+c=0ax^2 + bx + c = 0.
x2−kx+(5−k2)=0x^2 - kx + (5 - k^2) = 0
Standard form makes it easier to identify the coefficients a=1a = 1, b=−kb = -k, and c=5−k2c = 5 - k^2.
4
Set the discriminant b2−4acb^2 - 4ac equal to 0.
(−k)2−4(1)(5−k2)=0(-k)^2 - 4(1)(5 - k^2) = 0
A quadratic equation has exactly one real solution if and only if its discriminant is zero.
5
Simplify the discriminant equation and solve for kk.
5k2−20=0  ⟹  k2=4  ⟹  k=25k^2 - 20 = 0 \implies k^2 = 4 \implies k = 2 (since kk must be positive)
To find the positive constant kk that satisfies the condition.

Key Concept

Determining the number of solutions to a nonlinear system by substituting and setting the discriminant of the resulting quadratic equation to zero.

Alternative Method

Alternatively, one can rewrite the second equation by grouping: x2−3xy+y2=(x−y)2−xy=5x^2 - 3xy + y^2 = (x - y)^2 - xy = 5. Since x−y=kx - y = k, we have k2−xy=5k^2 - xy = 5, so xy=k2−5xy = k^2 - 5. We now have a system of x−y=kx - y = k and xy=k2−5xy = k^2 - 5. Substituting y=x−ky = x - k gives x(x−k)=k2−5x(x - k) = k^2 - 5, leading to x2−kx+(5−k2)=0x^2 - kx + (5 - k^2) = 0, which can be solved using the discriminant as shown in the primary method.
Estimated Time:2m 30s
Question 84Question →

For all x>0x > 0 and y>0y > 0, the expression (x2y3)a(xy2)2\frac{(x^2y^3)^a}{(xy^2)^2} is equivalent to x6y8x^6y^8, where aa is a constant. What is the value of aa?

Show answer & explanation

Answer: 4

Answer

4
Applying the rules of exponents, the expression (x2y3)a(xy2)2\frac{(x^2y^3)^a}{(xy^2)^2} simplifies to x2ay3ax2y4=x2a−2y3a−4\frac{x^{2a}y^{3a}}{x^2y^4} = x^{2a-2}y^{3a-4}. Setting this equal to the equivalent expression x6y8x^6y^8 gives the system of equations 2a−2=62a - 2 = 6 and 3a−4=83a - 4 = 8. Solving either equation yields a=4a = 4.

Step-by-Step Solution

1
Apply the power rule of exponents to the numerator and denominator.
Numerator: x2ay3ax^{2a}y^{3a}, Denominator: x2y4x^2y^4
To expand the parentheses by multiplying the outer exponent with the inner exponents.
2
Apply the quotient rule of exponents to divide the numerator by the denominator.
x2a−2y3a−4x^{2a-2}y^{3a-4}
To simplify the rational expression by subtracting the exponents in the denominator from the exponents in the numerator.
3
Equate the simplified exponent of xx to the exponent of xx in the given equivalent expression.
2a−2=62a - 2 = 6
Equivalent expressions must have identical exponents for corresponding variable bases.
4
Solve the linear equation for aa.
a=4a = 4
To find the constant value that satisfies the equation.
5
Verify the value of aa using the exponents of yy.
3(4)−4=83(4) - 4 = 8, which is true.
To ensure consistency across both variable exponents.

Key Concept

Equivalent algebraic expressions involving exponent rules
Question 85Question →

A parabola and a line intersect at exactly one point in the xyxy-plane. The parabola is defined by the equation y=−x2+6x−2y = -x^2 + 6x - 2 and the line is defined by the equation y=2x+ky = 2x + k, where kk is a constant. What is the value of kk?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.
To find the value of kk where the parabola and the line intersect at exactly one point, we equate the two equations: −x2+6x−2=2x+k-x^2 + 6x - 2 = 2x + k. Rearranging this into standard quadratic form yields x2−4x+(k+2)=0x^2 - 4x + (k + 2) = 0. For a quadratic equation to have exactly one real solution, its discriminant, b2−4acb^2 - 4ac, must be equal to zero. Substituting a=1a = 1, b=−4b = -4, and c=k+2c = k + 2 into the discriminant formula gives (−4)2−4(1)(k+2)=0(-4)^2 - 4(1)(k + 2) = 0, which simplifies to 16−4k−8=016 - 4k - 8 = 0, or 8−4k=08 - 4k = 0. Solving for kk gives k=2k = 2.

Step-by-Step Solution

1
Equate the equations of the parabola and the line to set up an equation for the x-coordinates of their intersection points.
−x2+6x−2=2x+k-x^2 + 6x - 2 = 2x + k
At the points of intersection, the y-values of both equations are equal.
2
Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−4x+(k+2)=0x^2 - 4x + (k + 2) = 0
This allows us to identify the coefficients a=1a = 1, b=−4b = -4, and c=k+2c = k + 2 to apply the quadratic discriminant.
3
Set the discriminant b2−4acb^2 - 4ac equal to zero.
(−4)2−4(1)(k+2)=0(-4)^2 - 4(1)(k + 2) = 0
A quadratic system has exactly one real solution (tangency) if and only if the discriminant of the resulting quadratic equation is zero.
4
Solve the linear equation for kk.
k=2k = 2
Simplifying the expression yields 16−4k−8=016 - 4k - 8 = 0, which simplifies to 8−4k=08 - 4k = 0, giving k=2k = 2.

