Advanced Math Alıştırma Soruları — SAT — Sayfa 8

Soru 141Soru →

In the quadratic equation $x^2 - bx + 64 = 0$ , $b$ is a constant. The two real solutions to the equation are $r$ and $s$ , where $r = s^2$ . What is the value of $b$ ?

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Cevap: 20

Cevap

20

By applying Vieta's formulas to the quadratic equation

x^2 - bx + 64 = 0

, we find that the product of the solutions is

r \cdot s = 64

. Substituting the given relation

r = s^2

gives

s^3 = 64

, which yields the real solution

s = 4

. Substituting this back into the relation gives the other solution

r = 16

. Finally, the sum of the solutions is

r + s = b

, so

b = 16 + 4 = 20

.

Adım Adım Çözüm

1

Apply Vieta's formula for the product of the roots

r \cdot s = 64

For a quadratic equation in the form

x^2 - bx + c = 0

, the product of the roots is equal to the constant term

c

.

2

Substitute the given root relationship into the product equation

s^3 = 64

, which solves to

s = 4

We are given that one root is the square of the other (

r = s^2

), so substituting

s^2

for

r

allows us to solve for

s

.

3

Calculate the value of the second root

r

r = 16

Using the relation

r = s^2

with

s = 4

, we find

r = 4^2 = 16

.

4

Apply Vieta's formula for the sum of the roots to find

b

b = 20

For the equation

x^2 - bx + 64 = 0

, the sum of the roots is

r + s = b

. Substituting

r = 16

and

s = 4

gives

16 + 4 = 20

.

Anahtar Kavram

Vieta's formulas relating the coefficients of a quadratic equation to its roots

Soru 142Soru →

If $3^{2x - 1} = 27$ , what is the value of $x$ ?

1

5

2

14

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Cevap: 2

Cevap

2

To solve the equation

3^{2x - 1} = 27

, we first express 27 as a power of 3, which is

3^3

. Since the bases are the same, we set the exponents equal to each other to get

2x - 1 = 3

. Adding 1 to both sides gives

2x = 4

, and dividing both sides by 2 gives the value of

x

as 2.

Adım Adım Çözüm

1

Express 27 with a base of 3

3^{2x - 1} = 3^3

To solve an exponential equation, we need to write both sides of the equation with a common base.

2

Set the exponents equal to each other

2x - 1 = 3

Since the bases are equal, their exponents must also be equal.

3

Solve the linear equation for

x

x = 2

Add 1 to both sides to get

2x = 4

, and then divide both sides by 2 to isolate

x

.

Anahtar Kavram

Solving exponential equations by finding a common base

Tahmini Süre:45s

Soru 143Soru →

In the $xy$ -plane, the graph of $y = x^2$ is shifted $4$ units to the right and $9$ units up to produce the graph of $y = h(x)$ . Which of the following equations defines the function $h$ ?

h(x) = (x + 4)^2 + 9

h(x) = (x - 4)^2 - 9

h(x) = (x - 4)^2 + 9

h(x) = (x + 4)^2 - 9

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Cevap: $h(x) = (x - 4)^2 + 9$

Cevap

h(x) = (x - 4)^2 + 9

The parent function

y = x^2

has its vertex at

(0, 0)

. Shifting this graph

4

units to the right moves the x-coordinate of the vertex to

4

, and shifting it

9

units up moves the y-coordinate of the vertex to

9

. The new vertex is at

(4, 9)

. A parabola with a vertex of

(h, k)

and a leading coefficient of

1

is written in vertex form as

y = (x - h)^2 + k

. Substituting

h = 4

and

k = 9

gives the equation

h(x) = (x - 4)^2 + 9

.

Adım Adım Çözüm

1

Identify the base function and the rules of graph translation.

The base function is

y = x^2

. Shifting a graph horizontally by

h

units changes the input

x

to

(x - h)

, and shifting vertically by

k

units adds

k

to the function value.

Applying transformations to the parent function changes its position in the coordinate plane while maintaining its shape.

2

Apply the horizontal translation of

4

units to the right.

Replacing

x

with

(x - 4)

in the parent function gives the intermediate equation

y = (x - 4)^2

.

A horizontal shift of

h

units to the right corresponds to replacing the input variable

x

with

(x - h)

.

3

Apply the vertical translation of

9

units up.

Adding

9

to the expression yields the final equation

h(x) = (x - 4)^2 + 9

.

A vertical shift of

k

units upward corresponds to adding

k

to the output of the function.

Anahtar Kavram

Quadratic graph transformations and translation rules

Soru 144Soru →

If $4^x = 5$ , what is the value of $2^{4x - 1}$ ?

