Quadratic Equations and Factoring Practice Questions — GRE General Test — Page 2

Question 21Question →

Let $f(x) = x^2 + bx + c$ , where $b$ and $c$ are integers. The equation $f(x) = 0$ has two distinct real roots, $\alpha$ and $\beta$ . If $\alpha + \beta = \alpha\beta$ and $c > 0$ , which of the following statements must be true? Indicate all such statements.

Select all that apply

The constant

c

is strictly greater than 4.

Both roots

\alpha

and

\beta

are positive.

The vertex of the parabola

y = f(x)

lies in Quadrant IV of the xy-plane.

The coefficient

b

is positive.

The difference between the two roots,

|\alpha - \beta|

, can equal 2.

Show answer & explanation

Answer: The constant $c$ is strictly greater than 4.; Both roots $\alpha$ and $\beta$ are positive.; The vertex of the parabola $y = f(x)$ lies in Quadrant IV of the xy-plane.

Answer

The statements asserting that the constant

c

is strictly greater than 4, both roots are positive, and the vertex lies in Quadrant IV must all be true.

By Vieta's formulas,

\alpha + \beta = -b

and

\alpha\beta = c

. Equating sum and product gives

b = -c

. The discriminant

\Delta = c^2 - 4c > 0

requires

c > 4

because

c > 0

. Since sum and product of the roots equal

c > 0

, both roots are positive. The vertex coordinates

(c/2, -c(c-4)/4)

have a positive x-value and a negative y-value, placing the vertex in Quadrant IV.

Step-by-Step Solution

1

Apply Vieta's formulas and establish the relationship between coefficients

\alpha + \beta = -b

and

\alpha\beta = c

. Setting them equal gives

-b = c \implies b = -c

.

The stem specifies that the sum of the roots equals their product.

2

Analyze the discriminant for distinct real roots

\Delta = b^2 - 4c = (-c)^2 - 4c = c(c - 4) > 0

. Since

c > 0

,

c - 4 > 0 \implies c > 4

.

Two distinct real roots require a strictly positive discriminant.

3

Determine the signs of the roots

\alpha + \beta = c > 0

and

\alpha\beta = c > 0

, implying

\alpha > 0

and

\beta > 0

.

If the sum and product of two real numbers are both positive, both numbers must be positive.

4

Find the location of the parabola's vertex

x_v = \frac{c}{2} > 0

and

y_v = -\frac{c(c-4)}{4} < 0

, placing the vertex in Quadrant IV.

A point with a positive x-coordinate and negative y-coordinate resides in the fourth quadrant.

5

Evaluate the remaining options regarding

b

and

|\alpha - \beta|

b = -c < -4

(negative), and

|\alpha - \beta| = 2 \implies c = 2 + 2\sqrt{2}

, which is not an integer.

These evaluations disprove the statements that

b

is positive and that

|\alpha - \beta|

can equal 2.

Key Concept

Quadratic Root Properties and Vieta's Formulas

Question 22Question →

The quadratic equation $x^2 - 2mx + (m^2 - 4m + 12) = 0$ , where $m$ is a real constant, has two distinct real roots $x_1$ and $x_2$ . If the distance between the two roots on the real number line is $4\sqrt{3}$ , what is the value of $x_1^2 + x_2^2$ ?

48

96

120

144

192

Show answer & explanation

Answer: 96

Answer

96

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

x_1 + x_2 = 2m

and the product of the roots is

x_1 x_2 = m^2 - 4m + 12

. The distance between the roots is given by

|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2}

. Squaring both sides gives

(4\sqrt{3})^2 = 48 = (2m)^2 - 4(m^2 - 4m + 12) = 16m - 48

. Solving

16m - 48 = 48

yields

m = 6

. Substituting

m = 6

into the expressions for the sum and product of the roots gives

x_1 + x_2 = 12

and

x_1 x_2 = 24

. Finally, applying the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

yields

12^2 - 2(24) = 144 - 48 = 96

.

Step-by-Step Solution

1

Apply Vieta's formulas to determine the sum and product of the roots in terms of

m

.

For

x^2 - 2mx + (m^2 - 4m + 12) = 0

, the sum of roots is

x_1 + x_2 = 2m

and the product of roots is

x_1 x_2 = m^2 - 4m + 12

.

Vieta's relations relate quadratic coefficients directly to root sums and products.

2

Express the distance between the roots

|x_1 - x_2|

in terms of

m

and solve for

m

.

|x_1 - x_2|^2 = (x_1 + x_2)^2 - 4x_1 x_2 = (2m)^2 - 4(m^2 - 4m + 12) = 16m - 48

. Given

|x_1 - x_2| = 4\sqrt{3}

, we have

16m - 48 = (4\sqrt{3})^2 = 48 \implies 16m = 96 \implies m = 6

.

The difference between two roots of a quadratic is linked to the discriminant via

|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}

.

3

Calculate the numerical values of the root sum

x_1 + x_2

and root product

x_1 x_2

using

m = 6

.

x_1 + x_2 = 2(6) = 12

and

x_1 x_2 = 6^2 - 4(6) + 12 = 24

.