Key Concept

Nonlinear Systems of Equations
Question 86Question →
y=x2−5x+8y = x^2 - 5x + 8
y=2x+2y = 2x + 2

The system of equations above has two real solutions, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). If y1>y2y_1 > y_2, what is the value of x1−x2x_1 - x_2?

Show answer & explanation

Answer: 5

Answer

The value of the difference between the two x-coordinates is 5.
To solve the system, substitute y=2x+2y = 2x + 2 into y=x2−5x+8y = x^2 - 5x + 8 to obtain 2x+2=x2−5x+82x + 2 = x^2 - 5x + 8. Subtracting 2x+22x + 2 from both sides gives the quadratic equation x2−7x+6=0x^2 - 7x + 6 = 0. Factoring this equation yields (x−6)(x−1)=0(x - 6)(x - 1) = 0, which gives the x-coordinates x=6x = 6 and x=1x = 1. Substituting these back into y=2x+2y = 2x + 2 gives the corresponding y-coordinates: y=14y = 14 when x=6x = 6, and y=4y = 4 when x=1x = 1. Thus, the two solutions are (6,14)(6, 14) and (1,4)(1, 4). Because y1>y2y_1 > y_2, we must have (x1,y1)=(6,14)(x_1, y_1) = (6, 14) and (x2,y2)=(1,4)(x_2, y_2) = (1, 4). The value of x1−x2x_1 - x_2 is therefore 6−1=56 - 1 = 5.

Step-by-Step Solution

1
Substitute the expression for yy from the second equation into the first equation.
2x+2=x2−5x+82x + 2 = x^2 - 5x + 8
This substitution reduces the system of two equations to a single quadratic equation in terms of xx.
2
Rearrange the quadratic equation into standard form by subtracting 2x2x and 22 from both sides.
x2−7x+6=0x^2 - 7x + 6 = 0
Putting the equation in standard form is necessary to factor it and find its roots.
3
Factor the quadratic equation to find the two possible values of xx.
(x−6)(x−1)=0(x - 6)(x - 1) = 0, which gives x=6x = 6 or x=1x = 1.
Factoring allows us to find the x-coordinates of the points where the two graphs intersect.
4
Substitute the x-values back into the linear equation y=2x+2y = 2x + 2 to find their corresponding y-values.
For x=6x = 6, y=2(6)+2=14y = 2(6) + 2 = 14. For x=1x = 1, y=2(1)+2=4y = 2(1) + 2 = 4. The two solutions are (6,14)(6, 14) and (1,4)(1, 4).
Finding the y-values helps identify which coordinate pair corresponds to (x1,y1)(x_1, y_1) and which to (x2,y2)(x_2, y_2) using the given condition.
5
Apply the condition y1>y2y_1 > y_2 to assign the variables and calculate x1−x2x_1 - x_2.
Since 14>414 > 4, the solution with the larger y-value is (x1,y1)=(6,14)(x_1, y_1) = (6, 14) and the other is (x2,y2)=(1,4)(x_2, y_2) = (1, 4). Thus, x1−x2=6−1=5x_1 - x_2 = 6 - 1 = 5.
This calculation yields the final requested value.

Key Concept

Solving a system of a linear equation and a quadratic equation by substitution.
Question 87Question →

If x(x−9)=−20x(x - 9) = -20, which of the following is a possible value of xx?

Show answer & explanation

Answer: 5

Answer

5
To solve the equation x(x−9)=−20x(x - 9) = -20, first distribute the xx to get x2−9x=−20x^2 - 9x = -20. Next, rewrite the equation in standard form by adding 2020 to both sides, which gives x2−9x+20=0x^2 - 9x + 20 = 0. Factoring the quadratic expression yields (x−4)(x−5)=0(x - 4)(x - 5) = 0. Applying the zero product property gives two possible solutions: x=4x = 4 and x=5x = 5. Therefore, the value 55 is a possible value of xx.

Step-by-Step Solution

1
Distribute xx on the left side of the equation.
x2−9x=−20x^2 - 9x = -20
Expanding the product allows the quadratic equation to be rewritten in standard form.
2
Add 2020 to both sides to set the equation to standard form ax2+bx+c=0ax^2 + bx + c = 0.
x2−9x+20=0x^2 - 9x + 20 = 0
Setting the quadratic equation to equal zero is necessary to solve by factoring or using the quadratic formula.
3
Factor the quadratic trinomial by finding two integers that multiply to 2020 and add to −9-9.
(x−4)(x−5)=0(x - 4)(x - 5) = 0
The numbers −4-4 and −5-5 satisfy these conditions, allowing the expression to be factored.
4
Set each factor equal to zero to find the possible values of xx.
x=4x = 4 or x=5x = 5
According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero.

