Quadratic Equations Practice Questions — SAT

Question 1Question →

In the quadratic equation $2x^2 - 16x + c = 0$ , $c$ is a constant. If one of the solutions to the equation is $4 - \sqrt{3}$ , what is the value of $c$ ?

Show answer & explanation

Answer: 26

Answer

26

Since the coefficients of the quadratic equation are real, any irrational roots must occur in conjugate pairs. Therefore, if one solution is

4 - \sqrt{3}

, the other solution must be

4 + \sqrt{3}

. According to Vieta's formulas, the product of the roots

r_1

and

r_2

for a quadratic equation

ax^2 + bx + c = 0

is

\frac{c}{a}

. In this equation,

a = 2

, so the product of the roots is

\frac{c}{2}

. Multiplying the two solutions gives

(4 - \sqrt{3})(4 + \sqrt{3}) = 16 - 3 = 13

. Setting

\frac{c}{2} = 13

and solving for

c

yields

c = 26

.

Step-by-Step Solution

1

Determine the second root of the quadratic equation.

The second root is

4 + \sqrt{3}

.

Since the quadratic equation has real coefficients, the irrational roots must be conjugate pairs.

2

Express the product of the roots using Vieta's formulas.

The product of the roots is

\frac{c}{2}

.

For any quadratic equation

ax^2 + bx + c = 0

, the product of the roots is equal to

\frac{c}{a}

. Here,

a = 2

.

3

Calculate the product of the two roots.

(4 - \sqrt{3})(4 + \sqrt{3}) = 16 - 3 = 13

.

Using the difference of squares formula,

(a-b)(a+b) = a^2 - b^2

.

4

Equate the product of the roots to the expression from step 2 and solve for

c

.

\frac{c}{2} = 13 \Rightarrow c = 26

.

By substituting the calculated product into the formula for the product of the roots, we find the value of the constant.

Key Concept

Using Vieta's formulas and the conjugate root theorem to solve for coefficients of a quadratic equation.

Alternative Method

Alternatively, substitute the given solution

x = 4 - \sqrt{3}

directly into the equation

2x^2 - 16x + c = 0

and solve for

c

. First, calculate

x^2 = (4 - \sqrt{3})^2 = 16 - 8\sqrt{3} + 3 = 19 - 8\sqrt{3}

. Then substitute this into the equation:

2(19 - 8\sqrt{3}) - 16(4 - \sqrt{3}) + c = 0

. Simplifying this yields

38 - 16\sqrt{3} - 64 + 16\sqrt{3} + c = 0 \Rightarrow -26 + c = 0 \Rightarrow c = 26

.

Estimated Time:1m 30s

Question 2Question →

What are the solutions to the quadratic equation $x^2 - 4x - 12 = 0$ ?

x = -6

and

x = 2

x = 6

and

x = -2

x = 6

and

x = 2

x = -6

and

x = -2

Show answer & explanation

Answer: $x = 6$ and $x = -2$

Answer

The solutions are

x = 6

and

x = -2

.

The correct answer is

x = 6

and

x = -2

. Factoring the quadratic equation

x^2 - 4x - 12 = 0

involves finding two numbers whose product is

-12

and whose sum is

-4

. These numbers are

-6

and

2

. Rewriting the equation as

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

and setting each factor to zero yields the solutions

x = 6

and

x = -2

.

Step-by-Step Solution

1

Identify the quadratic equation to factor.

The equation is

x^2 - 4x - 12 = 0

.

To find the solutions, we need to rewrite the quadratic expression in factored form.

2

Find two numbers that multiply to

-12

and add up to

-4

.

The numbers are

-6

and

2

.

These numbers satisfy both conditions:

(-6) \times 2 = -12

and

-6 + 2 = -4

.

3

Write the quadratic equation in factored form and solve for

x

.

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

, which means

x - 6 = 0

or

x + 2 = 0

. Thus,

x = 6

or

x = -2

.

Applying the zero product property allows us to solve for the individual values of

x

.

Key Concept

Solving quadratic equations by factoring

Question 3Question →

In the quadratic equation $x^2 - kx + 9 = 0$ , $k$ is a positive constant. If the equation has exactly one real solution, what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

6

For a quadratic equation

ax^2 + bx + c = 0

to have exactly one real solution, its discriminant must equal zero (

b^2 - 4ac = 0

). In the equation

x^2 - kx + 9 = 0

, the coefficients are

a = 1

,

b = -k

, and

c = 9

. Setting the discriminant to zero gives

(-k)^2 - 4(1)(9) = 0

, which simplifies to

k^2 - 36 = 0

. Solving this equation yields

k^2 = 36

, so

k = 6

or

k = -6

. Since

k

is a positive constant, we reject the negative root, leaving

k = 6

.

Step-by-Step Solution

1

Identify the condition for a quadratic equation to have exactly one real solution.

The discriminant of the quadratic equation must be equal to 0, which is represented by the formula

b^2 - 4ac = 0

.