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Cevap: 12.5

Cevap

12.5

By writing

4^x

as

(2^2)^x = 2^{2x}

, we find that

2^{2x} = 5

. The expression

2^{4x-1}

can be rewritten using exponent properties as

\frac{2^{4x}}{2^1} = \frac{(2^{2x})^2}{2}

. Substituting

2^{2x} = 5

into this expression yields

\frac{5^2}{2} = \frac{25}{2} = 12.5

. Thus, the correct numerical response is 12.5.

Adım Adım Çözüm

1

Express the given equation in terms of base 2.

2^{2x} = 5

Since

4 = 2^2

, we can rewrite

4^x

as

(2^2)^x

. Applying the power of a power rule,

(a^m)^n = a^{mn}

, gives

(2^2)^x = 2^{2x}

.

2

Rewrite the expression to be evaluated using exponent properties.

2^{4x - 1} = \frac{(2^{2x})^2}{2}

Using the division property of exponents,

a^{m-n} = \frac{a^m}{a^n}

, we can write

2^{4x - 1}

as

\frac{2^{4x}}{2}

. Then, using the power of a power rule in reverse,

2^{4x} = (2^{2x})^2

.

3

Substitute the known value of

2^{2x}

and simplify the numerical expression.

12.5

Substitute

2^{2x} = 5

into the expression

\frac{(2^{2x})^2}{2}

to obtain

\frac{5^2}{2} = \frac{25}{2}

, which simplifies to

12.5

.

Anahtar Kavram

Manipulating exponential equations by expressing bases in terms of common prime factors and applying properties of exponents.

Tahmini Süre:1m 30s

Soru 145Soru →

In the quadratic equation $x^2 - ax + b = 0$ , the roots $r$ and $s$ are positive integers. The quadratic equation $x^2 - (a+12)x + (b+12) = 0$ has roots $r^2$ and $s^2$ , where $a$ and $b$ are constants. What is the value of $a$ ?

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Cevap: 5

Cevap

The value of

a

is 5.

Applying Vieta's formulas to both equations and using the algebraic identity

r^2 + s^2 = (r+s)^2 - 2rs

yields the system of equations

a^2 - 2b = a + 12

and

b^2 = b + 12

. Solving the second equation gives

b = 4

(discarding the negative root since

r, s > 0

). Substituting

b = 4

into the first equation yields

a^2 - a - 20 = 0

, which solves to

a = 5

(discarding

a = -4

since

r, s > 0

).

Adım Adım Çözüm

1

Apply Vieta's formulas to the first quadratic equation.

r+s = a

and

rs = b

To express the sum and product of the roots of the first equation in terms of the coefficients

a

and

b

.

2

Apply Vieta's formulas to the second quadratic equation.

r^2+s^2 = a+12

and

r^2s^2 = b+12

To express the sum and product of the squared roots in terms of the coefficients of the second equation.

3

Express

r^2+s^2

in terms of

a

and

b

using the algebraic expansion of a binomial square.

a^2 - 2b = a+12

By rewriting

r^2+s^2

as

(r+s)^2 - 2rs

, we can substitute

a

and

b

to form a system of equations.

4

Solve for the constant

b

using the product relation.

b = 4

The relation

(rs)^2 = b^2 = b+12

simplifies to the quadratic equation

b^2 - b - 12 = 0

. Factoring yields

b = 4

or

b = -3

. Since

r

and

s

are positive integers, their product

b

must be positive, meaning

b = 4

.

5

Substitute

b = 4

into the sum relation to solve for

a

.

a = 5

Substituting

b = 4

yields

a^2 - 2(4) = a+12

, which simplifies to

a^2 - a - 20 = 0

. Factoring yields

a = 5

or

a = -4

. Since

r

and

s

are positive integers, their sum

a

must be positive, meaning

a = 5

.

Anahtar Kavram

Vieta's Formulas and Symmetric Polynomial Relationships

Tahmini Süre:3m 0s

Soru 146Soru →

An exponential function $g$ is defined by $g(x) = p \cdot q^x$ , where $p$ and $q$ are positive constants. If $g\left(\frac{1}{2}\right) = 6$ and $g\left(\frac{3}{2}\right) = 54$ , what is the value of $g(2)$ ?

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Cevap: 162

Cevap

162

The correct answer is 162. Substituting the coordinates into the function gives the system

p \cdot q^{1/2} = 6

and

p \cdot q^{3/2} = 54

. Dividing the second equation by the first yields

q = 9

. Substituting

q = 9

into the first equation yields

p \cdot 3 = 6

, so

p = 2

. The complete function is

g(x) = 2 \cdot 9^x

, and evaluating

g(2)

gives

2 \cdot 81 = 162

.

Adım Adım Çözüm

1

Substitute the given values into the function definition to set up a system of equations.

p \cdot q^{1/2} = 6

and

p \cdot q^{3/2} = 54

To represent the given coordinate relationships using the exponential model parameters.

