Quadratic Equations and Factoring Alıştırma Soruları — GRE General Test

Soru 1Soru →

The quadratic equation $x^2 - kx + 36 = 0$ , where $k$ is a positive constant, has two distinct real roots $r_1$ and $r_2$ such that $r_2 - r_1 = 5$ . What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The value of

k

is 13.

According to Vieta's formulas, for the quadratic equation

x^2 - kx + 36 = 0

, the sum of the roots is

r_1 + r_2 = k

and the product of the roots is

r_1 r_2 = 36

. Using the identity

(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2

, we substitute the known values

r_2 - r_1 = 5

,

r_1 + r_2 = k

, and

r_1 r_2 = 36

. This gives

5^2 = k^2 - 4(36)

, which simplifies to

25 = k^2 - 144

. Solving for

k^2

gives

k^2 = 169

. Since

k

is specified as a positive constant,

k = 13

.

Adım Adım Çözüm

1

Apply Vieta's formulas to the given quadratic equation

The sum of the roots is

r_1 + r_2 = k

and the product of the roots is

r_1 r_2 = 36

.

For any quadratic equation

ax^2 + bx + c = 0

, the sum of roots is

-b/a

and the product of roots is

c/a

.

2

Relate the difference of the roots to their sum and product

(r_2 - r_1)^2 = (r_1 + r_2)^2 - 4r_1 r_2

Expanding both sides shows that

r_2^2 - 2r_1 r_2 + r_1^2 = r_1^2 + 2r_1 r_2 + r_2^2 - 4r_1 r_2

, which is an algebraic identity.

3

Substitute the known values into the identity

5^2 = k^2 - 4(36) \implies 25 = k^2 - 144

We are given that

r_2 - r_1 = 5

,

r_1 r_2 = 36

, and

r_1 + r_2 = k

.

4

Solve for the positive constant

k

k^2 = 169 \implies k = 13

Adding 144 to both sides gives

k^2 = 169

. Taking the positive square root because

k > 0

yields

k = 13

.

Anahtar Kavram

Vieta's Formulas and Root Difference Identity

Soru 2Soru →

If $(2x - 3)^2 = 25$ and $x < 0$ , what is the value of $x$ ?

-1

4

-2

-4

1

Cevabı ve açıklamayı göster

Cevap: $-1$

Cevap

-1

Taking the square root of both sides of

(2x - 3)^2 = 25

yields two equations:

2x - 3 = 5

(which gives

x = 4

) and

2x - 3 = -5

(which gives

x = -1

). Because the problem specifies that

x < 0

, the correct value must be

-1

.

Adım Adım Çözüm

1

Take the square root of both sides of the equation

(2x - 3)^2 = 25

2x - 3 = \pm 5

Applying the square root property to a squared binomial requires considering both positive and negative principal roots.

2

Set up two separate linear equations corresponding to the two cases

Case 1:

2x - 3 = 5

or Case 2:

2x - 3 = -5

To find all possible solutions for

x

, evaluate both root possibilities.

3

Solve each linear equation for

x

Case 1 yields

2x = 8 \implies x = 4

. Case 2 yields

2x = -2 \implies x = -1

.

Isolate

x

using standard algebraic operations.

4

Apply the given constraint

x < 0

x = -1

Since

4 > 0

, the positive root is eliminated, leaving

x = -1

as the sole valid solution.

Anahtar Kavram

Quadratic Equations and Factoring

Soru 3Soru →

If $x = 5$ is a solution to the quadratic equation $x^2 - (k + 3)x + 3k + 1 = 0$ , where $k$ is a constant, what is the value of the other solution?

\frac{7}{2}

\frac{11}{2}

-\frac{7}{2}

-\frac{11}{2}

\frac{5}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{7}{2}$

Cevap

\frac{7}{2}

Substituting

x = 5

into

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

yields

25 - 5(k + 3) + 3k + 1 = 0

, which simplifies to

11 - 2k = 0

, so

k = \frac{11}{2}

. Substituting

k = \frac{11}{2}

back into the product of roots formula

x_1 \cdot x_2 = 3k + 1

gives

5 \cdot x_2 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}

. Dividing by

5

yields the second root

x_2 = \frac{7}{2}

.

Adım Adım Çözüm

1

Substitute the known solution

x = 5

into the quadratic equation to solve for

k

.

5^2 - (k + 3)(5) + 3k + 1 = 0 \implies 25 - 5k - 15 + 3k + 1 = 0 \implies 11 - 2k = 0 \implies k = \frac{11}{2}

.

Since

x = 5

is a solution, it must satisfy the equation.

2

Use Vieta's formulas to find the other solution

x_2

.

Product of roots

x_1 \cdot x_2 = 3k + 1 = 3\left(\frac{11}{2}\right) + 1 = \frac{35}{2}

. Since

x_1 = 5

,

5 \cdot x_2 = \frac{35}{2} \implies x_2 = \frac{7}{2}

.