Key Concept

Solving quadratic equations by factoring
Question 88Question →

For all x>0x > 0 and x≠1x \neq 1, the expression x5/2−x3/2−x+1x3/2−x−x1/2+1\frac{x^{5/2} - x^{3/2} - x + 1}{x^{3/2} - x - x^{1/2} + 1} is equivalent to x+xa+bx + x^a + b, where aa and bb are constants and a<1a < 1. What is the value of 2a+b2a + b?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.
Substituting u=x1/2u = x^{1/2} transforms the expression into u5−u3−u2+1u3−u2−u+1\frac{u^5 - u^3 - u^2 + 1}{u^3 - u^2 - u + 1}. Factoring the numerator by grouping yields (u3−1)(u2−1)(u^3 - 1)(u^2 - 1), and factoring the denominator yields (u2−1)(u−1)(u^2 - 1)(u - 1). Canceling the common factor (u2−1)(u^2 - 1) leaves u3−1u−1\frac{u^3 - 1}{u - 1}. Applying the difference of cubes formula to factor u3−1u^3 - 1 as (u−1)(u2+u+1)(u - 1)(u^2 + u + 1) allows us to cancel the (u−1)(u - 1) term, leaving u2+u+1u^2 + u + 1. Re-substituting u=x1/2u = x^{1/2} results in the equivalent expression x+x1/2+1x + x^{1/2} + 1. Comparing this to x+xa+bx + x^a + b with a<1a < 1 gives a=0.5a = 0.5 and b=1b = 1. Evaluating 2a+b2a + b yields 2(0.5)+1=22(0.5) + 1 = 2.

Step-by-Step Solution

1
Introduce a variable substitution to eliminate fractional exponents.
Let u=x1/2u = x^{1/2}, which implies x=u2x = u^2. The expression becomes u5−u3−u2+1u3−u2−u+1\frac{u^5 - u^3 - u^2 + 1}{u^3 - u^2 - u + 1}.
Substitution simplifies working with fractional exponents and makes factoring patterns easier to identify.
2
Factor the numerator by grouping terms.
The numerator factors as u3(u2−1)−(u2−1)=(u3−1)(u2−1)u^3(u^2 - 1) - (u^2 - 1) = (u^3 - 1)(u^2 - 1).
Grouping common terms helps simplify polynomials into products of lower-degree factors.
3
Factor the denominator by grouping terms.
The denominator factors as u2(u−1)−(u−1)=(u2−1)(u−1)u^2(u - 1) - (u - 1) = (u^2 - 1)(u - 1).
Factoring the denominator reveals common terms that can be canceled with the numerator.
4
Combine the factored forms and cancel the common factor.
The expression becomes (u3−1)(u2−1)(u2−1)(u−1)=u3−1u−1\frac{(u^3 - 1)(u^2 - 1)}{(u^2 - 1)(u - 1)} = \frac{u^3 - 1}{u - 1} for u≠1u \neq 1.
Canceling (u2−1)(u^2 - 1) simplifies the rational expression since x≠1x \neq 1 ensures u2−1≠0u^2 - 1 \neq 0.
5
Simplify the remaining rational expression using the difference of cubes formula.
Factoring u3−1u^3 - 1 as (u−1)(u2+u+1)(u - 1)(u^2 + u + 1) and canceling (u−1)(u - 1) yields u2+u+1u^2 + u + 1.
The difference of cubes formula allows cancellation of the linear term in the denominator.
6
Substitute back the original variable and determine the constants.
Substituting u=x1/2u = x^{1/2} back gives x+x1/2+1x + x^{1/2} + 1, meaning a=0.5a = 0.5 and b=1b = 1.
This puts the expression in the final requested form x+xa+bx + x^a + b to find aa and bb.
7
Calculate the requested value 2a+b2a + b.
2(0.5)+1=22(0.5) + 1 = 2.
This computes the final numeric answer requested by the question.

Key Concept

Simplifying rational expressions with fractional exponents by substitution and grouping
Question 89Question →

A bank account is opened with an initial deposit of 800.Theaccountbalanceincreasesby800. The account balance increases by 5\%eachyear.Ifnoothertransactionsaremade,whichofthefollowingfunctionsbestmodelstheaccountbalance, each year. If no other transactions are made, which of the following functions best models the account balance, B(t),indollars,after, in dollars, after t$ years?

Show answer & explanation

Answer: B(t)=800(1.05)tB(t) = 800(1.05)^t

Answer

The function B(t)=800(1.05)tB(t) = 800(1.05)^t best models the account balance.
The initial deposit of 800representstheinitialvalueoftheexponentialfunctionwhen800 represents the initial value of the exponential function when t = 0 .Sincetheaccountbalanceincreasesby. Since the account balance increases by 5\%eachyear,thevalueismultipliedbyagrowthfactorof each year, the value is multiplied by a growth factor of 1 + 0.05 = 1.05 eachyear.Thus,theexponentialfunctionthatmodelsthebalanceafter each year. Thus, the exponential function that models the balance after t yearsis years is B(t) = 800(1.05)^t$.

Step-by-Step Solution

1
Identify the initial value of the exponential growth function.
The initial deposit is 800,sotheinitialvalueat800, so the initial value at t = 0 is is 800$.
An exponential model is written in the form B(t)=a(b)tB(t) = a(b)^t, where aa represents the initial value.
2
Determine the growth factor based on the annual percentage increase.
The growth rate is r=5%=0.05r = 5\% = 0.05. The growth factor bb is 1+r=1+0.05=1.051 + r = 1 + 0.05 = 1.05.
To find the growth factor for an increasing quantity, add the growth rate as a decimal to 1.
3
Write the final exponential function by substituting the initial value and growth factor.
The function is B(t)=800(1.05)tB(t) = 800(1.05)^t.
Substitute a=800a = 800 and b=1.05b = 1.05 into the standard exponential form B(t)=a(b)tB(t) = a(b)^t.