The discriminant determines the number of real solutions of a quadratic equation. If the discriminant is 0, there is exactly one real solution.

2

Identify the coefficients

a

,

b

, and

c

of the given equation

x^2 - kx + 9 = 0

.

a = 1

,

b = -k

, and

c = 9

.

These coefficients are required to compute the value of the discriminant.

3

Set the discriminant equal to 0 and simplify the equation.

(-k)^2 - 4(1)(9) = 0 \Rightarrow k^2 - 36 = 0

.

Substituting the coefficients into the discriminant formula sets up the algebraic relationship to find

k

.

4

Solve for the positive constant

k

.

k^2 = 36 \Rightarrow k = 6

or

k = -6

. Since

k

must be positive,

k = 6

.

Solving the equation gives two possible values, but the negative solution is discarded because the problem specifies

k

is a positive constant.

Key Concept

Discriminant of a quadratic equation

Alternative Method

Alternatively, a quadratic equation has exactly one real solution if it can be written as a perfect square trinomial in the form

(x - d)^2 = 0

, which expands to

x^2 - 2dx + d^2 = 0

. Comparing this with

x^2 - kx + 9 = 0

, we get

d^2 = 9

and

2d = k

. Since

d^2 = 9

,

d

can be

3

or

-3

. Given that

k

is positive and

k = 2d

,

d

must also be positive, meaning

d = 3

. Substituting this back gives

k = 2(3) = 6

.

Estimated Time:45s

Question 4Question →

In the quadratic equation $3x^2 - kx + 12 = 0$ , $k$ is a constant. The equation has two distinct real solutions, and the difference between these two solutions is less than $2$ . Which of the following describes all possible values of $k$ ?

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

-6\sqrt{5} < k < 6\sqrt{5}

-12\sqrt{2} < k < -12

or

12 < k < 12\sqrt{2}

k < -12

or

k > 12

Show answer & explanation

Answer: $-6\sqrt{5} < k < -12$ or $12 < k < 6\sqrt{5}$

Answer

The possible values of

k

are

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

The correct answer is found by combining two constraints. First, for the quadratic equation to have two distinct real solutions, the discriminant must be strictly positive:

(-k)^2 - 4(3)(12) > 0 \Rightarrow k^2 > 144

, which means

|k| > 12

. Second, the difference between the roots of

ax^2 + bx + c = 0

is

\frac{\sqrt{b^2 - 4ac}}{|a|}

. Here, the difference is

\frac{\sqrt{k^2 - 144}}{3}

. Setting this difference to be less than 2 gives

\frac{\sqrt{k^2 - 144}}{3} < 2 \Rightarrow \sqrt{k^2 - 144} < 6 \Rightarrow k^2 - 144 < 36 \Rightarrow k^2 < 180

, which means

|k| < 6\sqrt{5}

. Combining these inequalities yields

12 < |k| < 6\sqrt{5}

, which translates to

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

.

Step-by-Step Solution

1

Find the condition for the quadratic equation to have two distinct real solutions.

k^2 > 144

, which means

k < -12

or

k > 12

.

For the equation

3x^2 - kx + 12 = 0

to have two distinct real solutions, its discriminant

\Delta = b^2 - 4ac

must be strictly positive. Here,

\Delta = (-k)^2 - 4(3)(12) = k^2 - 144 > 0

.

2

Express the difference between the two solutions using the quadratic formula.

|x_1 - x_2| = \frac{\sqrt{k^2 - 144}}{3}

.

The solutions to the quadratic equation are given by

x = \frac{k \pm \sqrt{k^2 - 144}}{6}

. The difference between these solutions is

x_1 - x_2 = \frac{(k + \sqrt{k^2 - 144}) - (k - \sqrt{k^2 - 144})}{6} = \frac{2\sqrt{k^2 - 144}}{6} = \frac{\sqrt{k^2 - 144}}{3}

.

3

Apply the condition that the difference between the solutions is less than 2.

k^2 < 180

, which means

-6\sqrt{5} < k < 6\sqrt{5}

.

We set the difference expression to be less than 2:

\frac{\sqrt{k^2 - 144}}{3} < 2 \Rightarrow \sqrt{k^2 - 144} < 6

. Squaring both sides gives

k^2 - 144 < 36 \Rightarrow k^2 < 180

. Since

\sqrt{180} = 6\sqrt{5}

, this yields

-6\sqrt{5} < k < 6\sqrt{5}

.

4

Combine the inequalities from Step 1 and Step 3 to find the final overlapping range for

k

.

-6\sqrt{5} < k < -12

or

12 < k < 6\sqrt{5}

.

We must satisfy both

k^2 > 144

(for real solutions) and

k^2 < 180

(for the difference to be less than 2). This gives

144 < k^2 < 180

, which corresponds to the union of intervals

-6\sqrt{5} < k < -12

and

12 < k < 6\sqrt{5}

.

Key Concept

Using the discriminant and quadratic formula to analyze properties of roots under inequality constraints.