By Vieta's formulas for a standard quadratic

x^2 + bx + c = 0

, the product of the roots equals

c

.

Anahtar Kavram

Quadratic equations, Vieta's formulas, and parameter evaluation

Soru 4Soru →

The quadratic equation $x^2 + bx + 45 = 0$ , where $b$ is a constant, has two negative integer roots $p$ and $q$ such that $p < q$ . If $q - p = 4$ , what is the value of $b$ ?

Cevabı ve açıklamayı göster

Cevap: 14

Cevap

The value of

b

is 14.

For the quadratic equation

x^2 + bx + 45 = 0

, the roots

p

and

q

must satisfy

pq = 45

and

p + q = -b

. The negative integer factor pairs of 45 with

p < q

are

(-45, -1)

,

(-15, -3)

, and

(-9, -5)

. Calculating the difference

q - p

for each pair yields 44, 12, and 4, respectively. The condition

q - p = 4

uniquely identifies the roots as

p = -9

and

q = -5

. Summing these roots gives

p + q = -14

, so

b = -(-14) = 14

.

Adım Adım Çözüm

1

Set up the relationships for the roots of the quadratic equation.

pq = 45

and

p + q = -b

.

For any quadratic equation

x^2 + bx + c = 0

, the product of the roots equals

c

and the sum of the roots equals

-b

.

2

Find all negative integer factor pairs

(p, q)

of 45 such that

p < q

.

The possible pairs are

(-45, -1)

,

(-15, -3)

, and

(-9, -5)

.

Since both roots are negative integers, their product is positive 45.

3

Determine the difference

q - p

for each pair to match the given condition

q - p = 4

.

For

(-45, -1)

,

q - p = -1 - (-45) = 44

. For

(-15, -3)

,

q - p = -3 - (-15) = 12

. For

(-9, -5)

,

q - p = -5 - (-9) = 4

.

The pair

(-9, -5)

satisfies

q - p = 4

, establishing

p = -9

and

q = -5

.

4

Calculate the coefficient

b

.

b = -(p + q) = -(-9 + (-5)) = -(-14) = 14

.

Substituting the root values into

b = -(p + q)

yields the final answer.

Anahtar Kavram

Factoring Quadratics and Relationships Between Roots and Coefficients

Soru 5Soru →

If $x$ and $y$ are non-zero real numbers such that $9x^2 - 12xy + 4y^2 = 0$ , what is the value of $\frac{3x + y}{2y}$ ?

\frac{3}{2}

\frac{5}{6}

3

\frac{1}{2}

\frac{5}{2}

Cevabı ve açıklamayı göster

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

Cevap

The value of the expression is

\frac{3}{2}

.

The given expression

9x^2 - 12xy + 4y^2

factors into the perfect square

(3x - 2y)^2 = 0

. Setting the base equal to zero gives

3x = 2y

. Substituting

2y

in place of

3x

in the target expression yields

\frac{2y + y}{2y} = \frac{3y}{2y} = \frac{3}{2}

.

Adım Adım Çözüm

1

Factor the quadratic expression

9x^2 - 12xy + 4y^2 = (3x - 2y)^2 = 0

The equation is a perfect square trinomial of the form

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

where

a = 3x

and

b = 2y

.

2

Solve for the relationship between

x

and

y

3x - 2y = 0 \implies 3x = 2y

Taking the square root of both sides gives a linear relationship between

3x

and

2y

.

3

Substitute

3x = 2y

into the targeted expression

\frac{3x + y}{2y} = \frac{2y + y}{2y} = \frac{3y}{2y} = \frac{3}{2}

Replacing

3x

with

2y

eliminates

x

and simplifies the expression directly.

Anahtar Kavram

Perfect Square Trinomial Factoring

Soru 6Soru →

The quadratic equation $2x^2 + px + q = 0$ , where $p$ and $q$ are constants, has roots $r$ and $s$ . The quadratic equation $x^2 + (p - 2)x + 24 = 0$ has roots $r + 2$ and $s + 2$ . What is the value of $q$ ?

Cevabı ve açıklamayı göster

Cevap: 32

Cevap

The value of

q

is 32.

Applying Vieta's formulas to

2x^2 + px + q = 0

gives

r + s = -p/2

and

rs = q/2

. For the second equation

x^2 + (p-2)x + 24 = 0

, the sum of roots is

(r+2) + (s+2) = -(p-2)

, which simplifies to

(r+s) + 4 = 2 - p

. Substituting

r+s = -p/2

yields

-p/2 + 4 = 2 - p

, solving to

p = -4

and

r+s = 2

. The product of roots for the second equation is

(r+2)(s+2) = rs + 2(r+s) + 4 = 24

. Substituting

rs = q/2

and

r+s = 2

gives

q/2 + 4 + 4 = 24

, which simplifies to

q/2 = 16

and

q = 32

.

Adım Adım Çözüm

1

Express the sum and product of roots

r

and

s

in terms of

p

and

q

using Vieta's formulas.

r + s = -\frac{p}{2}

and

rs = \frac{q}{2}

.