Key Concept

Writing and interpreting exponential growth functions from a real-world context.
Question 90Question →
y=x2−5y=4x\begin{aligned} y &= x^2 - 5 \\ y &= 4x \end{aligned}

If (x,y)(x, y) is a solution to the system of equations above and y>0y > 0, what is the value of yy?

Show answer & explanation

Answer: 20

Answer

20
Substituting y=4xy = 4x into the equation y=x2−5y = x^2 - 5 yields 4x=x2−54x = x^2 - 5. Rearranging this equation into standard form gives x2−4x−5=0x^2 - 4x - 5 = 0. Factoring the quadratic expression gives (x−5)(x+1)=0(x - 5)(x + 1) = 0, which means x=5x = 5 or x=−1x = -1. Substituting these values back into y=4xy = 4x gives the coordinates of the two solutions: (5,20)(5, 20) and (−1,−4)(-1, -4). Since the question specifies that y>0y > 0, the correct value is the positive yy-value, which is 20.

Step-by-Step Solution

1
Substitute the expression for yy from the second equation into the first equation.
4x=x2−54x = x^2 - 5
This eliminates yy and creates a single equation in terms of xx.
2
Subtract 4x4x from both sides to write the quadratic equation in standard form.
x2−4x−5=0x^2 - 4x - 5 = 0
Standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0 is required to solve by factoring.
3
Factor the quadratic equation.
(x−5)(x+1)=0(x - 5)(x + 1) = 0
Factoring helps find the values of xx that satisfy the equation.
4
Solve for xx by setting each factor to zero.
x=5x = 5 or x=−1x = -1
Applying the zero product property yields the possible xx-coordinates of the solutions.
5
Calculate the corresponding yy-values using the equation y=4xy = 4x.
For x=5x = 5, y=20y = 20. For x=−1x = -1, y=−4y = -4.
This determines the coordinates of the intersection points, which are (5,20)(5, 20) and (−1,−4)(-1, -4).
6
Apply the constraint y>0y > 0 to identify the correct value of yy.
y=20y = 20
Since −4-4 is not greater than zero, the only valid solution is (5,20)(5, 20), giving y=20y = 20.

Key Concept

Solving a system of linear and quadratic equations using substitution.
Question 91Question →

For all x>5x > 5, which of the following is equivalent to the expression 3x2−14x−53x+1−x2−25x−5\frac{3x^2 - 14x - 5}{3x + 1} - \frac{x^2 - 25}{x - 5}?

Show answer & explanation

Answer: −10-10

Answer

−10-10
Factoring the numerators allows both rational expressions to be simplified. The first expression, 3x2−14x−53x+1\frac{3x^2 - 14x - 5}{3x + 1}, factors as (3x+1)(x−5)3x+1\frac{(3x + 1)(x - 5)}{3x + 1}, which simplifies to x−5x - 5 since x>5x > 5. The second expression, x2−25x−5\frac{x^2 - 25}{x - 5}, factors as (x−5)(x+5)x−5\frac{(x - 5)(x + 5)}{x - 5}, which simplifies to x+5x + 5. Subtracting the second simplified expression from the first gives (x−5)−(x+5)=x−5−x−5=−10(x - 5) - (x + 5) = x - 5 - x - 5 = -10. Therefore, the equivalent expression is −10-10.

Step-by-Step Solution

1
Factor the numerator of the first rational expression, 3x2−14x−53x^2 - 14x - 5.
(3x+1)(x−5)(3x + 1)(x - 5)
To find common factors that can be simplified with the denominator 3x+13x + 1.
2
Simplify the first expression, 3x2−14x−53x+1\frac{3x^2 - 14x - 5}{3x + 1}, for x>5x > 5.
x−5x - 5
Since x>5x > 5, 3x+1≠03x + 1 \neq 0, allowing us to divide out the common factor (3x+1)(3x + 1).
3
Factor the numerator of the second rational expression, x2−25x^2 - 25.
(x−5)(x+5)(x - 5)(x + 5)
Using the difference of squares identity, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
4
Simplify the second expression, x2−25x−5\frac{x^2 - 25}{x - 5}, for x>5x > 5.
x+5x + 5
Since x>5x > 5, x−5≠0x - 5 \neq 0, allowing us to divide out the common factor (x−5)(x - 5).
5
Subtract the second simplified expression from the first.
(x−5)−(x+5)=x−5−x−5=−10(x - 5) - (x + 5) = x - 5 - x - 5 = -10
Distributing the negative sign through the parentheses and combining like terms yields the final simplified equivalent value.

Key Concept

Simplifying rational expressions by factoring and performing polynomial operations.
Question 92Question →

The quadratic equation x2−4x+2=0x^2 - 4x + 2 = 0 has roots r1r_1 and r2r_2. A second quadratic equation, x2−px+q=0x^2 - px + q = 0, has roots (r1+1r2)\left(r_1 + \frac{1}{r_2}\right) and (r2+1r1)\left(r_2 + \frac{1}{r_1}\right), where pp and qq are constants. What is the value of pp?