Substituting

m = 6

gives the exact values needed for algebraic evaluation.

4

Compute

x_1^2 + x_2^2

using the identity

x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2

.

x_1^2 + x_2^2 = 12^2 - 2(24) = 144 - 48 = 96

.

The sum of squares identity isolates

x_1^2 + x_2^2

without requiring explicit calculation of individual root values.

Key Concept

Quadratic Root Relationships and Vieta's Formulas

Estimated Time:2m 30s

Question 23Question →

What is the sum of all real solutions to the equation

(x^2 - 3x)^2 - 8(x^2 - 3x) + 12 = 0

?

3

6

8

12

-6

Show answer & explanation

Answer: $6$

Answer

The sum of all real solutions to the given equation is 6.

By substituting

u = x^2 - 3x

, the quartic equation reduces to the quadratic equation

u^2 - 8u + 12 = 0

, which factors as

(u - 2)(u - 6) = 0

. Setting

x^2 - 3x = 2

and

x^2 - 3x = 6

yields two distinct quadratic equations:

x^2 - 3x - 2 = 0

and

x^2 - 3x - 6 = 0

. Since both discriminants (

17

and

33

) are strictly positive, each equation has two distinct real solutions. By Vieta's formulas, the sum of the roots for a quadratic equation

ax^2 + bx + c = 0

is

-b/a

. Thus, the sum of the roots for

x^2 - 3x - 2 = 0

is

3

, and the sum of the roots for

x^2 - 3x - 6 = 0

is

3

. Adding these together gives a total sum of

3 + 3 = 6

.

Step-by-Step Solution

1

Apply algebraic substitution to reduce the equation to a standard quadratic form.

Let

u = x^2 - 3x

. The original equation becomes

u^2 - 8u + 12 = 0

.

Recognizing repeated quadratic expressions allows for simplification into a single quadratic in terms of

u

.

2

Factor the quadratic equation in

u

to find its roots.

(u - 2)(u - 6) = 0

, which gives

u = 2

and

u = 6

.

The factors of

12

that sum to

-8

are

-2

and

-6

.

3

Substitute

x^2 - 3x

back for

u

and check the discriminant of each resulting quadratic equation.

For

u = 2

:

x^2 - 3x - 2 = 0

has discriminant

\Delta_1 = (-3)^2 - 4(1)(-2) = 17 > 0

(2 distinct real roots).
For

u = 6

:

x^2 - 3x - 6 = 0

has discriminant

\Delta_2 = (-3)^2 - 4(1)(-6) = 33 > 0

(2 distinct real roots).

Verifying that the discriminant is positive ensures that all four roots are real numbers.

4

Calculate the sum of the roots for each quadratic equation using Vieta's formulas.

For

x^2 - 3x - 2 = 0

, the sum of roots is

-\frac{-3}{1} = 3

.
For

x^2 - 3x - 6 = 0

, the sum of roots is

-\frac{-3}{1} = 3

.

According to Vieta's formulas, for a quadratic

ax^2 + bx + c = 0

, the sum of the roots is

-b/a

.

5

Add the sums of the roots from both equations to find the total sum of all real solutions.

Total sum

= 3 + 3 = 6

.

Since all four roots are real and distinct, the overall sum is the sum of the roots of the two constituent quadratics.

Key Concept

Quadratic Substitution and Vieta's Formulas for Root Sums

Question 24Question →

The quadratic equation $(x - a)(x - b) = c$ , where $a$ , $b$ , and $c$ are real constants with $a < b$ and $c > 0$ , has two real roots $x_1$ and $x_2$ such that $x_1 < x_2$ . Which of the following statements must be true? Indicate all such statements.

Select all that apply

x_1 < a

x_2 > b

The distance between the roots,

x_2 - x_1

, is strictly greater than

b - a

x_1 + x_2 > a + b

The product of the roots,

x_1 x_2

, must be negative

Show answer & explanation

Answer: $x_1 < a$ ; $x_2 > b$ ; The distance between the roots, $x_2 - x_1$ , is strictly greater than $b - a$

Answer

The correct statements are that the smaller root is less than

a

(

x_1 < a

), the larger root is greater than

b

(

x_2 > b

), and the distance between the roots is greater than

b - a

(

x_2 - x_1 > b - a

).

Evaluating

f(x) = (x-a)(x-b) - c

at

x = a

and

x = b

yields negative values (

-c

). Because the parabola opens upward, the graph must cross the x-axis to the left of

a

and to the right of

b

. This establishes

x_1 < a

and

x_2 > b

. Combining these inequalities shows that the distance between the roots

x_2 - x_1

must exceed

b - a

.

Step-by-Step Solution

1

Formulate the quadratic function and analyze its values at key points

Let

f(x) = (x - a)(x - b) - c = 0

. Evaluating at

x = a

and

x = b

gives

f(a) = (0)(a-b) - c = -c < 0

and

f(b) = (b-a)(0) - c = -c < 0

.

Evaluating at

a

and

b

reveals the sign of the function inside the interval

[a, b]

.