For a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

and the product is

\frac{c}{a}

.

2

Relate the sum of the shifted roots

(r + 2)

and

(s + 2)

to the coefficients of the second quadratic equation.

(r + 2) + (s + 2) = -(p - 2) \implies (r + s) + 4 = 2 - p

.

The coefficient of

x

in

x^2 + (p - 2)x + 24 = 0

is

(p - 2)

, so the sum of its roots equals

-(p - 2)

.

3

Substitute

r + s = -\frac{p}{2}

into the sum relation to determine

p

.

-\frac{p}{2} + 4 = 2 - p \implies \frac{p}{2} = -2 \implies p = -4

.

Solving the linear equation for

p

yields

p = -4

, which means

r + s = 2

.

4

Expand the product of the shifted roots

(r + 2)(s + 2) = 24

and solve for

q

.

rs + 2(r + s) + 4 = 24 \implies \frac{q}{2} + 2(2) + 4 = 24 \implies \frac{q}{2} + 8 = 24 \implies q = 32

.

Substituting

rs = \frac{q}{2}

and

r + s = 2

isolates

q

, yielding

q = 32

.

Anahtar Kavram

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

Soru 7Soru →

If $x$ is a real number that satisfies the equation $(x - 3)^2 = 16$ , which of the following could be the value of $x^2 - 4x$ ? Select all such values.

Geçerli olan tümünü seçin

-5

5

21

45

77

Cevabı ve açıklamayı göster

Cevap: $5$ ; $21$

Cevap

The possible values of

x^2 - 4x

are

5

and

21

.

Taking the square root of both sides of

(x - 3)^2 = 16

yields

x - 3 = \pm 4

. This gives two solutions for

x

:

x = 7

and

x = -1

. Substituting

x = 7

into

x^2 - 4x

gives

49 - 28 = 21

. Substituting

x = -1

into

x^2 - 4x

gives

(-1)^2 - 4(-1) = 1 + 4 = 5

. Thus, both

5

and

21

are valid values.

Adım Adım Çözüm

1

Solve the quadratic equation for

x

x - 3 = 4 \implies x = 7

or

x - 3 = -4 \implies x = -1

Taking the square root of both sides of

(x - 3)^2 = 16

yields both positive and negative roots.

2

Substitute the first root into the expression

x^2 - 4x

7^2 - 4(7) = 49 - 28 = 21

Evaluating the target expression for

x = 7

.

3

Substitute the second root into the expression

x^2 - 4x

(-1)^2 - 4(-1) = 1 + 4 = 5

Evaluating the target expression for

x = -1

.

Anahtar Kavram

Quadratic Equations and Factoring

Soru 8Soru →

If $x$ is a real number such that $(x^2 - 4)^2 - 3(x^2 - 4) - 10 = 0$ , which of the following could be the value of $x$ ? Select all such values.

Geçerli olan tümünü seçin

-3

-\sqrt{2}

2

3

\sqrt{5}

Cevabı ve açıklamayı göster

Cevap: $-3$ ; $-\sqrt{2}$ ; $3$

Cevap

The values that satisfy the equation are

-3

,

-\sqrt{2}

, and

3

.

The given equation is in quadratic form with respect to

(x^2 - 4)

. Letting

u = x^2 - 4

, we factor the equation as

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

, giving

u = 5

or

u = -2

. Setting

x^2 - 4 = 5

gives

x^2 = 9

, which yields

x = 3

or

x = -3

. Setting

x^2 - 4 = -2

gives

x^2 = 2

, which yields

x = \sqrt{2}

or

x = -\sqrt{2}

. Thus,

-3

,

-\sqrt{2}

, and

3

are all valid solutions.

Adım Adım Çözüm

1

Use algebraic substitution to simplify the quadratic form.

Let

u = x^2 - 4

. The equation becomes

u^2 - 3u - 10 = 0

.

Recognizing the structure as a quadratic equation in terms of

(x^2 - 4)

simplifies factoring.

2

Factor the quadratic equation in terms of

u

.

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

, giving solutions

u = 5

or

u = -2

.

Finding the roots of the substituted variable

u

leads to the possible values for

x^2 - 4

.

3

Substitute back

x^2 - 4

for

u

and solve for

x

in each case.

For

u = 5

:

x^2 - 4 = 5 \implies x^2 = 9 \implies x = 3

or

x = -3

.
For

u = -2

:

x^2 - 4 = -2 \implies x^2 = 2 \implies x = \sqrt{2}

or

x = -\sqrt{2}

.

Taking both positive and negative square roots provides all valid real solutions.

Anahtar Kavram

Quadratic Form Substitution and Factoring

Tahmini Süre:1m 40s

Soru 9Soru →

If $x$ is a negative real number such that $(x + 5)^2 = 36$ , what is the value of $x^2 - 2x$ ?

-1

3

-99

143

99

Cevabı ve açıklamayı göster

Cevap: $143$

Cevap

The value of

x^2 - 2x

is

143

.