Show answer & explanation

Answer: 6

Answer

The value of pp is 6.
By Vieta's formulas, the sum of the roots of x2−4x+2=0x^2 - 4x + 2 = 0 is r1+r2=4r_1 + r_2 = 4 and the product of the roots is r1r2=2r_1 r_2 = 2. The sum of the roots of the second equation x2−px+q=0x^2 - px + q = 0 is pp. Therefore, p=(r1+1r2)+(r2+1r1)=(r1+r2)+r1+r2r1r2p = \left(r_1 + \frac{1}{r_2}\right) + \left(r_2 + \frac{1}{r_1}\right) = (r_1 + r_2) + \frac{r_1 + r_2}{r_1 r_2}. Substituting the known values yields p=4+42=6p = 4 + \frac{4}{2} = 6.

Step-by-Step Solution

1
Determine the sum and product of the roots of the first equation.
r1+r2=4r_1 + r_2 = 4 and r1r2=2r_1 r_2 = 2
Vieta's formulas state that for x2−Bx+C=0x^2 - Bx + C = 0, the sum of the roots is BB and the product is CC.
2
Express the sum of the roots of the second equation, which is pp.
p=(r1+r2)+(1r1+1r2)p = (r_1 + r_2) + \left(\frac{1}{r_1} + \frac{1}{r_2}\right)
The sum of the roots of x2−px+q=0x^2 - px + q = 0 is pp.
3
Simplify the fractional part of the equation.
1r1+1r2=r1+r2r1r2\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}
Finding a common denominator allows us to write the sum of reciprocals in terms of the sum and product of the roots.
4
Substitute the values of the sum and product into the expression for pp.
p=4+42=6p = 4 + \frac{4}{2} = 6
This evaluates the expression to find the final value of pp.

Key Concept

Relating the roots and coefficients of quadratic equations using Vieta's formulas.
Question 93Question →

If (2y−1)2=49(2y - 1)^2 = 49 and y>0y > 0, what is the value of yy?

Show answer & explanation

Answer: 4

Answer

The correct answer is 44.
By taking the square root of both sides of the equation (2y−1)2=49(2y - 1)^2 = 49, we get 2y−1=72y - 1 = 7 or 2y−1=−72y - 1 = -7. Solving 2y−1=72y - 1 = 7 gives 2y=82y = 8, or y=4y = 4. Solving 2y−1=−72y - 1 = -7 gives 2y=−62y = -6, or y=−3y = -3. Since it is given that y>0y > 0, the value of yy must be 44.

Step-by-Step Solution

1
Take the square root of both sides of the equation.
2y−1=72y - 1 = 7 or 2y−1=−72y - 1 = -7
Applying the square root property of equality to isolate the linear term.
2
Solve each linear equation for yy.
y=4y = 4 or y=−3y = -3
Isolating yy in each equation by adding 11 and then dividing by 22.
3
Apply the constraint y>0y > 0 to identify the final value.
y=4y = 4
The value y=−3y = -3 is negative, so it does not satisfy the constraint y>0y > 0.

Key Concept

Solving quadratic equations of the form (ax+b)2=c(ax + b)^2 = c using the square root property.
Question 94Question →

In the xyxy-plane, a circle has center (6,8)(6, 8) and is tangent to the yy-axis. A line with a positive slope mm passes through the origin and is tangent to the circle. If mm is written as a fraction in simplest form, pq\frac{p}{q}, what is the value of p+qp + q?

Show answer & explanation

Answer: 31

Answer

The value of p+qp + q is 31.
The radius of the circle is 6 since the center is (6,8)(6, 8) and it is tangent to the yy-axis. Substituting the line y=mxy = mx into the circle's equation (x−6)2+(y−8)2=36(x-6)^2 + (y-8)^2 = 36 yields a quadratic equation in xx: (1+m2)x2−(12+16m)x+64=0(1+m^2)x^2 - (12+16m)x + 64 = 0. Since the line is tangent to the circle, there is exactly one solution, meaning the discriminant of this quadratic must equal zero: (12+16m)2−4(1+m2)(64)=0(12+16m)^2 - 4(1+m^2)(64) = 0. Simplifying this equation yields 24m−7=024m - 7 = 0, which gives a slope of m=724m = \frac{7}{24}. Since the fraction is in simplest form, p=7p=7 and q=24q=24, and their sum is 3131.

Step-by-Step Solution

1
Determine the radius of the circle.
r=6r = 6
Since the center is at (6,8)(6, 8) and the circle is tangent to the yy-axis, the radius is the horizontal distance from the center to the yy-axis.
2
Write the equation of the circle.
(x−6)2+(y−8)2=36(x-6)^2 + (y-8)^2 = 36
A circle with center (h,k)(h, k) and radius rr has the equation (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.
3
Substitute y=mxy = mx into the circle's equation.
(1+m2)x2−(12+16m)x+64=0(1+m^2)x^2 - (12+16m)x + 64 = 0
Substituting the equation of the line y=mxy = mx into the circle's equation and expanding gives a quadratic equation in terms of xx.
4
Set the discriminant of the quadratic equation to zero.
Δ=(12+16m)2−256(1+m2)=0\Delta = (12+16m)^2 - 256(1+m^2) = 0
For the line to be tangent to the circle, the system must have exactly one real solution, meaning the quadratic equation must have a discriminant of zero.
5
Solve for the positive slope mm.
m=724m = \frac{7}{24}
Expanding and simplifying the discriminant equation yields 24m−7=024m - 7 = 0, which gives m=724m = \frac{7}{24}.
6
Calculate p+qp + q using the simplest form fraction.
3131
Since m=724m = \frac{7}{24} is in simplest form, p=7p = 7 and q=24q = 24. Therefore, p+q=7+24=31p + q = 7 + 24 = 31.