2

Divide the equation for

g\left(\frac{3}{2}\right)

by the equation for

g\left(\frac{1}{2}\right)

to eliminate

p

and solve for

q

.

q = 9

Using the exponent quotient rule,

\frac{q^{3/2}}{q^{1/2}} = q^{3/2 - 1/2} = q^1 = q

. Dividing the constants gives

\frac{54}{6} = 9

.

3

Substitute

q = 9

back into the first equation and solve for

p

.

p = 2

Evaluating the fractional exponent gives

9^{1/2} = \sqrt{9} = 3

. Substituting this in yields

3p = 6

, which simplifies to

p = 2

.

4

Evaluate the completed exponential function

g(x) = 2 \cdot 9^x

at

x = 2

.

162

Calculating

g(2) = 2 \cdot 9^2 = 2 \cdot 81 = 162

solves for the target value.

Anahtar Kavram

Solving a system of exponential equations using fractional exponent rules and base evaluation.

Soru 147Soru →

A radioactive substance decays according to the equation $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{8}}$ , where $N(t)$ is the amount of the substance remaining after $t$ days, and $N_0$ is the initial amount. If the substance decays to $\frac{1}{16}$ of its initial amount in $d$ days, what is the value of $d$ ?

32

4

2

128

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Cevap: 32

Cevap

32

The correct answer is 32. The remaining amount of the substance is given as

\frac{1}{16}

of the initial amount, so

\frac{N(d)}{N_0} = \frac{1}{16}

. Substituting

d

for

t

in the decay formula yields

\left(\frac{1}{2}\right)^{\frac{d}{8}} = \frac{1}{16}

. Since

\frac{1}{16} = \left(\frac{1}{2}\right)^4

, we can set the exponents equal to each other:

\frac{d}{8} = 4

. Solving for

d

gives

d = 32

.

Adım Adım Çözüm

1

Set up the equation for the remaining fraction of the substance.

\frac{N(d)}{N_0} = \frac{1}{16}

The problem states the substance decays to

\frac{1}{16}

of its initial amount

N_0

in

d

days.

2

Substitute the given decay formula into the equation.

\left(\frac{1}{2}\right)^{\frac{d}{8}} = \frac{1}{16}

By replacing

t

with

d

in the formula

\frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{8}}

, we get the expression for the remaining fraction.

3

Rewrite

\frac{1}{16}

as a power with base

\frac{1}{2}

to solve for the exponent.

\left(\frac{1}{2}\right)^4 = \frac{1}{16}

Since

2^4 = 16

, the fraction

\frac{1}{16}

is equivalent to

\left(\frac{1}{2}\right)^4

.

4

Equate the exponents and solve for

d

.

\frac{d}{8} = 4 \implies d = 32

Since the bases are equal, their exponents must be equal. Multiplying both sides by 8 isolates

d

.

Anahtar Kavram

Solving exponential equations by equating bases

Soru 148Soru →

In the system of equations below, $x$ and $y$ are real numbers.

4^x \cdot 8^y = 128

9^x \cdot 3^y = 243

What is the value of $x$ ?

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Cevap: 2

Cevap

The value of

x

is

2

.

To solve the system of equations, express all terms in each equation using common bases. In the first equation,

4^x \cdot 8^y = 128

can be written as

(2^2)^x \cdot (2^3)^y = 2^7

, which simplifies to

2^{2x+3y} = 2^7

, meaning

2x + 3y = 7

. In the second equation,

9^x \cdot 3^y = 243

can be written as

(3^2)^x \cdot 3^y = 3^5

, which simplifies to

3^{2x+y} = 3^5

, meaning

2x + y = 5

. Subtracting the two equations gives

(2x + 3y) - (2x + y) = 7 - 5 \implies 2y = 2 \implies y = 1

. Substituting

y = 1

back into

2x + y = 5

gives

2x + 1 = 5 \implies 2x = 4 \implies x = 2

.

Adım Adım Çözüm

1

Rewrite the first equation using a common base of 2.

2x + 3y = 7

By converting

4^x

to

(2^2)^x = 2^{2x}

,

8^y

to

(2^3)^y = 2^{3y}

, and

128

to

2^7

, we can equate the exponents:

2x + 3y = 7

.

2

Rewrite the second equation using a common base of 3.

2x + y = 5

By converting

9^x

to

(3^2)^x = 3^{2x}

, keeping

3^y

, and converting

243

to

3^5

, we can equate the exponents:

2x + y = 5

.

3

Solve the system of linear equations for

y

.

y = 1

Subtracting

2x + y = 5

from

2x + 3y = 7

eliminates

x

, leaving

2y = 2

, which gives

y = 1

.

4

Substitute

y = 1

into one of the linear equations to solve for

x

.

x = 2

Substituting

y = 1

into

2x + y = 5

yields

2x + 1 = 5

, which simplifies to

2x = 4

, so

x = 2

.

Anahtar Kavram

Solving systems of exponential equations by converting to a common base and applying exponent laws.