Taking the square root of both sides of

(x + 5)^2 = 36

gives

x + 5 = 6

or

x + 5 = -6

. Solving for

x

yields

x = 1

or

x = -11

. Because

x

is given as a negative real number,

x = -11

. Evaluating the target expression at

x = -11

yields

(-11)^2 - 2(-11) = 121 + 22 = 143

.

Adım Adım Çözüm

1

Solve the quadratic equation for

x

by applying the square root property.

x + 5 = \pm \sqrt{36} = \pm 6

Taking the square root of both sides yields both positive and negative root options.

2

Determine the two potential solutions for

x

.

x = 6 - 5 = 1

or

x = -6 - 5 = -11

Isolating

x

gives two real solutions.

3

Select the valid solution based on the problem constraints.

x = -11

The stem specifies that

x

must be a negative real number, so

x = 1

is rejected.

4

Substitute

x = -11

into the expression

x^2 - 2x

.

(-11)^2 - 2(-11) = 121 + 22 = 143

Squaring a negative number yields a positive result, and subtracting a negative quantity adds to the total.

Anahtar Kavram

Solving quadratic equations using square roots and evaluating algebraic expressions

Soru 10Soru →

If $x$ is a solution to the quadratic equation $x^2 - 11x + 28 = 0$ such that $x > 5$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

x

is 7.

Factoring

x^2 - 11x + 28 = 0

yields

(x - 7)(x - 4) = 0

, giving solutions

x = 7

and

x = 4

. Since the problem specifies that

x > 5

, the only valid value for

x

is 7.

Adım Adım Çözüm

1

Factor the quadratic expression

(x - 7)(x - 4) = 0

Find two numbers that multiply to 28 and sum to -11, which are -7 and -4.

2

Find the roots of the equation

x = 7

or

x = 4

Set each linear factor equal to 0 and solve for

x

.

3

Apply the given constraint

x > 5

x = 7

Out of the two roots 4 and 7, only 7 is strictly greater than 5.

Anahtar Kavram

Factoring quadratic equations of the form

x^2 + bx + c = 0

Soru 11Soru →

If $x$ is a negative real number such that $x^2 + 5x - 14 = 0$ , what is the value of $x + 3$ ?

-10

-7

-4

2

5

Cevabı ve açıklamayı göster

Cevap: $-4$

Cevap

The value of

x + 3

is

-4

.

Factoring

x^2 + 5x - 14 = 0

yields

(x + 7)(x - 2) = 0

, giving solutions

x = -7

and

x = 2

. Given that

x

must be negative,

x = -7

. Substituting this into the target expression gives

-7 + 3 = -4

.

Adım Adım Çözüm

1

Factor the quadratic equation

x^2 + 5x - 14 = 0

.

(x + 7)(x - 2) = 0

Find two numbers that multiply to

-14

and add to

5

, which are

7

and

-2

.

2

Solve for the possible values of

x

.

x = -7

or

x = 2

Set each linear factor equal to zero.

3

Apply the condition that

x

is a negative real number.

x = -7

The value

x = 2

is positive and must be discarded.

4

Evaluate the target expression

x + 3

.

-7 + 3 = -4

Substitute

x = -7

into

x + 3

.

Anahtar Kavram

Factoring Quadratic Equations and Evaluating Expressions

Soru 12Soru →

Which of the following values are solutions to the equation $(x + 4)^2 = 49$ ? Select all that apply.

Geçerli olan tümünü seçin

-11

-3

3

7

\sqrt{33}

Cevabı ve açıklamayı göster

Cevap: $-11$ ; $3$

Cevap

The correct solutions are

-11

and

3

.

Taking the square root of both sides of

(x + 4)^2 = 49

yields

x + 4 = 7

or

x + 4 = -7

. Solving these two linear equations gives

x = 3

and

x = -11

. Both values satisfy the original quadratic equation.

Adım Adım Çözüm

1

Apply the square root property to both sides of the quadratic equation.

x + 4 = \pm \sqrt{49} = \pm 7

Taking the square root of both sides accounts for both the positive and negative roots of

49

.

2

Solve the first linear equation for the positive case.

x + 4 = 7 \implies x = 7 - 4 = 3

Subtract

4

from both sides when the expression equals

+7

.

3

Solve the second linear equation for the negative case.

x + 4 = -7 \implies x = -7 - 4 = -11

Subtract

4

from both sides when the expression equals

-7

.

Anahtar Kavram

Square Root Property of Quadratic Equations

Soru 13Soru →

What is the sum of all real solutions to the equation $x - 3 = \sqrt{x + 3}$ ?

1

4

5

6

7

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

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

Squaring both sides of

x - 3 = \sqrt{x + 3}

gives

x^2 - 6x + 9 = x + 3

, which simplifies to

x^2 - 7x + 6 = 0

. Factoring yields

(x - 6)(x - 1) = 0

, giving candidate solutions

x = 6

and

x = 1

. Substituting

x = 6

into the original equation gives

3 = \sqrt{9} = 3

, which is valid. Substituting

x = 1

gives

-2 = \sqrt{4} = 2

, which is false, making

x = 1

an extraneous solution. Therefore,

x = 6

is the single valid solution, and its sum is 6.