Question 5Question →

If $(2x - 3)(x + 4) = 0$ and $x > 0$ , what is the value of $x$ ?

-4

\frac{3}{2}

-\frac{3}{2}

4

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

The value of

x

is

\frac{3}{2}

.

To find the solutions to the equation

(2x - 3)(x + 4) = 0

, we set each factor equal to zero. Setting

2x - 3 = 0

gives

x = \frac{3}{2}

, and setting

x + 4 = 0

gives

x = -4

. The problem states that

x > 0

, meaning the value of

x

must be positive. Therefore, the only valid solution is

\frac{3}{2}

.

Step-by-Step Solution

1

Set each factor of the quadratic equation equal to zero using the zero product property.

2x - 3 = 0

or

x + 4 = 0

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

2

Solve each linear equation to find the possible values of

x

.

x = \frac{3}{2}

or

x = -4

Isolating

x

in each equation gives the roots of the quadratic equation.

3

Apply the given constraint

x > 0

to identify the correct solution.

x = \frac{3}{2}

is the only valid solution because

\frac{3}{2} > 0

and

-4 \ngtr 0

.

The problem restricts the solution to values of

x

that are strictly greater than zero, so the negative root must be discarded.

Key Concept

Solving factored quadratic equations with inequality constraints

Question 6Question →

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

-5

-4

5

9

Show answer & explanation

Answer: 5

Answer

5

To solve the equation

x(x - 9) = -20

, first distribute the

x

to get

x^2 - 9x = -20

. Next, rewrite the equation in standard form by adding

20

to both sides, which gives

x^2 - 9x + 20 = 0

. Factoring the quadratic expression yields

(x - 4)(x - 5) = 0

. Applying the zero product property gives two possible solutions:

x = 4

and

x = 5

. Therefore, the value

5

is a possible value of

x

.

Step-by-Step Solution

1

Distribute

x

on the left side of the equation.

x^2 - 9x = -20

Expanding the product allows the quadratic equation to be rewritten in standard form.

2

Add

20

to both sides to set the equation to standard form

ax^2 + bx + c = 0

.

x^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

20

and add to

-9

.

(x - 4)(x - 5) = 0

The numbers

-4

and

-5

satisfy these conditions, allowing the expression to be factored.

4

Set each factor equal to zero to find the possible values of

x

.

x = 4

or

x = 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 7Question →

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

Show answer & explanation

Answer: 6

Answer

The value of

p

is 6.

By Vieta's formulas, the sum of the roots of

x^2 - 4x + 2 = 0

is

r_1 + r_2 = 4

and the product of the roots is

r_1 r_2 = 2

. The sum of the roots of the second equation

x^2 - px + q = 0

is

p

. Therefore,

p = \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 + \frac{4}{2} = 6

.

Step-by-Step Solution

1

Determine the sum and product of the roots of the first equation.

r_1 + r_2 = 4

and

r_1 r_2 = 2

Vieta's formulas state that for

x^2 - Bx + C = 0

, the sum of the roots is

B

and the product is

C

.

2

Express the sum of the roots of the second equation, which is

p

.

p = (r_1 + r_2) + \left(\frac{1}{r_1} + \frac{1}{r_2}\right)

The sum of the roots of

x^2 - px + q = 0

is

p

.

3

Simplify the fractional part of the equation.

\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

p

.

p = 4 + \frac{4}{2} = 6

This evaluates the expression to find the final value of

p

.

Key Concept

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

Question 8Question →

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

Show answer & explanation

Answer: 4

Answer

The correct answer is

4

.

By taking the square root of both sides of the equation

(2y - 1)^2 = 49

, we get

2y - 1 = 7

or

2y - 1 = -7

. Solving

2y - 1 = 7

gives

2y = 8

, or

y = 4

. Solving

2y - 1 = -7

gives

2y = -6

, or

y = -3

. Since it is given that

y > 0

, the value of

y

must be

4

.

Step-by-Step Solution

1

Take the square root of both sides of the equation.

2y - 1 = 7

or

2y - 1 = -7

Applying the square root property of equality to isolate the linear term.

2

Solve each linear equation for

y

.

y = 4

or

y = -3

Isolating

y

in each equation by adding

1

and then dividing by

2

.

3

Apply the constraint

y > 0

to identify the final value.

y = 4

The value

y = -3

is negative, so it does not satisfy the constraint

y > 0

.

Key Concept

Solving quadratic equations of the form

(ax + b)^2 = c

using the square root property.

Question 9Question →

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

\frac{3}{4}

\frac{3}{2}

-\frac{3}{2}

3

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

The sum of all possible values of

a

is \frac{3}{2}.