Soru 149Soru →

Which of the following is equivalent to the expression $\frac{3x^2 - 5x - 4}{x - 2}$ for all $x \neq 2$ ?

3x - 1 - \frac{6}{x - 2}

3x + 1 - \frac{2}{x - 2}

3x + 1 + \frac{2}{x - 2}

3x - 5 - \frac{4}{x - 2}

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Cevap: $3x + 1 - \frac{2}{x - 2}$

Cevap

3x + 1 - \frac{2}{x - 2}

The correct answer represents the equivalent expression obtained by performing polynomial division on the rational expression. Dividing the numerator

3x^2 - 5x - 4

by the denominator

x - 2

yields a quotient of

3x + 1

and a remainder of

-2

. This can be written in the form of the quotient plus the remainder over the divisor, resulting in

3x + 1 - \frac{2}{x - 2}

.

Adım Adım Çözüm

1

Divide the leading term of the numerator,

3x^2

, by the leading term of the denominator,

x

.

The first term of the quotient is

3x

.

This starts the polynomial long division process.

2

Multiply

3x

by the divisor

(x - 2)

and subtract the result from the numerator.

(3x^2 - 5x - 4) - (3x^2 - 6x) = x - 4

.

Subtracting the multiplied term helps find the remainder of the first division step.

3

Divide the leading term of the remaining expression,

x

, by the leading term of the divisor,

x

.

The second term of the quotient is

1

.

To continue the division process with the remaining terms.

4

Multiply

1

by the divisor

(x - 2)

and subtract the result from

x - 4

.

(x - 4) - (x - 2) = -2

.

This step determines the final remainder of

-2

because the degree of the remainder is now less than the degree of the divisor.

5

Express the final result as the sum of the quotient and the remainder divided by the divisor.

3x + 1 - \frac{2}{x - 2}

To construct the equivalent algebraic expression.

Anahtar Kavram

Equivalent Algebraic Expressions

Soru 150Soru →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + bx + c$ has a vertex at $(h, k)$ . The graph of $f$ is translated $4$ units to the right and $5$ units up to form the graph of a quadratic function $g$ . If the graph of $g$ passes through the point $(1, 16)$ and has a $y$ -intercept of $13$ , what is the value of $k$ ?

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Cevap: 12

Cevap

The value of

k

is

12

.

By writing the function in vertex form as

f(x) = -(x-h)^2 + k

, we can apply the transformations directly. Translating the graph

4

units to the right and

5

units up gives the function

g(x) = f(x-4) + 5 = -(x-h-4)^2 + k+5

. Using the given points

g(0) = 13

and

g(1) = 16

, we set up a system of equations:

-(h+4)^2 + k+5 = 13

and

-(h+3)^2 + k+5 = 16

. Solving these simultaneously gives

h = -2

and

k = 12

.

Adım Adım Çözüm

1

Write the function

f(x)

in vertex form.

f(x) = -(x-h)^2 + k

Since the leading coefficient of

f(x) = -x^2 + bx + c

is

-1

, the vertex form must have a coefficient of

-1

.

2

Represent the translated function

g(x)

.

g(x) = -(x-h-4)^2 + k+5

Translating a function

f(x)

by

4

units to the right and

5

units up yields

g(x) = f(x-4) + 5

.

3

Use the

y

-intercept of

g(x)

to write the first equation.

k - (h+4)^2 = 8

The

y

-intercept is the value of the function at

x = 0

, so

g(0) = 13

.

4

Use the point

(1, 16)

on

g(x)

to write the second equation.

k - (h+3)^2 = 11

The graph passing through

(1, 16)

means

g(1) = 16

.

5

Solve the system of equations for

h

and

k

.

h = -2

and

k = 12

Equating the expressions for

k

gives

(h+4)^2 + 8 = (h+3)^2 + 11

, which simplifies to

2h = -4

, so

h = -2

. Substituting

h = -2

into

k = (h+4)^2 + 8

yields

k = 12

.

Anahtar Kavram

Translating quadratic functions and solving systems of quadratic equations.

Soru 151Soru →

\begin{aligned} (x - 4)^2 + (y + 1)^2 &= 18 \\ x + y &= k \end{aligned}

In the system of equations above, $k$ is a constant. If the system has exactly one real solution, what is the sum of all possible values of $k$ ?

-6

2

6

10

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Cevap: 6

Cevap

The sum of all possible values of

k

is

6

.

The system of equations has exactly one real solution when the line is tangent to the circle. The circle

(x - 4)^2 + (y + 1)^2 = 18

has center

(4, -1)

and radius

\sqrt{18} = 3\sqrt{2}

. The distance from the center of the circle to the line

x + y - k = 0

is given by

\frac{|4 - 1 - k|}{\sqrt{1^2 + 1^2}} = \frac{|3 - k|}{\sqrt{2}}

. Setting this distance equal to the radius yields

\frac{|3 - k|}{\sqrt{2}} = 3\sqrt{2} \implies |3 - k| = 6

. Solving this equation gives

k = 9

and

k = -3

. The sum of these possible values is

9 + (-3) = 6

.