2

Use the sign of the leading coefficient and intermediate value properties to locate the roots

The coefficient of

x^2

is

1 > 0

, so the parabola opens upward. Since

f(a) < 0

and

f(b) < 0

, and

f(x) \to +\infty

as

x \to \pm\infty

, the two real roots

x_1

and

x_2

must satisfy

x_1 < a < b < x_2

.

A continuous upward-opening parabola must cross the x-axis outside any interval where its values are negative.

3

Analyze the distance between the roots

Subtracting

x_1 < a

from

x_2 > b

gives

x_2 - x_1 > b - a

.

Since

a

and

b

fall strictly between

x_1

and

x_2

, the distance between

x_1

and

x_2

exceeds the distance between

a

and

b

.

4

Check the remaining statements using Vieta's formulas

Expanding

f(x) = x^2 - (a+b)x + (ab - c) = 0

yields sum of roots

x_1 + x_2 = a + b

and product of roots

x_1 x_2 = ab - c

. Thus

x_1 + x_2 > a+b

is false, and

x_1 x_2 < 0

is not necessarily true.

Vieta's formulas give exact values for the sum and product of roots in terms of coefficients.

Key Concept

Geometric interpretation of quadratic functions and root locations relative to evaluated points

Question 25Question →

For a nonzero real constant $m$ , the quadratic equation $x^2 - \left(m + \frac{6}{m}\right)x + 6 = 0$ has real roots $a$ and $b$ . If $a^2 + b^2 = 25$ , what is the sum of all possible values of $m^2$ ?

13

25

37

49

5

Show answer & explanation

Answer: $25$

Answer

The sum of all possible values of

m^2

is

25

.

Using Vieta's formulas on

x^2 - \left(m + \frac{6}{m}\right)x + 6 = 0

, the sum of the roots is

a + b = m + \frac{6}{m}

and the product is

ab = 6

. Substituting these into the identity

a^2 + b^2 = (a+b)^2 - 2ab = 25

yields

\left(m + \frac{6}{m}\right)^2 - 12 = 25

, which simplifies to

m^2 + \frac{36}{m^2} = 25

. Setting

u = m^2

gives

u^2 - 25u + 36 = 0

. Since this quadratic has two distinct positive real roots for

u

, the sum of all possible values of

m^2

is given by Vieta's formula as

25

.

Step-by-Step Solution

1

Apply Vieta's formulas to relate the roots

a

and

b

to the coefficients of the given quadratic equation.

a + b = m + \frac{6}{m}

and

ab = 6

.

For any quadratic equation

x^2 - Sx + P = 0

, the sum of the roots is

S

and the product of the roots is

P

.

2

Express

a^2 + b^2

using the algebraic identity for the sum of squares.

a^2 + b^2 = (a + b)^2 - 2ab = \left(m + \frac{6}{m}\right)^2 - 2(6) = \left(m + \frac{6}{m}\right)^2 - 12

.

Rewriting the sum of squares allows substitution of the known sum and product of the roots.

3

Set the expression equal to the given value

25

and expand the square.

m^2 + 2(m)\left(\frac{6}{m}\right) + \frac{36}{m^2} - 12 = 25 \implies m^2 + 12 + \frac{36}{m^2} - 12 = 25 \implies m^2 + \frac{36}{m^2} = 25

.

Expanding

\left(m + \frac{6}{m}\right)^2

yields

m^2 + 12 + \frac{36}{m^2}

, which simplifies directly after subtracting

12

.

4

Formulate a quadratic equation in terms of

u = m^2

.

u + \frac{36}{u} = 25 \implies u^2 - 25u + 36 = 0

.

Multiplying through by

u = m^2

transforms the rational equation into a standard polynomial quadratic equation.

5

Find the sum of all valid solutions for

u = m^2

using Vieta's formulas.

The discriminant of

u^2 - 25u + 36 = 0

is

\Delta = (-25)^2 - 4(1)(36) = 625 - 144 = 481 > 0

, and the product of the roots is

36 > 0

, so both roots

u_1, u_2

are positive real numbers. By Vieta's formulas,

u_1 + u_2 = 25

.

Since both roots

u_1

and

u_2

are positive real numbers, they correspond to valid real values of

m^2

, making their sum equal to

25

.

Key Concept

Quadratic Equations, Vieta's Formulas, and Factoring Substitution

Estimated Time:2m 0s

Question 26Question →

The quadratic function $f(x) = -2x^2 + kx - 18$ has a maximum value of $14$ , where $k$ is a positive constant. What is the value of $k$ ?

Show answer & explanation

Answer: 16

Answer

The value of

k

is

16

.

The vertex of the parabola

f(x) = -2x^2 + kx - 18

is located at

x = \frac{k}{4}

. Evaluating

f\left(\frac{k}{4}\right)

gives the maximum value

\frac{k^2}{8} - 18

. Setting this expression equal to

14

leads to

\frac{k^2}{8} = 32

, so

k^2 = 256

. Taking the positive root as required by the problem statement yields

k = 16

.

Step-by-Step Solution

1

Find the

x

-coordinate of the vertex of the quadratic function.

For

f(x) = -2x^2 + kx - 18

, we have

a = -2

,

b = k

, and

c = -18

. The vertex occurs at

x = -\frac{b}{2a} = -\frac{k}{2(-2)} = \frac{k}{4}

.