Adım Adım Çözüm

1

Eliminate the radical by squaring both sides of the equation.

(x - 3)^2 = (\sqrt{x + 3})^2 \implies x^2 - 6x + 9 = x + 3

Squaring both sides converts the radical equation into a standard quadratic equation.

2

Rearrange the terms into standard quadratic form

ax^2 + bx + c = 0

and factor.

x^2 - 7x + 6 = 0 \implies (x - 6)(x - 1) = 0

Subtracting

x

and

3

from both sides isolates zero on one side so the quadratic expression can be factored.

3

Solve for potential values of

x

.

x = 6

or

x = 1

Setting each factor to zero yields the algebraic solutions to the quadratic equation.

4

Check each potential solution in the original radical equation

x - 3 = \sqrt{x + 3}

.

For

x = 6

:

6 - 3 = 3

and

\sqrt{6 + 3} = 3

(Valid). For

x = 1

:

1 - 3 = -2

and

\sqrt{1 + 3} = 2

(Invalid, since

-2 \neq 2

).

Squaring an equation can introduce extraneous solutions that do not satisfy the original principal square root relationship.

5

Calculate the sum of all valid real solutions.

The only valid real solution is

6

, so the sum is

6

.

Extraneous solutions must be discarded before computing the final sum.

Anahtar Kavram

Solving radical equations using quadratic factoring and identifying extraneous solutions

Tahmini Süre:2m 0s

Soru 14Soru →

The quadratic equation $x^2 - px + q = 0$ , where $p$ and $q$ are real constants with $p > 0$ , has two real roots $r_1$ and $r_2$ . If $r_1^2 + r_2^2 = 29$ and $r_1^3 + r_2^3 = 117$ , what is the value of $q$ ?

Cevabı ve açıklamayı göster

Cevap: -10

Cevap

The value of

q

is

-10

.

By Vieta's formulas,

r_1 + r_2 = p

and

r_1 r_2 = q

. Using the identity

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = 29

, we get

q = \frac{p^2 - 29}{2}

. Using the sum of cubes identity

r_1^3 + r_2^3 = (r_1 + r_2)(r_1^2 - r_1 r_2 + r_2^2) = 117

, we obtain

p(29 - q) = 117

. Substituting

q

into this equation yields

p^3 - 87p + 234 = 0

. Factoring this cubic equation yields

p = 3

as a valid root, leading to

q = -10

. Evaluating the discriminant

p^2 - 4q = 9 - 4(-10) = 49 > 0

confirms that real roots exist. The second positive root for

p

gives a negative discriminant, making

-10

the unique correct value of

q

.

Adım Adım Çözüm

1

Express symmetric sums of roots in terms of equation coefficients

p

and

q

using Vieta's formulas.

r_1 + r_2 = p

and

r_1 r_2 = q

. Therefore,

r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = p^2 - 2q = 29

, which gives

q = \frac{p^2 - 29}{2}

.

Relating root sums and products directly to coefficients simplifies the algebraic system.

2

Express the sum of cubes in terms of

p

and

q

and substitute

q

.

r_1^3 + r_2^3 = (r_1 + r_2)(r_1^2 - r_1 r_2 + r_2^2) = p(29 - q) = 117

. Substituting

q = \frac{p^2 - 29}{2}

yields

p\left(29 - \frac{p^2 - 29}{2}\right) = 117 \implies p(87 - p^2) = 234 \implies p^3 - 87p + 234 = 0

.

Combining symmetric expressions forms a single polynomial equation in variable

p

.

3

Factor the cubic equation

p^3 - 87p + 234 = 0

to find candidate values for

p > 0

.

Testing factor

p = 3

gives

3^3 - 87(3) + 234 = 27 - 261 + 234 = 0

. Factoring out

(p - 3)

gives

(p - 3)(p^2 + 3p - 78) = 0

.

Finding the roots of the polynomial identifies candidate values for coefficient

p

.

4

Verify the real root constraint

p^2 - 4q > 0

for candidates of

p

.

For

p = 3

,

q = \frac{3^2 - 29}{2} = -10

, giving discriminant

p^2 - 4q = 9 - 4(-10) = 49 > 0

(real roots

5

and

-2

). For the positive root

p = \frac{\sqrt{321}-3}{2} \approx 7.458

of

p^2 + 3p - 78 = 0

, the discriminant is

p^2 - 4q = 3p - 29 \approx -6.63 < 0

, which yields non-real complex roots.

The question specifies real roots, so candidate solutions resulting in negative discriminants must be discarded.

Anahtar Kavram

Quadratic Equations and Factoring (Vieta's Formulas, Symmetric Polynomials, and Real Root Conditions)

Soru 15Soru →

For what values of the real constant $k$ does the quadratic equation $(k-2)x^2 - 2kx + (2k - 3) = 0$ have two distinct real roots $r_1$ and $r_2$ such that $r_1 < 1 < r_2$ ?