Adım Adım Çözüm

1

Relate the algebraic condition of having exactly one solution to the geometric relationship between the circle and the line.

The line must be tangent to the circle, meaning the perpendicular distance from the center of the circle to the line must equal the radius of the circle.

A system consisting of a circle and a line has exactly one real solution if and only if the line is tangent to the circle.

2

Identify the center and radius of the circle from the equation

(x - 4)^2 + (y + 1)^2 = 18

.

The center of the circle is

(4, -1)

and the radius is

r = \sqrt{18} = 3\sqrt{2}

.

The standard form of a circle equation is

(x - h)^2 + (y - k)^2 = r^2

, where

(h, k)

is the center and

r

is the radius.

3

Express the line equation in standard form

Ax + By + C = 0

and set up the perpendicular distance formula.

The line equation is

x + y - k = 0

. The distance from the center

(4, -1)

to the line is

d = \frac{|(1)(4) + (1)(-1) - k|}{\sqrt{1^2 + 1^2}} = \frac{|3 - k|}{\sqrt{2}}

.

The perpendicular distance from a point

(x_0, y_0)

to a line

Ax + By + C = 0

is given by

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

.

4

Set the distance equal to the radius of the circle and solve the absolute value equation for

k

.

\frac{|3 - k|}{\sqrt{2}} = 3\sqrt{2} \implies |3 - k| = 6

. This yields two equations:

3 - k = 6 \implies k = -3

, and

3 - k = -6 \implies k = 9$.

Setting the distance equal to the radius ensures that the line is tangent to the circle, resulting in exactly one point of intersection.

5

Calculate the sum of all possible values of

k

.

Sum =

-3 + 9 = 6

.

The question asks for the sum of all values of

k

that satisfy the tangency condition.

Anahtar Kavram

Determining the relationship between a line and a circle in a system of equations by setting the perpendicular distance from the center of the circle equal to its radius.

Alternatif Yöntem

Substitute

y = k - x

into the circle equation to obtain a quadratic equation in

x

:

(x - 4)^2 + (k - x + 1)^2 = 18

. Expanding and combining like terms yields

2x^2 - 2(k + 5)x + (k^2 + 2k - 1) = 0

. For the system to have exactly one real solution, the discriminant

\Delta

of this quadratic equation must be zero:

\Delta = [-2(k + 5)]^2 - 4(2)(k^2 + 2k - 1) = 0

. Simplifying this equation leads to

4(k^2 + 10k + 25) - 8(k^2 + 2k - 1) = 0 \implies k^2 - 6k - 27 = 0

. By Vieta's formulas, the sum of the solutions to this quadratic equation for

k

is given by

-\frac{b}{a} = -\frac{-6}{1} = 6

.

Tahmini Süre:2m 30s

Soru 152Soru →

A system of equations consists of the quadratic function $f(x) = -x^2 + 6x - 2$ and the linear function $g(x) = 2x + 1$ . If the graphs of $f$ and $g$ intersect at the points $(x_1, y_1)$ and $(x_2, y_2)$ in the $xy$ -plane, what is the value of $y_1 + y_2$ ?

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Cevap: 10

Cevap

10

By setting the two equations equal to each other, we obtain

-x^2 + 6x - 2 = 2x + 1

. Rearranging terms to one side gives

x^2 - 4x + 3 = 0

, which factors as

(x-1)(x-3) = 0

. The x-coordinates of the intersection points are

x = 1

and

x = 3

. Substituting these into the linear equation gives the y-coordinates: for

x = 1

,

y = 2(1) + 1 = 3

; for

x = 3

,

y = 2(3) + 1 = 7

. Summing these y-coordinates gives

3 + 7 = 10

.

Adım Adım Çözüm

1

Set the quadratic function equal to the linear function to find the x-coordinates of the intersection points.

-x^2 + 6x - 2 = 2x + 1

At the points of intersection, the y-values of both functions must be equal.

2

Rearrange the equation into standard quadratic form and solve for

x

by factoring.

x^2 - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0 \Rightarrow x = 1 \text{ or } x = 3

Factoring the quadratic equation gives the x-coordinates of the intersection points.

3

Substitute the x-coordinates back into the linear equation to find the corresponding y-coordinates.

For

x_1 = 1

,

y_1 = 2(1) + 1 = 3

. For

x_2 = 3

,

y_2 = 2(3) + 1 = 7

.

The linear equation

g(x) = 2x + 1

is simpler to use for finding the y-coordinates than the quadratic equation.

4

Sum the two y-coordinates.

y_1 + y_2 = 3 + 7 = 10

The question asks for the sum of the y-coordinates of the intersection points.