Key Concept

Nonlinear Systems of Equations
Question 95Question →

The quadratic function ff is defined by f(x)=(x−4)2+7f(x) = (x - 4)^2 + 7. What are the coordinates of the vertex of the graph of y=f(x)y = f(x) in the xyxy-plane?

Show answer & explanation

Answer: (4,7)(4, 7)

Answer

The coordinates of the vertex are (4,7)(4, 7).
The quadratic function is given in vertex form, f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, where the vertex of the graph of the function in the xyxy-plane is (h,k)(h, k). Comparing f(x)=(x−4)2+7f(x) = (x - 4)^2 + 7 to this form, we get h=4h = 4 and k=7k = 7. Therefore, the coordinates of the vertex are (4,7)(4, 7).

Step-by-Step Solution

1
Identify the standard vertex form of a quadratic function.
The vertex form of a quadratic function is f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
This standard form allows direct extraction of the vertex coordinates by matching the variables.
2
Compare the given function f(x)=(x−4)2+7f(x) = (x - 4)^2 + 7 to the standard vertex form.
Comparing the two equations, we identify that a=1a = 1, h=4h = 4, and k=7k = 7.
Matching coefficients and constants determines the values of the vertex parameters.
3
Write the coordinates of the vertex (h,k)(h, k) using the identified values.
The vertex is at (4,7)(4, 7).
Substituting h=4h = 4 and k=7k = 7 into (h,k)(h, k) yields the final vertex coordinates.

Key Concept

Identifying the vertex of a quadratic function from its vertex form.
Question 96Question →

For all x>0x > 0, which of the following is equivalent to the expression (x−1x)2(x+1+1x)\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \left(x + 1 + \frac{1}{x}\right)?

Show answer & explanation

Answer: (x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}

Answer

(x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}
The correct expression is obtained by first converting x−1x\sqrt{x} - \frac{1}{\sqrt{x}} to x−1x\frac{x-1}{\sqrt{x}}, and then squaring it to get (x−1)2x\frac{(x-1)^2}{x}. Next, the term x+1+1xx + 1 + \frac{1}{x} is converted to x2+x+1x\frac{x^2+x+1}{x}. Multiplying these two results yields (x−1)2(x2+x+1)x2\frac{(x-1)^2(x^2+x+1)}{x^2}. Grouping (x−1)(x2+x+1)(x-1)(x^2+x+1) and substituting it with the difference of cubes identity, x3−1x^3-1, gives the final equivalent expression (x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}.

Step-by-Step Solution

1
Simplify the first factor by writing x−1x\sqrt{x} - \frac{1}{\sqrt{x}} under a common denominator and squaring it.
(x−1x)2=(x−1)2x\left(\frac{x-1}{\sqrt{x}}\right)^2 = \frac{(x-1)^2}{x}
This allows us to combine the terms in the first factor before multiplication.
2
Write the second factor x+1+1xx + 1 + \frac{1}{x} under a common denominator.
x2+x+1x\frac{x^2+x+1}{x}
This allows us to multiply the two fractional factors easily.
3
Multiply the two simplified factors together.
(x−1)2(x2+x+1)x2=(x−1)(x−1)(x2+x+1)x2\frac{(x-1)^2(x^2+x+1)}{x^2} = \frac{(x-1)(x-1)(x^2+x+1)}{x^2}
Multiplying the numerators and denominators of the fractions gives the product of the terms.
4
Apply the difference of cubes factorization formula: (x−1)(x2+x+1)=x3−1(x-1)(x^2+x+1) = x^3-1.
(x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}
This simplifies the product to its final equivalent form.

Key Concept

Simplifying rational expressions and applying algebraic identities (specifically the binomial square and difference of cubes).

Alternative Method

Alternatively, you can substitute a convenient value for xx, such as x=4x = 4, into the original expression and each of the options, then compare the results.
Estimated Time:2m 0s
Question 97Question →

For a real solution (x,y)(x, y) to the system of equations below, which of the following is a possible value of x+yx + y?

x2+3y2=21x2−2y=5\begin{aligned} x^2 + 3y^2 &= 21 \\ x^2 - 2y &= 5 \end{aligned}
Show answer & explanation

Answer: 5

Answer

The correct answer is 5, which is the sum of the coordinates of the real solution (3, 2) of the system.
Substituting x2=2y+5x^2 = 2y + 5 into the first equation yields the quadratic equation 3y2+2y−16=03y^2 + 2y - 16 = 0. Solving this quadratic equation gives y=2y = 2 or y=−8/3y = -8/3. Since x2x^2 must be non-negative for real solutions, y=−8/3y = -8/3 yields no real solutions because x2=2(−8/3)+5=−1/3<0x^2 = 2(-8/3) + 5 = -1/3 < 0. For y=2y = 2, we have x2=2(2)+5=9x^2 = 2(2) + 5 = 9, which gives x=3x = 3 or x=−3x = -3. The real solutions to the system are (3,2)(3, 2) and (−3,2)(-3, 2). The possible values for x+yx + y are 3+2=53 + 2 = 5 and −3+2=−1-3 + 2 = -1. Therefore, 5 is a possible value of x+yx + y.