1 < k < 6

2 < k < 5

k < 2

or

k > 5

1 < k < 5

k > 2

Cevabı ve açıklamayı göster

Cevap: $2 < k < 5$

Cevap

The correct inequality expressing all possible values of

k

is

2 < k < 5

.

The correct inequality

2 < k < 5

is obtained by requiring

(k-2)f(1) < 0

, which ensures

x = 1

falls between the two distinct real roots. Evaluating

f(1) = k - 5

yields

(k-2)(k-5) < 0

, giving

2 < k < 5

. The discriminant condition

\Delta = -4(k-1)(k-6) > 0

gives

1 < k < 6

, which fully encompasses

(2, 5)

.

Adım Adım Çözüm

1

Define the quadratic function and state the conditions for

r_1 < 1 < r_2

.

Let

f(x) = (k-2)x^2 - 2kx + (2k - 3)

. For a quadratic function to have two real roots with

x = 1

located between them, the product of the leading coefficient

(k-2)

and

f(1)

must be strictly negative, i.e.,

(k-2)f(1) < 0

.

If a parabola opens upwards (

k-2 > 0

), its value at a point between its roots must be negative (

f(1) < 0

). If it opens downwards (

k-2 < 0

), its value at a point between its roots must be positive (

f(1) > 0

).

2

Evaluate

f(1)

in terms of

k

.

f(1) = (k-2)(1)^2 - 2k(1) + (2k - 3) = k - 2 - 2k + 2k - 3 = k - 5

.

Substitute

x = 1

directly into the expression for

f(x)

.

3

Solve the inequality

(k-2)f(1) < 0

.

(k-2)(k-5) < 0 \implies 2 < k < 5

.

The product of two linear factors

(k-2)

and

(k-5)

is negative between their roots,

k = 2

and

k = 5

.

4

Verify discriminant condition

\Delta > 0

for real roots.

\Delta = (-2k)^2 - 4(k-2)(2k-3) = 4k^2 - 4(2k^2 - 7k + 6) = -4k^2 + 28k - 24 = -4(k-1)(k-6) > 0 \implies 1 < k < 6

.

Since the interval

(2, 5)

is entirely contained within

(1, 6)

, any

k \in (2, 5)

automatically guarantees two distinct real roots.

Anahtar Kavram

Location of roots of quadratic equations and sign analysis of quadratic functions.

Tahmini Süre:2m 0s

Soru 16Soru →

Let $P(x) = x^2 - mx + n$ be a quadratic polynomial with real coefficients $m$ and $n$ , having two distinct real roots $\alpha$ and \beta. If the roots satisfy the system of equations $\alpha^3 + \beta^3 = m(n + 7)$ and $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}$ , which of the following statements MUST be true? Select all such statements.

Geçerli olan tümünü seçin

The product of the roots,

n

, is equal to

1.5

.

The sum of the squares of the roots,

\alpha^2 + \beta^2

, is equal to

10

.

The discriminant of the polynomial

P(x)

is equal to

7

.

The absolute value of the sum of the roots,

|\alpha + \beta|

, is equal to

5

.

The roots

\alpha

and

\beta

have opposite signs.

Cevabı ve açıklamayı göster

Cevap: The product of the roots, $n$ , is equal to $1.5$ .; The sum of the squares of the roots, $\alpha^2 + \beta^2$ , is equal to $10$ .; The discriminant of the polynomial $P(x)$ is equal to $7$ .

Cevap

The correct statements are those asserting that the product of the roots is

1.5

, the sum of the squares of the roots is

10

, and the discriminant of

P(x)

is

7

.

Using Vieta's formulas and algebraic identity expansions for

\alpha^3 + \beta^3

and

\frac{1}{\alpha^2} + \frac{1}{\beta^2}

establishes the system of equations

m^2 = 4n + 7

and

m^2 = 2n + 10

. Solving this system gives

n = 1.5

,

m^2 = 13

, and a discriminant

\Delta = m^2 - 4n = 7

. Thus, the product of roots is

1.5

, the sum of squares

\alpha^2 + \beta^2 = m^2 - 2n = 10

, and the discriminant is

7

.

Adım Adım Çözüm

1

Apply Vieta's formulas to express sum and product of roots.

\alpha + \beta = m \quad \text{and} \quad \alpha\beta = n

Vieta's relations link polynomial coefficients directly to symmetrical root expressions.

2

Expand

\alpha^3 + \beta^3

in terms of

m

and

n

.

\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = m^3 - 3mn = m(m^2 - 3n)

Using algebraic identities converts root powers into functions of

m

and

n

.

3

Equate the expression from Step 2 to

m(n + 7)

to find an equation for

m^2

.

m(m^2 - 3n) = m(n + 7) \implies m^2 - 3n = n + 7 \implies m^2 = 4n + 7

Since the roots are distinct,

m \neq 0

, allowing division by

m

.