Anahtar Kavram

Solving a system consisting of a linear equation and a quadratic equation by substitution or equating them.

Soru 153Soru →

The quadratic function $f$ is defined by $f(x) = 3(x - 2)^2 + 5$ . In the $xy$ -plane, the graph of $y = f(x)$ intersects the $y$ -axis at the point $(0, c)$ . What is the value of $c$ ?

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

The correct answer is 17.

To find the

y

-intercept of the graph of

y = f(x)

, we substitute

x = 0

into the equation:

f(0) = 3(0 - 2)^2 + 5

. Simplifying the expression inside the parentheses gives

3(-2)^2 + 5

. Squaring

-2

gives

4

, making the expression

3(4) + 5 = 12 + 5 = 17

. Therefore, the graph intersects the

y

-axis at the point

(0, 17)

, which means the value of

c

is 17.

Adım Adım Çözüm

1

Set

x = 0

to find the point where the graph intersects the

y

-axis.

f(0) = 3(0 - 2)^2 + 5

The

y

-intercept of any graph in the

xy

-plane occurs when the

x

-coordinate is equal to

0

.

2

Perform the arithmetic evaluation to find the value of

f(0)

.

f(0) = 17

First, evaluate the term inside the parentheses:

0 - 2 = -2

. Squaring

-2

gives

4

. Then, multiply

3

by

4

to get

12

. Finally, add

5

to obtain

17

.

Anahtar Kavram

Evaluating a quadratic function at

x = 0

to determine its

y

-intercept.

Soru 154Soru →

In the quadratic equation $x^2 + bx + c = 0$ , $b$ and $c$ are integers such that $b + c = 10$ . If the equation has no real solutions, what is the maximum possible value of $c$ ?

18

19

14

6

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The maximum possible value of

c

is 18.

For the quadratic equation to have no real solutions, the discriminant

b^2 - 4ac

must be negative. Given

a=1

, this simplifies to

b^2 - 4c < 0

. Using the constraint

b + c = 10

, we can express

c

as

10 - b

. Substituting this into the inequality gives

b^2 - 4(10 - b) < 0

, which simplifies to

b^2 + 4b - 40 < 0

. Solving the boundary equation

b^2 + 4b - 40 = 0

using the quadratic formula gives

b = -2 \pm \sqrt{44}

. Since

\sqrt{44} \approx 6.63

, the boundary values of

b

are approximately

-8.63

and

4.63

. Thus, the inequality holds for

-8.63 < b < 4.63

. To maximize

c = 10 - b

, we need the smallest integer value for

b

in this range, which is

b = -8

. Substituting

b = -8

gives the maximum value of

c = 10 - (-8) = 18

.

Adım Adım Çözüm

1

State the condition for the quadratic equation to have no real solutions.

The discriminant must be negative:

b^2 - 4ac < 0

. Since

a = 1

, this simplifies to

b^2 - 4c < 0

.

A quadratic equation has no real solutions if and only if its discriminant is strictly less than zero.

2

Express

c

in terms of

b

using the given equation

b + c = 10

, and substitute it into the discriminant inequality.

Since

c = 10 - b

, substituting gives

b^2 - 4(10 - b) < 0

, which simplifies to

b^2 + 4b - 40 < 0

.

Substituting one variable in terms of the other allows us to solve a single-variable quadratic inequality.

3

Find the boundary values of

b

by solving the equation

b^2 + 4b - 40 = 0

using the quadratic formula.

b = \frac{-4 \pm \sqrt{4^2 - 4(1)(-40)}}{2} = \frac{-4 \pm \sqrt{176}}{2} = -2 \pm \sqrt{44}

. Since

\sqrt{44} \approx 6.63

, the boundaries are approximately

b \approx -8.63

and

b \approx 4.63

.

The roots of the quadratic equation define the boundaries of the interval that satisfies the inequality.

4

Determine the interval for

b

that satisfies the inequality and identify the value of

b

that maximizes

c

.

The inequality

b^2 + 4b - 40 < 0

is satisfied when

-8.63 < b < 4.63

. Since

c = 10 - b

, maximizing

c

requires selecting the smallest possible integer value of

b

in this range, which is

b = -8

.

Subtracting a smaller number from 10 results in a larger value for

c

.

5

Calculate the maximum value of

c

using the optimal value of

b

.

c = 10 - (-8) = 18

.

Substituting the minimum integer value of

b

gives the maximum integer value of

c

.

Anahtar Kavram

Analyzing the discriminant of a quadratic equation to determine the nature of its roots and using substitution to solve optimization problems under constraints.

Soru 155Soru →

If $z$ is a positive number such that $z^2 - 4z - 12 = 0$ , what is the value of $z$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The correct answer is 6.