The correct answer is derived by expressing the sum of the squares of the roots,

x_1^2 + x_2^2

, as

(x_1 + x_2)^2 - 2x_1x_2

. Applying Vieta's formulas to the equation

3x^2 - 2ax + a - 4 = 0

yields

x_1 + x_2 = \frac{2a}{3}

and

x_1 x_2 = \frac{a - 4}{3}

. Substituting these values into the identity gives

\frac{4a^2}{9} - \frac{2(a - 4)}{3} = \frac{28}{9}

. Multiplying by 9 and simplifying leads to the quadratic equation

2a^2 - 3a - 2 = 0

. Since this equation has real solutions and the original equation always has real roots for any real

a

, the sum of all possible values of

a

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

x_1^2 + x_2^2

in terms of the sum and product of the roots.

x_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

3x^2 - 2ax + a - 4 = 0

to find the sum and product of the roots.

x_1 + x_2 = \frac{2a}{3}

and

x_1 x_2 = \frac{a - 4}{3}

For any quadratic equation

Ax^2 + Bx + C = 0

, the sum of the roots is

-\frac{B}{A}

and the product is

\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.

\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

a

.

4

Simplify the algebraic equation by multiplying both sides by 9 and expanding terms.

4a^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

a

.

5

Verify that the solutions for

a

are real and that they yield real roots for

x

in the original equation.

For

2a^2 - 3a - 2 = 0

, the discriminant

\Delta_a = 25 > 0

, so

a

has real solutions. The discriminant of the original equation is

\Delta_x = 4a^2 - 12a + 48

, which is positive for all real values of

a

since its own discriminant is negative.

This confirms that the solutions for

a

are valid and satisfy the condition that

x_1

and

x_2

are real numbers.

6

Use Vieta's formula on the quadratic equation

2a^2 - 3a - 2 = 0

to find the sum of all possible values of

a

.

Sum of values

= -\frac{-3}{2} = \frac{3}{2}

The sum of the roots of a quadratic equation in the form

Aa^2 + Ba + C = 0

is equal to

-\frac{B}{A}

.

Key Concept

Vieta's Formulas and Algebraic Identities for Quadratic Equations

Question 10Question →

If $x^2 - 9x + 18 = 0$ and $x > 4$ , what is the value of $x$ ?

Show answer & explanation

Answer: 6

Answer

The value of

x

is 6.

Factoring the equation

x^2 - 9x + 18 = 0

gives

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

, which yields the solutions

x = 3

and

x = 6

. Applying the constraint

x > 4

, the only valid solution is 6.

Step-by-Step Solution

1

Factor the quadratic equation

x^2 - 9x + 18 = 0

.

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

Finding two numbers that multiply to 18 and add to -9 allows us to write the quadratic expression in its factored form.

2

Solve for the roots of the equation.

x = 3

or

x = 6

By the zero product property, setting each factor to zero yields the possible solutions for the equation.

3

Apply the given constraint

x > 4

.

x = 6

Since the question specifies that

x

must be greater than 4, the root

x = 3

is discarded, leaving

x = 6

as the only valid solution.

Key Concept

Solving quadratic equations by factoring and applying constraints.

Question 11Question →

The quadratic equation $3x^2 + 12x - 15 = 0$ has solutions $r$ and $s$ , where $r > s$ . What is the value of $r - s$ ?

4

6

-4

-6

Show answer & explanation

Answer: 6

Answer

6

The correct answer is 6. Dividing the given quadratic equation

3x^2 + 12x - 15 = 0

by 3 simplifies it to

x^2 + 4x - 5 = 0

. Factoring this expression gives

(x + 5)(x - 1) = 0

, which yields the solutions

x = 1

and

x = -5

. Since we are given that

r > s

, we define

r = 1

and

s = -5

. The difference between the roots is

r - s = 1 - (-5) = 1 + 5 = 6

.

Step-by-Step Solution

1

Divide the entire equation

3x^2 + 12x - 15 = 0

by 3 to simplify it.

x^2 + 4x - 5 = 0

Simplifying the equation makes it easier to factor by reducing the coefficients.

2

Factor the simplified quadratic equation

x^2 + 4x - 5 = 0

.

(x + 5)(x - 1) = 0

We look for two numbers that multiply to -5 and add to 4, which are 5 and -1.

3

Solve for the roots by setting each factor to zero, and assign the variables

r

and

s

such that

r > s

.

x = 1

and

x = -5

, meaning

r = 1

and

s = -5

.

Setting the factors to zero gives the solutions. Since the problem specifies r > s, the larger solution (1) is assigned to r and the smaller solution (-5) is assigned to s.

4

Calculate the value of

r - s

by substituting the values of

r

and

s

.

r - s = 1 - (-5) = 6

Subtracting s from r gives the final value of 6.

Key Concept

Solving quadratic equations by factoring and finding the difference between roots.

Alternative Method

Alternatively, Vieta's formulas can be used. For any quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

r + s = -b/a

and the product of the roots is

rs = c/a

. For

3x^2 + 12x - 15 = 0

, we find

r + s = -12/3 = -4

and

rs = -15/3 = -5

. The relationship between the sum, product, and difference of two numbers is given by the algebraic identity

(r - s)^2 = (r + s)^2 - 4rs

. Substituting our values gives

(r - s)^2 = (-4)^2 - 4(-5) = 16 + 20 = 36

. Since

r > s

, the difference

r - s

must be positive, so we take the positive square root:

r - s = \sqrt{36} = 6

.