The maximum or minimum of any quadratic function

ax^2 + bx + c

occurs at its vertex, where

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

.

2

Evaluate the function at the vertex to determine the maximum value in terms of

k

.

f\left(\frac{k}{4}\right) = -2\left(\frac{k}{4}\right)^2 + k\left(\frac{k}{4}\right) - 18 = -2\left(\frac{k^2}{16}\right) + \frac{k^2}{4} - 18 = -\frac{k^2}{8} + \frac{k^2}{4} - 18 = \frac{k^2}{8} - 18

.

Substituting the vertex

x

-coordinate into

f(x)

yields the maximum value of the downward-opening parabola.

3

Set the maximum value expression equal to

14

and solve for

k^2

.

\frac{k^2}{8} - 18 = 14 \implies \frac{k^2}{8} = 32 \implies k^2 = 256.

The problem states that the maximum value of

f(x)

is

14

.

4

Solve for the positive constant

k

.

k = \sqrt{256} = 16.

Taking the square root of

256

gives

k = 16

or

k = -16

. Since

k

is given as a positive constant,

k = 16

.

Key Concept

Finding the extreme value of a quadratic function by locating its vertex

Question 27Question →

For a real constant $k$ , the quadratic equation $x^2 - 2kx + (k^2 - k - 6) = 0$ has two distinct real roots $r$ and $s$ such that $r < 0 < s$ and $|r| < |s|$ . Which of the following inequalities expresses all possible values of $k$ ?

0 < k < 3

-2 < k < 3

-6 < k < 3

k < -2

or

k > 3

-2 < k < 0

Show answer & explanation

Answer: $0 < k < 3$

Answer

The inequality expressing all possible values of

k

is

0 < k < 3

.

The condition that one root is negative and one root is positive (

r < 0 < s

) requires the product of the roots

rs = k^2 - k - 6

to be negative, which resolves to

-2 < k < 3

. Furthermore, since the positive root

s

has a larger absolute magnitude than the negative root

r

(

|r| < |s|

), the sum of the roots

r + s = 2k

must be positive, requiring

k > 0

. Taking the intersection of

-2 < k < 3

and

k > 0

yields

0 < k < 3

.

Step-by-Step Solution

1

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

k

.

For

x^2 - 2kx + (k^2 - k - 6) = 0

, the sum of roots is

r + s = 2k

and the product of roots is

rs = k^2 - k - 6

.

Vieta's formulas directly relate the coefficients of a quadratic equation to the sum and product of its roots.

2

Analyze the condition

r < 0 < s

.

Since one root is negative and the other is positive, their product must be negative:

rs = k^2 - k - 6 < 0

. Factoring gives

(k - 3)(k + 2) < 0

, which yields

-2 < k < 3

.

A positive number multiplied by a negative number produces a negative product.

3

Analyze the condition

|r| < |s|

.

Since

r < 0

,

|r| = -r

. Since

s > 0

,

|s| = s

. The inequality

|r| < |s|

becomes

-r < s

, which simplifies to

r + s > 0

. Substituting

r + s = 2k

gives

2k > 0

, or

k > 0

.

The positive root having greater magnitude than the absolute value of the negative root means the sum of the roots must be positive.

4

Combine the conditions to find the valid range for

k

.

Combining

-2 < k < 3

and

k > 0

yields the intersection

0 < k < 3

. (The discriminant condition

\Delta = 4(k + 6) > 0 \implies k > -6

is satisfied for all

k \in (0, 3)

).

The parameter

k

must satisfy both root sign constraints simultaneously.

Key Concept

Using Vieta's formulas and root magnitude conditions to solve quadratic parameter inequality problems.

Question 28Question →

If $x$ is a positive real number such that $2x^2 - 5x - 3 = 0$ , what is the value of $x^2 + 1$ ?

4

\frac{5}{4}

9

10

26

Show answer & explanation

Answer: 10

Answer

10

Factoring the quadratic equation

2x^2 - 5x - 3 = 0

yields

(2x + 1)(x - 3) = 0

, giving solutions

x = -\frac{1}{2}

and

x = 3

. Because

x

is specified as a positive real number, we choose

x = 3

. Substituting

x = 3

into

x^2 + 1

yields

3^2 + 1 = 10

.

Step-by-Step Solution

1

Factor the quadratic equation

2x^2 - 5x - 3 = 0

.

(2x + 1)(x - 3) = 0

Find two linear factors whose product gives

2x^2 - 5x - 3

.

2

Solve for the roots of the equation.

x = -\frac{1}{2}

or

x = 3

Set each factor equal to zero:

2x + 1 = 0 \implies x = -\frac{1}{2}

and

x - 3 = 0 \implies x = 3

.

3

Apply the constraint

x > 0

to identify the valid root.

x = 3

The root

x = -\frac{1}{2}

is negative and thus violates the given condition that

x

is positive.

4

Substitute

x = 3

into the target expression

x^2 + 1

.

3^2 + 1 = 9 + 1 = 10

Evaluate the expression using the valid positive value of

x

.