To find the positive value of

z

that satisfies the equation

z^2 - 4z - 12 = 0

, we factor the quadratic expression. Finding two numbers that multiply to

-12

and add to

-4

gives

-6

and

2

. Thus, the equation can be factored as

(z - 6)(z + 2) = 0

. Setting each factor to zero yields the solutions

z = 6

and

z = -2

. Since the problem states that

z

is a positive number, we discard the negative solution, leaving

6

as the final answer.

Adım Adım Çözüm

1

Factor the quadratic equation

z^2 - 4z - 12 = 0

.

(z - 6)(z + 2) = 0

To solve the quadratic equation by finding two numbers that multiply to -12 and add to -4.

2

Solve for the roots of the equation.

z = 6

or

z = -2

Set each factor equal to zero and solve for

z

.

3

Apply the constraint that

z

must be positive.

z = 6

Discard the negative solution

z = -2

as the problem specifies

z

must be a positive number.

Anahtar Kavram

Solving quadratic equations by factoring

Soru 156Soru →

Which of the following is a solution to the equation $x^2 - 2x - 8 = 0$ ?

-4

2

4

-8

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The correct solution is 4.

To find the solutions to the equation

x^2 - 2x - 8 = 0

, we factor the quadratic expression on the left side of the equation. We need two numbers that multiply to

-8

and add to

-2

. These numbers are

-4

and

2

. Thus, the equation can be rewritten in factored form as

(x - 4)(x + 2) = 0

. Setting each factor to zero yields the solutions

x = 4

and

x = -2

. Among the given options,

4

is the only correct solution.

Adım Adım Çözüm

1

Factor the quadratic expression

x^2 - 2x - 8

.

The expression factors into

(x - 4)(x + 2)

.

We look for two numbers that multiply to the constant term

-8

and add to the linear coefficient

-2

. These numbers are

-4

and

2

.

2

Set each factor equal to zero to solve for

x

.

x - 4 = 0

or

x + 2 = 0

, which gives

x = 4

or

x = -2

.

By the zero product property, if the product of two factors is zero, then at least one of the factors must be zero.

3

Identify the solution that appears in the options.

x = 4

is a solution.

Checking the given options,

4

is listed while

-2

is not.

Anahtar Kavram

Solving quadratic equations by factoring

Soru 157Soru →

If $27^{2x - 1} \cdot 9^{y + 2} = 3^{13}$ and $8^{x - y} = 2^{y - 9}$ , what is the value of $x + y$ ?

2

4

6

9

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

To solve the system, we convert all exponential terms to their prime bases. The first equation becomes

3^{6x - 3} \cdot 3^{2y + 4} = 3^{13}

, which simplifies to

3^{6x + 2y + 1} = 3^{13}

. Equating the exponents yields

6x + 2y + 1 = 13

, which simplifies to

3x + y = 6

. The second equation becomes

2^{3x - 3y} = 2^{y - 9}

. Equating these exponents yields

3x - 4y = -9

. Solving this system of equations gives

x = 1

and

y = 3

. The sum of these values is

1 + 3 = 4

.

Adım Adım Çözüm

1

Convert the bases of the first equation to a common base of 3.

Since

27 = 3^3

and

9 = 3^2

, we can rewrite the equation

27^{2x - 1} \cdot 9^{y + 2} = 3^{13}

as

(3^3)^{2x - 1} \cdot (3^2)^{y + 2} = 3^{13}

. Applying the power of a power rule, this simplifies to

3^{6x - 3} \cdot 3^{2y + 4} = 3^{13}

. Using the product rule of exponents, we add the exponents to get

3^{6x + 2y + 1} = 3^{13}

.

This allows us to equate the exponents and form a linear equation.

2

Set the exponents equal to each other to obtain the first linear equation.

6x + 2y + 1 = 13 \implies 6x + 2y = 12 \implies 3x + y = 6

.

Since the bases are identical on both sides, their exponents must be equal.

3

Convert the bases of the second equation to a common base of 2 and equate their exponents.

Since

8 = 2^3

, rewrite

8^{x - y} = 2^{y - 9}

as

(2^3)^{x - y} = 2^{y - 9} \implies 2^{3x - 3y} = 2^{y - 9}

. Equating the exponents gives

3x - 3y = y - 9 \implies 3x - 4y = -9

.

This gives us a second linear equation to form a system of equations.

4

Solve the system of equations for the variables.

From the first equation, we have

3x = 6 - y

. Substituting this into the second equation gives

(6 - y) - 4y = -9 \implies 6 - 5y = -9 \implies -5y = -15 \implies y = 3

. Substituting

y = 3

back into the first equation yields

3x + 3 = 6 \implies 3x = 3 \implies x = 1

.

Solving the system of linear equations provides the individual values of the variables.

5

Calculate the sum of the variables.

x + y = 1 + 3 = 4

.

This answers the question asking for the sum of the variables.

Anahtar Kavram

Expressing exponential terms with a common base to form and solve a system of linear equations.