Step-by-Step Solution

1
Rearrange the second equation to express x2x^2 in terms of yy.
x2=2y+5x^2 = 2y + 5
This allows for direct substitution into the first equation to eliminate the xx variable.
2
Substitute the expression for x2x^2 into the first equation and simplify the resulting quadratic equation in terms of yy.
3y2+2y−16=03y^2 + 2y - 16 = 0
Substituting 2y+52y + 5 for x2x^2 in x2+3y2=21x^2 + 3y^2 = 21 gives (2y+5)+3y2=21(2y + 5) + 3y^2 = 21, which simplifies to 3y2+2y−16=03y^2 + 2y - 16 = 0.
3
Factor or use the quadratic formula to solve for yy.
y=2y = 2 or y=−8/3y = -8/3
The quadratic equation factors as (3y+8)(y−2)=0(3y + 8)(y - 2) = 0, giving the roots y=2y = 2 and y=−8/3y = -8/3.
4
Find the corresponding values of xx for each yy and determine the real solutions.
For y=2y = 2, x2=2(2)+5=9  ⟹  x=±3x^2 = 2(2) + 5 = 9 \implies x = \pm 3. For y=−8/3y = -8/3, x2=2(−8/3)+5=−1/3x^2 = 2(-8/3) + 5 = -1/3, which has no real solutions.
A real number squared cannot be negative, so only y=2y = 2 yields real solutions: (3,2)(3, 2) and (−3,2)(-3, 2).
5
Calculate the possible values of x+yx + y for the real solutions.
For (3,2)(3, 2), x+y=3+2=5x + y = 3 + 2 = 5. For (−3,2)(-3, 2), x+y=−3+2=−1x + y = -3 + 2 = -1.
Evaluating the target expression for both valid real solutions gives 5 and -1. Since 5 is the only value present in the options, it is the correct answer.

Key Concept

Solving systems of nonlinear equations algebraically using substitution and analyzing real solutions.
Question 98Question →

For a certain constant aa, the quadratic equation 3x2−2ax+a−4=03x^2 - 2ax + a - 4 = 0 has real roots x1x_1 and x2x_2. If x12+x22=289x_1^2 + x_2^2 = \frac{28}{9}, what is the sum of all possible values of aa?

Show answer & explanation

Answer: 32\frac{3}{2}

Answer

The sum of all possible values of aa is \frac{3}{2}.
The correct answer is derived by expressing the sum of the squares of the roots, x12+x22x_1^2 + x_2^2, as (x1+x2)2−2x1x2(x_1 + x_2)^2 - 2x_1x_2. Applying Vieta's formulas to the equation 3x2−2ax+a−4=03x^2 - 2ax + a - 4 = 0 yields x1+x2=2a3x_1 + x_2 = \frac{2a}{3} and x1x2=a−43x_1 x_2 = \frac{a - 4}{3}. Substituting these values into the identity gives 4a29−2(a−4)3=289\frac{4a^2}{9} - \frac{2(a - 4)}{3} = \frac{28}{9}. Multiplying by 9 and simplifying leads to the quadratic equation 2a2−3a−2=02a^2 - 3a - 2 = 0. Since this equation has real solutions and the original equation always has real roots for any real aa, the sum of all possible values of aa is given by the negative linear coefficient divided by the leading coefficient, which is \frac{3}{2}.

Step-by-Step Solution

1
Express the sum of squares x12+x22x_1^2 + x_2^2 in terms of the sum and product of the roots.
x12+x22=(x1+x2)2−2x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2
This allows the use of Vieta's formulas to relate the roots of the quadratic equation to its coefficients.
2
Apply Vieta's formulas to the quadratic equation 3x2−2ax+a−4=03x^2 - 2ax + a - 4 = 0 to find the sum and product of the roots.
x1+x2=2a3x_1 + x_2 = \frac{2a}{3} and x1x2=a−43x_1 x_2 = \frac{a - 4}{3}
For any quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0, the sum of the roots is −BA-\frac{B}{A} and the product is CA\frac{C}{A}.
3
Substitute the expressions for the sum and product of the roots into the identity from Step 1 and set it equal to the given value.
(2a3)2−2(a−43)=289\left(\frac{2a}{3}\right)^2 - 2\left(\frac{a - 4}{3}\right) = \frac{28}{9}
This sets up a solvable equation for the parameter aa.
4
Simplify the algebraic equation by multiplying both sides by 9 and expanding terms.
4a2−6(a−4)=28  ⟹  4a2−6a−4=0  ⟹  2a2−3a−2=04a^2 - 6(a - 4) = 28 \implies 4a^2 - 6a - 4 = 0 \implies 2a^2 - 3a - 2 = 0
Eliminating denominators and grouping terms yields a standard form quadratic equation in aa.
5
Verify that the solutions for aa are real and that they yield real roots for xx in the original equation.
For 2a2−3a−2=02a^2 - 3a - 2 = 0, the discriminant Δa=25>0\Delta_a = 25 > 0, so aa has real solutions. The discriminant of the original equation is Δx=4a2−12a+48\Delta_x = 4a^2 - 12a + 48, which is positive for all real values of aa since its own discriminant is negative.
This confirms that the solutions for aa are valid and satisfy the condition that x1x_1 and x2x_2 are real numbers.
6
Use Vieta's formula on the quadratic equation 2a2−3a−2=02a^2 - 3a - 2 = 0 to find the sum of all possible values of aa.
Sum of values =−−32=32= -\frac{-3}{2} = \frac{3}{2}
The sum of the roots of a quadratic equation in the form Aa2+Ba+C=0Aa^2 + Ba + C = 0 is equal to −BA-\frac{B}{A}.