Key Concept

Solving quadratic equations by factoring and applying domain constraints

Estimated Time:45s

Question 29Question →

Which of the following values are solutions to the equation $(x - 3)^2 = 16$ ? Select all that apply.

Select all that apply

-7

-1

5

7

19

Show answer & explanation

Answer: $-1$ ; $7$

Answer

The solutions to the equation are

-1

and

7

.

Taking the square root of both sides of

(x - 3)^2 = 16

gives

x - 3 = 4

or

x - 3 = -4

. Solving these equations yields

x = 7

and

x = -1

.

Step-by-Step Solution

1

Take the square root of both sides of the quadratic equation

(x - 3)^2 = 16

.

x - 3 = \pm \sqrt{16} = \pm 4

Applying the square root property yields both a positive and a negative root.

2

Solve the equation corresponding to the positive root:

x - 3 = 4

.

x = 4 + 3 = 7

Adding

3

to both sides isolates

x

.

3

Solve the equation corresponding to the negative root:

x - 3 = -4

.

x = -4 + 3 = -1

Adding

3

to both sides isolates

x

for the negative root.

Key Concept

Solving quadratic equations of the form

(x - a)^2 = k

using the square root property.

Estimated Time:1m 0s

Question 30Question →

If $x > 0$ and $x^2 - 6x - 16 = 0$ , what is the value of $x$ ?

Show answer & explanation

Answer: 8

Answer

The correct value of

x

is 8.

Factoring

x^2 - 6x - 16 = 0

gives

(x - 8)(x + 2) = 0

. Setting each factor to zero yields

x = 8

and

x = -2

. Given the constraint

x > 0

, the only valid answer is 8.

Step-by-Step Solution

1

Factor the quadratic expression

(x - 8)(x + 2) = 0

Find two numbers that multiply to

-16

and add to

-6

, which are

-8

and

2

.

2

Solve for possible values of

x

x = 8

or

x = -2

By the zero-product property, if

(x - 8)(x + 2) = 0

, then either

x - 8 = 0

or

x + 2 = 0

.

3

Apply the given constraint

x > 0

x = 8

The solution

x = -2

is rejected because

x

must be positive.

Key Concept

Factoring Quadratic Equations

Estimated Time:45s

Question 31Question →

A quadratic function is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are real constants with $a > 0$ . If the vertex of the parabola $y = f(x)$ lies in the third quadrant of the $xy$ -plane, which of the following statements must be true? Select all that apply.

Select all that apply

b > 0

The equation

f(x) = 0

has two distinct real solutions.

c

must be negative.

The sum of the two solutions of

f(x) = 0

is positive.

The discriminant

b^2 - 4ac

is less than zero.

Show answer & explanation

Answer: $b > 0$ ; The equation $f(x) = 0$ has two distinct real solutions.

Answer

The statements '

b > 0

' and 'The equation

f(x) = 0

has two distinct real solutions' must be true.

The vertex of the parabola

y = ax^2 + bx + c

is

(h, k) = \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)

. Because the vertex is in the third quadrant,

h < 0

and

k < 0

. With

a > 0

, the inequality

-\frac{b}{2a} < 0

directly requires

b > 0

. Furthermore, because

a > 0

(parabola opens upward) and the minimum value

k < 0

lies below the x-axis, the graph must cross the x-axis at two distinct points, establishing that

f(x) = 0

has two distinct real solutions.

Step-by-Step Solution

1

Analyze the coordinates of the vertex

(h, k)

relative to the third quadrant.

The third quadrant requires

h < 0

and

k < 0

.

Points in Quadrant III have negative x-coordinates and negative y-coordinates.

2

Evaluate the sign of

b

using the x-coordinate formula

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

.

b > 0

.

Since

h < 0

and

a > 0

,

-\frac{b}{2a} < 0 \implies \frac{b}{2a} > 0 \implies b > 0

.

3

Determine the number of real solutions using the vertex y-coordinate

k

and orientation

a > 0

.

The discriminant

b^2 - 4ac > 0

, giving two distinct real solutions.

For an upward-opening parabola (

a > 0

), having a vertex below the x-axis (

k < 0

) guarantees the curve crosses the x-axis twice.

4

Test counterexamples for

c

and evaluate the root sum

-\frac{b}{a}

.

c

is not strictly constrained in sign, and the root sum is negative.

The function

f(x) = (x+2)^2 - 1 = x^2 + 4x + 3

has vertex

(-2, -1) \in \text{QIII}

with

c = 3 > 0

. Also, since

a > 0

and

b > 0

, root sum

-\frac{b}{a} < 0

.

Key Concept

Parabola Vertex and Quadratic Discriminant Analysis

Question 32Question →

The length of a rectangular plot of land is $3$ meters less than twice its width. If the area of the plot is $90$ square meters, what is the perimeter of the plot, in meters?

Show answer & explanation

Answer: 39

Answer

The perimeter of the plot of land is 39 meters.

Setting the length to

2w - 3

gives an area equation of

w(2w - 3) = 90

, which expands and rearranges to

2w^2 - 3w - 90 = 0

. Factoring this quadratic equation yields

(2w - 15)(w + 6) = 0

. Since width must be positive,

w = 7.5

meters, which means the length is

12

meters. The perimeter is

2(12 + 7.5) = 39

meters.