Soru 158Soru →

A certain substance decays radioactively such that the mass of the substance, in grams, remaining after $t$ days is modeled by the function $M(t) = 80 \cdot 2^{-\frac{t}{5}}$ . After how many days will the mass of the substance be $10$ grams?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

15

To find the number of days after which the mass of the substance is

10

grams, we substitute

10

for

M(t)

in the given function, yielding

10 = 80 \cdot 2^{-\frac{t}{5}}

. Dividing both sides by

80

gives

\frac{10}{80} = 2^{-\frac{t}{5}}

, which simplifies to

\frac{1}{8} = 2^{-\frac{t}{5}}

. Since

\frac{1}{8}

can be written as

2^{-3}

, the equation becomes

2^{-3} = 2^{-\frac{t}{5}}

. Because the bases are the same, we equate the exponents:

-3 = -\frac{t}{5}

. Multiplying both sides by

-5

gives

t = 15

.

Adım Adım Çözüm

1

Set the mass

M(t)

equal to

10

in the given function.

10 = 80 \cdot 2^{-\frac{t}{5}}

We want to find the value of

t

when the remaining mass of the substance is

10

grams.

2

Divide both sides of the equation by

80

.

\frac{1}{8} = 2^{-\frac{t}{5}}

To isolate the exponential expression.

3

Express both sides of the equation with a common base of

2

.

2^{-3} = 2^{-\frac{t}{5}}

Since

8 = 2^3

, the fraction

\frac{1}{8}

can be written as

2^{-3}

. Having the same base on both sides allows us to equate the exponents.

4

Set the exponents equal to each other and solve for

t

.

t = 15

Because the bases are equal, the exponents must be equal:

-3 = -\frac{t}{5}

.

Anahtar Kavram

Solving exponential equations by finding a common base.

Soru 159Soru →

If $x > 0$ and $x^2 - 3x - 18 = 0$ , what is the value of $x + 2$ ?

5

6

8

-1

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

Factoring the quadratic equation

x^2 - 3x - 18 = 0

yields

(x - 6)(x + 3) = 0

. This gives two solutions:

x = 6

and

x = -3

. Since the condition specifies

x > 0

, we choose

x = 6

. The question asks for the value of

x + 2

, which is

6 + 2 = 8

.

Adım Adım Çözüm

1

Factor the quadratic equation

x^2 - 3x - 18 = 0

by finding two numbers that multiply to

-18

and add to

-3

.

(x - 6)(x + 3) = 0

Factoring allows us to find the individual roots of the quadratic equation.

2

Solve for the possible values of

x

by setting each factor equal to zero.

x = 6

or

x = -3

If the product of two factors is zero, then at least one of the factors must equal zero.

3

Apply the given constraint that

x > 0

.

x = 6

The problem specifies that

x

must be a positive number, so the negative root

x = -3

must be discarded.

4

Substitute the value of

x

into the expression

x + 2

.

8

The question asks for the value of

x + 2

, so we evaluate

6 + 2

.

Anahtar Kavram

Solving quadratic equations by factoring and applying constraints on roots

Tahmini Süre:1m 0s

Soru 160Soru →

For all $x > 1$ , the expression $\frac{2x^2 + 7x - 4}{x^2 - 1} \cdot \frac{x - 1}{2x - 1}$ is equivalent to $\frac{x+k}{x+1}$ , where $k$ is a constant. What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of the constant

k

is 4.

Factoring the numerator

2x^2 + 7x - 4

yields

(2x - 1)(x + 4)

and factoring the denominator

x^2 - 1

yields

(x - 1)(x + 1)

. Substituting these factored forms into the given product gives

\frac{(2x - 1)(x + 4)}{(x - 1)(x + 1)} \cdot \frac{x - 1}{2x - 1}

. Canceling the common factors

(2x - 1)

and

(x - 1)

simplifies the expression to

\frac{x + 4}{x + 1}

. Comparing this to

\frac{x + k}{x + 1}

shows that

k = 4

.

Adım Adım Çözüm

1

Factor the quadratic expression in the numerator:

2x^2 + 7x - 4

.

(2x - 1)(x + 4)

Factoring the numerator helps identify common factors that can be simplified.

2

Factor the difference of squares in the denominator:

x^2 - 1

.

(x - 1)(x + 1)

Factoring the denominator helps identify common factors that can be simplified.

3

Multiply the rational expressions and cancel out the common factors.

\frac{x + 4}{x + 1}

Since

x > 1

, the terms

(2x - 1)

and

(x - 1)

are not equal to zero and can be canceled.

4

Compare the resulting expression with

\frac{x + k}{x + 1}

to find the value of

k

.

k = 4

By matching the numerators of the equivalent expressions,

x + 4 = x + k

, which gives

k = 4

.

Anahtar Kavram

Factoring and simplifying products of rational expressions

Advanced Math

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