4

Simplify the second given equation

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{10}{n^2}

.

\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{m^2 - 2n}{n^2} = \frac{10}{n^2} \implies m^2 - 2n = 10 \implies m^2 = 2n + 10

Combining fractions over a common denominator

(\alpha\beta)^2 = n^2

isolates

m^2 - 2n

.

5

Solve for

n

,

m^2

, and the discriminant

\Delta

.

4n + 7 = 2n + 10 \implies 2n = 3 \implies n = 1.5; \quad m^2 = 13; \quad \Delta = m^2 - 4n = 13 - 6 = 7

Equating the two expressions for

m^2

yields unique values for

n

,

m^2

, and

\Delta

.

Anahtar Kavram

Quadratic Equations, Vieta's Formulas, and Symmetric Polynomial Expressions

Soru 17Soru →

The quadratic equation $x^2 - (k - 2)x + (k - 5) = 0$ has two distinct real roots, $\alpha$ and $\beta$ . If $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 1$ , what is the value of the constant $k$ ?

\frac{9}{4}

\frac{11}{4}

\frac{7}{2}

\frac{31}{8}

\frac{31}{4}

Cevabı ve açıklamayı göster

Cevap: $\frac{11}{4}$

Cevap

The value of the constant

k

is

\frac{11}{4}

.

Using Vieta's formulas for the quadratic equation

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

, we have

\alpha + \beta = k - 2

and

\alpha\beta = k - 5

. The given condition

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 1

simplifies to

\frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 1

. Substituting the expressions in terms of

k

yields

\frac{(k - 2)^2 - 2(k - 5)}{(k - 5)^2} = 1

, which expands to

k^2 - 6k + 14 = k^2 - 10k + 25

. Subtracting

k^2

from both sides gives

4k = 11

, so

k = \frac{11}{4}

.

Adım Adım Çözüm

1

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

k

\alpha + \beta = k - 2

and

\alpha\beta = k - 5

For any standard quadratic

ax^2 + bx + c = 0

, the sum of roots is

-\frac{b}{a}

and the product of roots is

\frac{c}{a}

.

2

Rewrite the given sum of reciprocal squares using algebraic identities

\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}

Combining fractions over a common denominator allows substitution of the known sum

\alpha+\beta

and product

\alpha\beta

.

3

Substitute the expressions for sum and product into the equation

\frac{(k - 2)^2 - 2(k - 5)}{(k - 5)^2} = 1

Set the algebraic expression equal to the given target value of 1.

4

Expand both numerator and denominator and solve for

k

\frac{k^2 - 4k + 4 - 2k + 10}{k^2 - 10k + 25} = 1 \implies k^2 - 6k + 14 = k^2 - 10k + 25 \implies 4k = 11 \implies k = \frac{11}{4}

Equating numerator and denominator eliminates the quadratic

k^2

terms, yielding a linear equation in

k

.

5

Verify that

k = \frac{11}{4}

yields real, distinct, non-zero roots

Discriminant

D = (k-2)^2 - 4(k-5) = 0.5625 + 9 = 9.5625 > 0

, and

\alpha\beta = -2.25 \neq 0

Ensures the quadratic has two distinct real roots as required by the problem prompt.

Anahtar Kavram

Vieta's Formulas and Symmetric Polynomial Transformations of Quadratic Roots

Soru 18Soru →

A parabola defined by the quadratic function $f(x) = ax^2 + bx + c$ , where $a, b,$ and $c$ are real constants with $a > 0$ , has its vertex at a minimum value of $-16$ . The distance between the two $x$ -intercepts of the parabola is $8$ . If $f(1) = -7$ and the $x$ -coordinate of the vertex is positive, what is the value of $f(-2)$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The value of

f(-2)

is 20.

By converting the parabola into vertex form

f(x) = a(x - h)^2 - 16

, the

x

-intercepts are found at

h \pm \frac{4}{\sqrt{a}}

. Equating their difference to

8

yields

a = 1

. Substituting

f(1) = -7

gives

(1 - h)^2 = 9

, which yields

h = 4

under the condition

h > 0

. Evaluating

f(-2) = (-2 - 4)^2 - 16

produces

20

.

Adım Adım Çözüm

1

Express the quadratic function in vertex form using the minimum value

f(x) = a(x - h)^2 - 16

, where

(h, -16)

is the vertex.

Since

a > 0

, the parabola opens upwards and its minimum value occurs at the vertex

y

-coordinate,

k = -16

.

2

Determine the leading coefficient

a

using the distance between

x

-intercepts

a = 1

Setting

f(x) = 0

yields

a(x - h)^2 - 16 = 0 \implies (x - h)^2 = \frac{16}{a} \implies x = h \pm \frac{4}{\sqrt{a}}

. The distance between roots is

\frac{8}{\sqrt{a}} = 8

, which gives

\sqrt{a} = 1 \implies a = 1

.