Step-by-Step Solution

1

Express the length in terms of width and set up the area equation.

Let

w

be the width of the rectangle. Length

l = 2w - 3

. Area equation:

w(2w - 3) = 90

.

The area of a rectangle is equal to length multiplied by width.

2

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

2w^2 - 3w - 90 = 0

Distributing

w

and subtracting

90

from both sides puts the equation in standard quadratic form.

3

Factor the quadratic equation.

(2w - 15)(w + 6) = 0

Finding two numbers with a product of

2 \times (-90) = -180

and a sum of

-3

gives

-15

and

12

.

4

Determine the valid physical dimensions.

w = 7.5

meters and

l = 12

meters.

The root

w = -6

is discarded because physical length cannot be negative. Thus

w = \frac{15}{2} = 7.5

meters.

5

Calculate the perimeter.

Perimeter

= 2(l + w) = 2(12 + 7.5) = 39

meters.

The perimeter of a rectangle is given by

2 \times (\text{length} + \text{width})

.

Key Concept

Solving quadratic word problems via factoring

Estimated Time:1m 30s

Question 33Question →

If $r_1$ and $r_2$ are the two distinct real solutions to the quadratic equation $x^2 - 4x - 21 = 0$ , where $r_1 > r_2$ , what is the value of $r_1 + 2r_2$ ?

1

13

-11

-5

17

Show answer & explanation

Answer: $1$

Answer

The correct value of

r_1 + 2r_2

is

1

.

Factoring the quadratic polynomial

x^2 - 4x - 21

gives

(x - 7)(x + 3) = 0

, which yields solutions

x = 7

and

x = -3

. Applying the constraint

r_1 > r_2

identifies

r_1 = 7

and

r_2 = -3

. Substituting these values into

r_1 + 2r_2

yields

7 + 2(-3) = 7 - 6 = 1

.

Step-by-Step Solution

1

Factor the quadratic equation

(x - 7)(x + 3) = 0

Find two numbers that multiply to

-21

and add up to

-4

, which are

-7

and

3

.

2

Solve for the roots of the equation

x = 7

or

x = -3

Set each linear factor equal to zero:

x - 7 = 0 \implies x = 7

and

x + 3 = 0 \implies x = -3

.

3

Assign values to

r_1

and

r_2

based on the condition

r_1 > r_2

r_1 = 7

and

r_2 = -3

Since

7 > -3

,

r_1

must be

7

and

r_2

must be

-3

.

4

Evaluate the expression

r_1 + 2r_2

7 + 2(-3) = 7 - 6 = 1

Substitute

r_1 = 7

and

r_2 = -3

into the given expression.

Key Concept

Solving quadratic equations by factoring and evaluating expressions involving roots.

Estimated Time:1m 30s

Question 34Question →

If $m$ and $n$ are the two distinct real solutions to the quadratic equation $2x^2 - 8x + c = 0$ , where $c$ is a constant, and $m^2 + n^2 = 12$ , what is the value of $c$ ?

-4

4

8

2

26

Show answer & explanation

Answer: 4

Answer

The value of

c

is

4

.

By Vieta's formulas for the quadratic equation

2x^2 - 8x + c = 0

, the sum of the roots is

m + n = 4

and the product of the roots is

mn = c/2

. Squaring the sum gives

(m + n)^2 = m^2 + 2mn + n^2 = 16

. Substituting

m^2 + n^2 = 12

gives

12 + 2(c/2) = 16

, which simplifies to

12 + c = 16

, so

c = 4

.

Step-by-Step Solution

1

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

For

2x^2 - 8x + c = 0

, the sum of roots is

m + n = -\frac{-8}{2} = 4

, and the product of roots is

m \cdot n = \frac{c}{2}

.

For a quadratic equation

ax^2 + bx + c = 0

, the sum of roots is

-b/a

and the product of roots is

c/a

.

2

Relate

m^2 + n^2

to the sum and product of roots using the identity

(m + n)^2 = m^2 + n^2 + 2mn

.

m^2 + n^2 = (m + n)^2 - 2mn

Rewriting the sum of squares in terms of

(m+n)

and

mn

allows substitution of the known values.

3

Substitute

m + n = 4

,

mn = \frac{c}{2}

, and

m^2 + n^2 = 12

into the equation and solve for

c

.

12 = (4)^2 - 2\left(\frac{c}{2}\right) \implies 12 = 16 - c \implies c = 4

.

Direct algebraic simplification yields the constant

c

.

Key Concept

Relationship between roots and coefficients of a quadratic equation (Vieta's Formulas) and algebraic identities.

Question 35Question →

For what positive value of the constant $k$ does the quadratic equation $x^2 - 2kx + 3k + 10 = 0$ have exactly one real solution?

Show answer & explanation

Answer: 5

Answer

The positive value of

k

is 5.