Estimated Time:1m 0s

Question 12Question →

In the quadratic equation $x^2 - kx + 36 = 0$ , $k$ is a positive constant. If the equation has exactly one real solution, what is the value of $k$ ?

Show answer & explanation

Answer: 12

Answer

The value of

k

is

12

.

The quadratic equation

x^2 - kx + 36 = 0

has exactly one real solution when its discriminant is equal to zero. The discriminant of

ax^2 + bx + c = 0

is

b^2 - 4ac

. For this equation,

a = 1

,

b = -k

, and

c = 36

. Setting the discriminant to zero gives

(-k)^2 - 4(1)(36) = 0

, which simplifies to

k^2 - 144 = 0

. Solving for

k

gives

k^2 = 144

, so

k = 12

or

k = -12

. Since the problem states that

k

is a positive constant, the correct value of

k

is

12

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation

x^2 - kx + 36 = 0

.

a = 1

,

b = -k

, and

c = 36

.

To apply the discriminant formula, we need the values of

a

,

b

, and

c

.

2

Set the discriminant

\Delta = b^2 - 4ac

to

0

.

(-k)^2 - 4(1)(36) = 0

.

A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

3

Solve the equation for

k

.

k^2 - 144 = 0 \implies k^2 = 144 \implies k = 12

(since

k

is a positive constant).

Solving the equation yields

k = \pm 12

, and we select the positive value as specified in the problem.

Key Concept

Discriminant of a quadratic equation

Question 13Question →

In the quadratic equation $8x^2 - kx + 27 = 0$ , where $k$ is a positive constant, one of the roots is the square of the other root. What is the value of $k$ ?

Show answer & explanation

Answer: 30

Answer

30

For the quadratic equation

8x^2 - kx + 27 = 0

, Vieta's formulas state that the product of the roots is

\frac{27}{8}

and the sum of the roots is

\frac{k}{8}

. Letting the roots be

r

and

r^2

, we obtain the equation

r^3 = \frac{27}{8}

, which gives

r = \frac{3}{2}

. The other root is

r^2 = \frac{9}{4}

. Summing these roots gives

\frac{3}{2} + \frac{9}{4} = \frac{15}{4}

. Equating this sum to the formula for the sum of the roots yields

\frac{k}{8} = \frac{15}{4}

, which simplifies to

k = 30

.

Step-by-Step Solution

1

Determine the product of the roots using Vieta's formulas.

The product of the roots is

r_1 r_2 = \frac{27}{8}

.

For any quadratic equation

ax^2 + bx + c = 0

, the product of its roots is given by

\frac{c}{a}

.

2

Use the root relationship to solve for one of the roots.

r_2^3 = \frac{27}{8} \implies r_2 = \frac{3}{2}

.

Substituting

r_1 = r_2^2

into the product equation yields

r_2^3 = \frac{27}{8}

.

3

Determine the value of the second root.

r_1 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

.

The first root is the square of the second root.

4

Calculate the sum of the two roots.

r_1 + r_2 = \frac{9}{4} + \frac{3}{2} = \frac{15}{4}

.

The sum of the roots is needed to relate back to the linear coefficient of the quadratic equation.

5

Solve for the constant

k

using the sum of the roots.

\frac{k}{8} = \frac{15}{4} \implies k = 30

.

By Vieta's formulas, the sum of the roots is

-\frac{b}{a} = \frac{k}{8}

.

Key Concept

Quadratic Equations and Vieta's Formulas

Question 14Question →

In the quadratic equation $x^2 + px + q = 0$ , $p$ and $q$ are prime numbers. If the equation has two distinct integer roots, what is the value of $p + q$ ?

5

6

7

9

Show answer & explanation

Answer: 5

Answer

5

The correct answer is

5

. By Vieta's formulas, the sum of the roots is

-p

and the product of the roots is

q

. Since the roots are integers and

q

is prime, the roots must be

\{-1, -q\}

or

\{1, q\}

. The case of positive roots leads to

p + q = -1

, which is impossible for prime numbers. The case of negative roots leads to

p - q = 1

. The only prime numbers that differ by

1

are

3

and

2

. Therefore,

p = 3

and

q = 2

, and their sum is

5

.

Step-by-Step Solution

1

Apply Vieta's formulas to express the sum and product of the roots in terms of the coefficients.

For the equation

x^2 + px + q = 0

with distinct integer roots

r

and

s

, we have

r + s = -p

and

rs = q

.

Vieta's formulas state that for any quadratic equation

x^2 + bx + c = 0

with roots

r

and

s

,

r+s = -b

and

rs = c

.

2

Analyze the product equation

rs = q

using the properties of prime numbers.