Key Concept

Vieta's Formulas and Algebraic Identities for Quadratic Equations
Question 99Question →

A system of two equations is given:

y=x2+kx+4y = x^2 + kx + 4
y=4x−ky = 4x - k

In these equations, kk represents a positive constant. If the system has a single real solution, what is the value of kk?

Show answer & explanation

Answer: 12

Answer

12
Setting the two equations equal yields the quadratic equation x2+(k−4)x+(k+4)=0x^2 + (k - 4)x + (k + 4) = 0. For the system to have exactly one real solution, the discriminant of this quadratic equation must be equal to 00. The discriminant is (k−4)2−4(1)(k+4)=k2−12k(k - 4)^2 - 4(1)(k + 4) = k^2 - 12k. Solving k2−12k=0k^2 - 12k = 0 yields k=0k = 0 or k=12k = 12. Since kk must be positive, the value of kk is 1212.

Step-by-Step Solution

1
Set the two equations equal to each other to form a single quadratic equation in terms of xx.
x2+(k−4)x+(k+4)=0x^2 + (k - 4)x + (k + 4) = 0
Equating the expressions for yy allows us to find the xx-coordinates where the graphs of the two equations intersect.
2
Set the discriminant of the quadratic equation to zero.
(k−4)2−4(1)(k+4)=0(k - 4)^2 - 4(1)(k + 4) = 0
A system of equations consisting of a line and a parabola has a single real solution if and only if the line is tangent to the parabola, which corresponds to a quadratic equation with a discriminant of zero.
3
Expand and simplify the equation for kk.
k2−12k=0k^2 - 12k = 0
Expanding (k−4)2(k - 4)^2 yields k2−8k+16k^2 - 8k + 16, and distributing −4-4 yields −4k−16-4k - 16. Combining like terms simplifies the relation.
4
Solve for kk and apply the constraint that k>0k > 0.
k=12k = 12
Factoring k(k−12)=0k(k - 12) = 0 gives k=0k = 0 or k=12k = 12. Since kk is specified to be positive, k=12k = 12 is the correct value.

Key Concept

Solving nonlinear systems of equations by substitution and using the discriminant to find conditions for a single real solution.
Question 100Question →

If (x,y)(x, y) is a solution to the system of equations below and x>0x > 0, what is the value of x+yx + y?

x2+y2=50x−y=6\begin{aligned} x^2 + y^2 &= 50 \\ x - y &= 6 \end{aligned}
Show answer & explanation

Answer: 8

Answer

8
Solving the linear equation for xx gives x=y+6x = y + 6. Substituting this into the circle equation yields (y+6)2+y2=50(y + 6)^2 + y^2 = 50, which simplifies to 2y2+12y−14=02y^2 + 12y - 14 = 0. Dividing by 22 gives y2+6y−7=0y^2 + 6y - 7 = 0, which factors as (y+7)(y−1)=0(y + 7)(y - 1) = 0. This yields y=1y = 1 (with x=7x = 7) or y=−7y = -7 (with x=−1x = -1). The constraint x>0x > 0 requires selecting the solution (7,1)(7, 1). The sum of xx and yy for this solution is 7+1=87 + 1 = 8.

Step-by-Step Solution

1
Express xx in terms of yy using the linear equation.
x=y+6x = y + 6
This allows for substitution into the second equation to reduce the system to a single variable.
2
Substitute the expression for xx into the equation of the circle and expand.
2y2+12y−14=02y^2 + 12y - 14 = 0
Substituting x=y+6x = y + 6 into x2+y2=50x^2 + y^2 = 50 yields (y+6)2+y2=50(y + 6)^2 + y^2 = 50. Expanding gives y2+12y+36+y2=50y^2 + 12y + 36 + y^2 = 50.
3
Simplify and solve the quadratic equation by factoring.
y=1y = 1 or y=−7y = -7
Dividing the equation by 2 gives y2+6y−7=0y^2 + 6y - 7 = 0, which factors as (y+7)(y−1)=0(y + 7)(y - 1) = 0.
4
Calculate the corresponding values of xx for each yy.
For y=1y = 1, x=7x = 7; for y=−7y = -7, x=−1x = -1
Using the relation x=y+6x = y + 6 yields the coordinates of the two intersection points.
5
Apply the constraint x>0x > 0 and calculate x+yx + y.
x+y=8x + y = 8
Since x>0x > 0, the only valid solution is (7,1)(7, 1). The sum of the coordinates is 7+1=87 + 1 = 8.

Key Concept

Solving a nonlinear system of equations representing a circle and a line using substitution and factoring.
PreviousPage 5 / 22Next