A quadratic equation has exactly one real solution when its discriminant

b^2 - 4ac

equals

0

. For

x^2 - 2kx + (3k + 10) = 0

, substituting

a = 1

,

b = -2k

, and

c = 3k + 10

gives

(-2k)^2 - 4(1)(3k + 10) = 0

, which simplifies to

4k^2 - 12k - 40 = 0

. Dividing the entire equation by

4

gives

k^2 - 3k - 10 = 0

. Factoring this quadratic yields

(k - 5)(k + 2) = 0

, giving solutions

k = 5

and

k = -2

. Since the question requires the positive value of

k

, the correct answer is

5

.

Step-by-Step Solution

1

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

The discriminant must equal zero (

b^2 - 4ac = 0

).

A quadratic equation

ax^2 + bx + c = 0

has a single repeated real root if and only if its discriminant is zero.

2

Set up the discriminant equation using the coefficients from

x^2 - 2kx + (3k + 10) = 0

.

Discriminant

= (-2k)^2 - 4(1)(3k + 10) = 4k^2 - 12k - 40 = 0

.

Here

a = 1

,

b = -2k

, and

c = 3k + 10

.

3

Divide by 4 and solve the quadratic in terms of

k

by factoring.

k^2 - 3k - 10 = 0 \Rightarrow (k - 5)(k + 2) = 0

, so

k = 5

or

k = -2

.

Factoring the simplified quadratic yields the two potential values of

k

.

4

Choose the value of

k

that satisfies the constraint in the stem.

k = 5

.

The question asks specifically for the positive value of

k

, so

k = -2

is rejected.

Key Concept

Quadratic Discriminant and Factoring

Estimated Time:1m 30s

Question 36Question →

If $x$ is a real number that satisfies the equation $x^4 - 13x^2 + 36 = 0$ , which of the following values could be the value of $x$ ? Select all such values.

Select all that apply

-3

-2

3

4

9

Show answer & explanation

Answer: $-3$ ; $-2$ ; $3$

Answer

The correct values are

-3

,

-2

, and

3

.

Factoring the equation yields

(x^2 - 4)(x^2 - 9) = (x - 2)(x + 2)(x - 3)(x + 3) = 0

, giving the four distinct real solutions

x = -3, -2, 2, 3

. The options corresponding to

-3

,

-2

, and

3

represent valid values of

x

.

Step-by-Step Solution

1

Substitute a variable

u = x^2

to express the equation in quadratic form.

u^2 - 13u + 36 = 0

This reduces the fourth-degree polynomial into a standard quadratic equation.

2

Factor the quadratic equation to solve for

u

.

(u - 4)(u - 9) = 0 \implies u = 4 \text{ or } u = 9

Factoring determines the values of

x^2

.

3

Substitute back

x^2 = u

and solve for

x

by taking both positive and negative square roots.

x^2 = 4 \implies x = \pm 2

, and

x^2 = 9 \implies x = \pm 3

Each positive value of

u

yields two real solutions for

x

.

Key Concept

Solving quadratic-form equations by factoring and taking positive and negative square roots.

Question 37Question →

If the quadratic equation $2x^2 - kx + 18 = 0$ has two positive real roots $x_1$ and $x_2$ such that $x_1 = 4x_2$ , what is the value of the constant $k$ ?

15

\frac{15}{2}

30

9

24

Show answer & explanation

Answer: $15$

Answer

15

By Vieta's formulas for

ax^2 + bx + c = 0

, the product of the roots is

x_1 x_2 = \frac{c}{a} = \frac{18}{2} = 9

. Given

x_1 = 4x_2

, substituting gives

4x_2^2 = 9

, which yields

x_2 = \frac{3}{2}

since roots are positive. Then

x_1 = 6

. The sum of the roots is

x_1 + x_2 = \frac{15}{2}

. By Vieta's formulas,

x_1 + x_2 = -\frac{b}{a} = \frac{k}{2}

. Setting

\frac{k}{2} = \frac{15}{2}

gives

k = 15

.

Step-by-Step Solution

1

Apply Vieta's formula for the product of the roots

x_1 \cdot x_2 = \frac{18}{2} = 9

For a quadratic equation

ax^2 + bx + c = 0

, the product of the roots is

\frac{c}{a}

.

2

Substitute the given relationship

x_1 = 4x_2

into the product equation

(4x_2)(x_2) = 9 \implies 4x_2^2 = 9 \implies x_2^2 = \frac{9}{4} \implies x_2 = \frac{3}{2}

Since the roots are positive, we take the positive square root of

\frac{9}{4}

.

3

Calculate the larger root

x_1

x_1 = 4 \left(\frac{3}{2}\right) = 6

Using the relationship

x_1 = 4x_2

.

4

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

k

x_1 + x_2 = 6 + \frac{3}{2} = \frac{15}{2} = \frac{k}{2} \implies k = 15

For

ax^2 + bx + c = 0

, the sum of the roots is

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

.

Key Concept

Relating roots of a quadratic equation to its coefficients using Vieta's formulas and factoring relationships.

Question 38Question →

If $a$ and $b$ are the two real solutions to the quadratic equation $2x^2 - 11x + 12 = 0$ , such that $a > b$ , what is the value of $a - 2b$ ?

Show answer & explanation

Answer: 1

Answer

The value of

a - 2b

is 1.