Since

q

is prime, its only integer factors are

\pm 1

and

\pm q

. Thus, the integer roots

r

and

s

must be either

\{-1, -q\}

or

\{1, q\}

.

A prime number has no positive integer divisors other than 1 and itself.

3

Evaluate the first case where the roots are

1

and

q

.

If

r = 1

and

s = q

, then

r + s = 1 + q = -p

, which simplifies to

p + q = -1

. Since

p

and

q

are prime numbers, they must be positive (

p, q \ge 2

). Therefore,

p + q = -1

has no solution.

Prime numbers are positive integers greater than 1, so their sum cannot be negative.

4

Evaluate the second case where the roots are

-1

and

-q

.

If

r = -1

and

s = -q

, then

r + s = -1 - q = -p

, which simplifies to

p - q = 1

.

Substituting the negative roots into the sum equation yields a positive relationship between

p

and

q

.

5

Find the prime numbers

p

and

q

that satisfy

p - q = 1

.

The only consecutive prime numbers are

2

and

3

. Therefore,

q = 2

and

p = 3

. Both are prime numbers, and the equation

x^2 + 3x + 2 = 0

has distinct integer roots

-1

and

-2

.

Since all primes except 2 are odd, any two primes with a difference of 1 must include the only even prime, 2.

6

Calculate the sum of

p

and

q

.

The sum

p + q = 3 + 2 = 5

.

To answer the question, we add the two identified prime values.

Key Concept

Using Vieta's formulas and number theory properties of prime numbers to solve for coefficients of a quadratic equation.

Estimated Time:3m 0s

Question 15Question →

In the quadratic equation $3x^2 - kx + 24 = 0$ , $k$ is a positive constant. If one of the roots of the equation is twice the other root, what is the value of $k$ ?

6

12

18

36

Show answer & explanation

Answer: 18

Answer

18

The correct answer is 18. Let the roots of the equation be

r

and

2r

. The product of the roots is given by

\frac{c}{a} = \frac{24}{3} = 8

. Therefore,

2r^2 = 8

, which means

r^2 = 4

. Since

k > 0

, the roots must be positive, so

r = 2

and the two roots are 2 and 4. The sum of the roots is

2 + 4 = 6

. According to Vieta's formulas, the sum of the roots is also equal to

-\frac{b}{a} = \frac{k}{3}

. Setting the two expressions for the sum of the roots equal to each other gives

\frac{k}{3} = 6

, which simplifies to

k = 18

.

Step-by-Step Solution

1

Define the roots and set up the product of the roots using Vieta's formulas.

Let the roots be

r

and

2r

. The product of the roots is

r \times 2r = 2r^2

. From the equation

3x^2 - kx + 24 = 0

, the product of the roots is also

\frac{c}{a} = \frac{24}{3} = 8

.

This establishes a relationship between the given ratio of the roots and the coefficients of the equation.

2

Solve for the root variable

r

.

Setting

2r^2 = 8

gives

r^2 = 4

. Since

k

is a positive constant, the sum of the roots must be positive, which means the roots themselves must be positive. Thus,

r = 2

, and the roots are

2

and

4

.

Finding the actual values of the roots is necessary to calculate their sum.

3

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

k

.

The sum of the roots is

2 + 4 = 6

. According to Vieta's formulas, the sum of the roots is

-\frac{b}{a} = \frac{k}{3}

. Setting them equal gives

\frac{k}{3} = 6

, which simplifies to

k = 18

.

This directly isolates and solves for the unknown constant

k

.

Key Concept

Using Vieta's formulas to relate the roots of a quadratic equation

ax^2 + bx + c = 0

to its coefficients: the sum of the roots is

-\frac{b}{a}

and the product of the roots is

\frac{c}{a}

.

Question 16Question →

A projectile is launched from a platform. Its height $h(t)$ , in meters, $t$ seconds after launch is modeled by the function $h(t) = -5t^2 + bt + 12$ , where $b$ is a positive constant. If the projectile reaches a maximum height of $32$ meters, what is the value of $b$ ?

Show answer & explanation

Answer: 20

Answer

The correct answer is 20.

To find the maximum height of the projectile, we locate the vertex of the quadratic function

h(t) = -5t^2 + bt + 12

. The

t

-coordinate of the vertex is given by

t = -\frac{B}{2A} = -\frac{b}{2(-5)} = \frac{b}{10}

. Substituting

t = \frac{b}{10}

into

h(t)

and setting the height to

32

meters gives

-5\left(\frac{b}{10}\right)^2 + b\left(\frac{b}{10}\right) + 12 = 32

. Simplifying this equation results in

-\frac{b^2}{20} + \frac{b^2}{10} = 20

, which reduces to

\frac{b^2}{20} = 20

. Solving for

b

gives

b^2 = 400

. Because

b

must be a positive constant,

b = 20

.

Step-by-Step Solution

1

Identify the time

t

at which the maximum height occurs using the vertex formula

t = -\frac{B}{2A}

for the quadratic function

h(t) = At^2 + Bt + C

.

t = \frac{b}{10}

The vertex of a downward-opening parabola represents its maximum value. For

h(t) = -5t^2 + bt + 12

, the coefficients are

A = -5

and

B = b

.

2

Substitute the time

t = \frac{b}{10}

back into the height function

h(t)

and set the expression equal to the maximum height of

32

meters.

-5\left(\frac{b}{10}\right)^2 + b\left(\frac{b}{10}\right) + 12 = 32

At the maximum height, the height of the projectile is

32

meters, which corresponds to the value of the function at the vertex time.

3

Simplify the equation to solve for

b^2

.

\frac{b^2}{20} = 20

Squaring the fraction yields

-5\left(\frac{b^2}{100}\right) + \frac{b^2}{10} + 12 = 32

. Simplifying the coefficients leads to

-\frac{b^2}{20} + \frac{b^2}{10} = 20

, which simplifies to

\frac{b^2}{20} = 20

.

4

Solve the equation for

b

.

b = 20

Multiplying both sides by

20

gives

b^2 = 400

. Taking the square root of both sides gives

b = \pm 20

. Since

b

is specified to be a positive constant,

b = 20

.

Key Concept

Finding the vertex of a quadratic function to determine maximum values in a real-world context.

Estimated Time:2m 0s

Question 17Question →

For a constant $k$ , the quadratic equation $x^2 - 2kx + k^2 - 4k + 3 = 0$ has two distinct real solutions. If both solutions are positive, which of the following describes all possible values of $k$ ?

k > 3

or

\frac{3}{4} < k < 1

k > 3

k > \frac{3}{4}

k < 1

or

k > 3

Show answer & explanation

Answer: $k > 3$ or $\frac{3}{4} < k < 1$

Answer

k > 3

or

\frac{3}{4} < k < 1

The correct answer is the option that includes both the interval between

\frac{3}{4}

and

1

and the interval greater than

3

. To ensure the quadratic equation has two distinct real solutions, the discriminant must be positive, which yields

k > \frac{3}{4}

. To ensure both solutions are positive, their sum and product must both be positive. The sum

2k > 0

requires

k > 0

. The product

k^2 - 4k + 3 > 0

requires

k < 1

or

k > 3

. The intersection of all these inequalities is

\frac{3}{4} < k < 1

or

k > 3

.

Step-by-Step Solution

1

Determine the condition for two distinct real solutions using the discriminant.

k > \frac{3}{4}

For the quadratic equation

x^2 - 2kx + k^2 - 4k + 3 = 0

to have two distinct real solutions, its discriminant

\Delta

must be strictly greater than

0

. Here,

a = 1

,

b = -2k

, and

c = k^2 - 4k + 3

. Thus,

\Delta = (-2k)^2 - 4(1)(k^2 - 4k + 3) = 4k^2 - 4k^2 + 16k - 12 = 16k - 12

. Setting

\Delta > 0

gives

16k - 12 > 0

, which simplifies to

k > \frac{3}{4}

.

2

Set up conditions for both solutions to be positive using the sum and product of the roots.

2k > 0

and

k^2 - 4k + 3 > 0

If a quadratic equation has two positive solutions, their sum must be positive and their product must also be positive. By Vieta's formulas, the sum of the solutions is

-\frac{b}{a} = 2k

and the product of the solutions is

\frac{c}{a} = k^2 - 4k + 3

.

3

Solve the inequalities for the sum and product of the roots.

k > 0

and either

k < 1

or

k > 3

The sum condition

2k > 0

simplifies to

k > 0

. The product condition

k^2 - 4k + 3 > 0

factors as

(k-1)(k-3) > 0

, which is satisfied when

k < 1

or

k > 3

.

4

Find the intersection of all the derived inequalities.

\frac{3}{4} < k < 1

or

k > 3

We must find the values of

k

that satisfy all three conditions:

k > \frac{3}{4}

,

k > 0

, and (

k < 1

or

k > 3

). Combining

k > \frac{3}{4}

and

k > 0

yields

k > \frac{3}{4}

. Combining

k > \frac{3}{4}

with the constraint that

k

must be less than

1

or greater than

3

gives the combined solution of

\frac{3}{4} < k < 1

or

k > 3

.

Key Concept

Using the discriminant and Vieta's formulas to determine the signs of the roots of a quadratic equation dependent on a parameter.

Alternative Method

Alternatively, you can write the solutions using the quadratic formula:

x = \frac{2k \pm \sqrt{16k - 12}}{2} = k \pm \sqrt{4k - 3}

. For both roots to be positive and distinct, we first need the term under the radical to be positive, so

4k - 3 > 0 \implies k > \frac{3}{4}

. Then, we need the smaller root to be positive:

k - \sqrt{4k - 3} > 0 \implies k > \sqrt{4k - 3}

. Since

k > \frac{3}{4} > 0

, we can square both sides:

k^2 > 4k - 3 \implies k^2 - 4k + 3 > 0 \implies (k - 1)(k - 3) > 0

. This yields

k < 1

or

k > 3

. Combining this with

k > \frac{3}{4}

gives the same result:

\frac{3}{4} < k < 1

or

k > 3

.