Factoring

2x^2 - 11x + 12 = 0

yields

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

, giving solutions

x = 1.5

and

x = 4

. Given that

a > b

, we must set

a = 4

and

b = 1.5

. Substituting these values into

a - 2b

gives

4 - 2(1.5) = 1

.

Step-by-Step Solution

1

Factor the quadratic equation

2x^2 - 11x + 12 = 0

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

Find two linear factors whose product expands to

2x^2 - 11x + 12

.

2

Find the roots of the equation

x = \frac{3}{2} = 1.5

and

x = 4

Apply the zero product property:

2x - 3 = 0 \Rightarrow x = 1.5

and

x - 4 = 0 \Rightarrow x = 4

.

3

Assign values to

a

and

b

based on the inequality

a > b

a = 4

and

b = 1.5

Since

4 > 1.5

,

a

must be 4 and

b

must be 1.5.

4

Evaluate the targeted expression

a - 2b

4 - 2(1.5) = 1

Substitute

a = 4

and

b = 1.5

into

a - 2b

.

Key Concept

Factoring Quadratic Equations

Question 39Question →

If $m$ and $n$ are positive integers such that the quadratic equation $x^2 - mx + n = 0$ has two roots that are distinct prime numbers, and the discriminant of the equation is equal to $1$ , what is the value of $m + n$ ?

5

7

11

13

1

Show answer & explanation

Answer: 11

Answer

11

For the quadratic equation

x^2 - mx + n = 0

with roots

p_1

and

p_2

, the discriminant is

D = (p_1 - p_2)^2

. Given

D = 1

, the difference between the two prime roots is

|p_1 - p_2| = 1

. The only prime numbers with a difference of 1 are 2 and 3. Thus,

m = 2 + 3 = 5

and

n = 2 \times 3 = 6

, giving

m + n = 11

.

Step-by-Step Solution

1

Express the discriminant of the quadratic equation in terms of its roots.

If

p_1

and

p_2

are the roots of

x^2 - mx + n = 0

, Vieta's formulas give

p_1 + p_2 = m

and

p_1 p_2 = n

. The discriminant is

D = m^2 - 4n = (p_1 + p_2)^2 - 4p_1 p_2 = (p_1 - p_2)^2

.

Relating the discriminant directly to the difference of the roots simplifies the constraint.

2

Determine the roots using the given discriminant value.

Since

D = 1

, we have

(p_1 - p_2)^2 = 1

, which implies

|p_1 - p_2| = 1

.

Taking the square root of both sides indicates the two roots differ by 1.

3

Identify the prime numbers that satisfy this condition.

The only pair of prime numbers that differ by 1 is 2 and 3, because 2 is the only even prime and all other primes are odd.

Consecutive integers that are both prime must be 2 and 3.

4

Calculate

m

,

n

, and their sum

m + n

.

m = 2 + 3 = 5

and

n = 2 \times 3 = 6

. Therefore,

m + n = 5 + 6 = 11

.

Substitute the root values into the sum and product formulas.

Key Concept

Quadratic discriminant and root relationships (Vieta's Formulas)

Estimated Time:1m 30s

Question 40Question →

If the quadratic equation $3x^2 - 12x + c = 0$ has two real roots, $r_1$ and $r_2$ , such that $r_1^2 + r_2^2 = 10$ , what is the value of the constant $c$ ?

3

6

9

18

-9

Show answer & explanation

Answer: 9

Answer

The value of the constant

c

is 9.

The correct answer is 9. By Vieta's formulas, the sum of the roots is

r_1 + r_2 = -\frac{-12}{3} = 4

and the product of the roots is

r_1 r_2 = \frac{c}{3}

. Using the identity

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2

, we substitute the given values to get

10 = 16 - \frac{2c}{3}

, which simplifies to

\frac{2c}{3} = 6

, giving

c = 9

.

Step-by-Step Solution

1

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

r_1 + r_2 = -\frac{-12}{3} = 4

and

r_1 r_2 = \frac{c}{3}

.

For any quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

and the product of the roots is

\frac{c}{a}

.

2

Relate the sum of squares

r_1^2 + r_2^2

to

(r_1 + r_2)

and

r_1 r_2

.

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2

.

Expanding

(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2

and rearranging gives the identity for the sum of squares.

3

Substitute the known values into the identity and solve for

c

.

10 = 4^2 - 2\left(\frac{c}{3}\right) \implies 10 = 16 - \frac{2c}{3} \implies \frac{2c}{3} = 6 \implies c = 9

.

Substituting

r_1^2 + r_2^2 = 10

and

r_1 + r_2 = 4

isolates the single variable

c

.

Key Concept

Vieta's Formulas and Algebraic Identities for Quadratic Equations

Alternative Method

Alternatively, factor

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

directly once

r_1

and

r_2

are identified. Since

r_1 + r_2 = 4

and

r_1^2 + r_2^2 = 10

, solving the system of equations for

r_1

and

r_2

yields roots of 1 and 3. The product of these roots is

(1)(3) = 3

, so

\frac{c}{3} = 3

, which gives

c = 9

.

Estimated Time:1m 30s

Quadratic Equations and Factoring

Answer

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Key Concept

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