Estimated Time:3m 0s

Question 18Question →

A toy rocket is launched vertically upward from ground level. Its height, $h$ , in meters, $t$ seconds after launch is modeled by the equation $h(t) = -5t^2 + 40t$ . How many seconds after launch does the rocket reach its maximum height?

8

4

80

20

Show answer & explanation

Answer: 4

Answer

4 seconds

The height of the rocket is modeled by the quadratic function

h(t) = -5t^2 + 40t

. Because the coefficient of

t^2

is negative (

a = -5

), the graph of this function is a parabola that opens downward, meaning its vertex represents the maximum height. The time

t

at which the vertex occurs is given by

t = -\frac{b}{2a}

. Substituting

a = -5

and

b = 40

gives

t = -\frac{40}{2(-5)} = 4

seconds.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation.

a = -5

,

b = 40

, and

c = 0

The equation is given in standard quadratic form

h(t) = at^2 + bt + c

.

2

Apply the vertex formula to find the time at which the maximum height occurs.

t = -\frac{b}{2a}

For a downward-opening parabola, the maximum value occurs at the vertex.

3

Substitute the values of

a

and

b

and simplify.

t = -\frac{40}{2(-5)} = 4

This calculation yields the time in seconds when the rocket reaches its maximum height.

Key Concept

Finding the vertex of a quadratic function to determine the maximum or minimum value in context.

Alternative Method

Find the times when the rocket is at ground level by solving

h(t) = 0

. Factoring

-5t^2 + 40t = 0

gives

-5t(t - 8) = 0

, so the rocket is on the ground at

t = 0

seconds and

t = 8

seconds. Since a parabola is symmetric, the maximum height must occur exactly halfway between the launch and landing times, which is

\frac{0 + 8}{2} = 4

seconds.

Estimated Time:45s

Question 19Question →

The quadratic equation $3x^2 - 12x + c = 0$ , where $c$ is a constant, has two real solutions, $r$ and $s$ . If $\frac{1}{r} + \frac{1}{s} = \frac{3}{2}$ , what is the value of $c$ ?

-8

8

18

24

Show answer & explanation

Answer: 8

Answer

8

The correct answer is 8. By Vieta's formulas, the sum of the solutions is

r + s = -\frac{-12}{3} = 4

and their product is

rs = \frac{c}{3}

. The given equation

\frac{1}{r} + \frac{1}{s} = \frac{3}{2}

can be rewritten as

\frac{r+s}{rs} = \frac{3}{2}

. Substituting the expressions for the sum and product gives

\frac{4}{c/3} = \frac{12}{c} = \frac{3}{2}

. Solving this equation yields

c = 8

. Checking the discriminant of the resulting quadratic equation

3x^2 - 12x + 8 = 0

gives

\Delta = (-12)^2 - 4(3)(8) = 48

, which is positive and confirms the existence of two real solutions.

Step-by-Step Solution

1

Use Vieta's formulas to express the sum and product of the solutions in terms of the coefficients of the quadratic equation.

The sum of the solutions is

r + s = -\frac{b}{a} = -\frac{-12}{3} = 4

, and the product of the solutions is

rs = \frac{c}{a} = \frac{c}{3}

.

Vieta's formulas relate the sum and product of the roots of a quadratic equation

ax^2 + bx + c = 0

directly to its coefficients.

2

Express the given algebraic relation

\frac{1}{r} + \frac{1}{s}

in terms of

r+s

and

rs

by finding a common denominator.

\frac{1}{r} + \frac{1}{s} = \frac{s + r}{rs} = \frac{r + s}{rs}

.

Finding a common denominator allows us to rewrite the sum of reciprocals using the sum and product of the solutions.

3

Substitute the sum and product from Step 1 into the rewritten equation and solve for

c

.

Substituting the values gives

\frac{4}{\frac{c}{3}} = \frac{3}{2}

, which simplifies to

\frac{12}{c} = \frac{3}{2}

. Solving for

c

yields

3c = 24

, or

c = 8

.

This substitution turns the given relation into a single linear equation in terms of the variable

c

.

4

Verify that

c = 8

yields two real solutions by checking the discriminant of the quadratic equation.

The discriminant is

\Delta = b^2 - 4ac = (-12)^2 - 4(3)(8) = 144 - 96 = 48

. Since

\Delta > 0

, there are indeed two real solutions.

The problem states that the equation has two real solutions, so we must confirm that

c = 8

satisfies this condition.

Key Concept

Applying Vieta's formulas and algebraic manipulation to solve for constants in a quadratic equation.

Estimated Time:2m 0s

Question 20Question →

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 $b$ ?

Show answer & explanation

Answer: 20

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

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

Quadratic Equations

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