3

Determine the vertex

x

-coordinate

h

using the point

f(1) = -7

h = 4

Substituting

a = 1

and

x = 1

into the vertex form gives

(1 - h)^2 - 16 = -7 \implies (1 - h)^2 = 9

. Taking square roots gives

1 - h = 3 \implies h = -2

or

1 - h = -3 \implies h = 4

. Since

h > 0

, we select

h = 4

.

4

Evaluate

f(-2)

using the fully specified function

f(-2) = 20

With

f(x) = (x - 4)^2 - 16

, substituting

x = -2

yields

f(-2) = (-2 - 4)^2 - 16 = (-6)^2 - 16 = 36 - 16 = 20

.

Anahtar Kavram

Quadratic Vertex Form, Root Separation, and Evaluation

Soru 19Soru →

A rectangular plot of land has a perimeter of 56 meters and an area of 180 square meters. A border of uniform width $x$ meters is constructed around the inside edge of the plot, reducing the remaining inner area to 84 square meters. What is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The border width x is 2 meters.

Solving the system

L + W = 28

and

LW = 180

gives outer dimensions of 18 and 10 meters. The inner area equation

(18 - 2x)(10 - 2x) = 84

simplifies to

x^2 - 14x + 24 = 0

, which factors into

(x - 2)(x - 12) = 0

. Since

2x

must be less than the total width of 10 meters (

x < 5

), the extraneous root

x = 12

is discarded, leaving

x = 2

.

Adım Adım Çözüm

1

Determine the outer dimensions of the plot.

Length = 18 meters, Width = 10 meters

The semi-perimeter is L + W = 56 / 2 = 28, and the area is L * W = 180. Forming the quadratic equation t^2 - 28t + 180 = 0 yields roots t = 18 and t = 10.

2

Write the quadratic expression for the inner area in terms of x.

(18 - 2x)(10 - 2x) = 84

A uniform border of width x along all four sides reduces both the length and width by 2x.

3

Expand and simplify the quadratic equation into standard form.

x^2 - 14x + 24 = 0

Expanding yields 180 - 56x + 4x^2 = 84. Subtracting 84 gives 4x^2 - 56x + 96 = 0, which divides by 4 to give x^2 - 14x + 24 = 0.

4

Factor the quadratic equation to solve for possible values of x.

x = 2 or x = 12

Factoring x^2 - 14x + 24 = 0 gives (x - 2)(x - 12) = 0.

5

Eliminate the extraneous root based on physical geometry constraints.

x = 2

The total width is 10 meters, so 2x must be less than 10 (x < 5). Thus, x = 12 is physically impossible and discarded.

Anahtar Kavram

Quadratic Equations, Factoring, and Extraneous Solutions

Soru 20Soru →

Consider the quadratic equation $x^2 - 2(k + 1)x + (k^2 + 4k + 1) = 0$ , where $k$ is a real constant. Let $\alpha$ and $\beta$ be the real roots of this equation. If $\alpha^2 + \beta^2 = 18$ , what is the sum of all possible values of $k$ ?

-2\sqrt{2}

0

2\sqrt{2}

2 - 2\sqrt{3}

-4\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $-2\sqrt{2}$

Cevap

The sum of all possible real values of

k

is

-2\sqrt{2}

.

Using Vieta's formulas,

\alpha + \beta = 2(k + 1)

and

\alpha\beta = k^2 + 4k + 1

. Expanding

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

gives

2k^2 + 2

. Setting

2k^2 + 2 = 18

yields

k^2 = 8

, so

k = \pm 2\sqrt{2}

. Furthermore, for the roots

\alpha

and

\beta

to be real numbers, the discriminant

\Delta = [-2(k+1)]^2 - 4(k^2+4k+1) = -8k

must be non-negative, requiring

k \le 0

. Only

k = -2\sqrt{2}

satisfies this requirement. Hence, the sum of all valid values of

k

is

-2\sqrt{2}

.

Adım Adım Çözüm

1

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

k

.

\alpha + \beta = 2(k + 1)

and

\alpha\beta = k^2 + 4k + 1

.

For any quadratic equation

x^2 - bx + c = 0

, the sum of roots is

b

and the product of roots is

c

.

2

Express

\alpha^2 + \beta^2

in terms of

k

using the algebraic identity

(\alpha + \beta)^2 - 2\alpha\beta

.

\alpha^2 + \beta^2 = [2(k + 1)]^2 - 2(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 2k^2 - 8k - 2 = 2k^2 + 2

.

Expanding and simplifying allows us to relate the given sum of squares to the parameter

k

.

3

Set the simplified expression equal to

18

and solve for

k

.

2k^2 + 2 = 18 \implies 2k^2 = 16 \implies k^2 = 8 \implies k = \pm 2\sqrt{2}

.

This yields two candidate real values for the parameter

k

.

4

Analyze the discriminant

\Delta

to enforce the condition that roots

\alpha

and

\beta

are real numbers.

\Delta = [-2(k + 1)]^2 - 4(1)(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 4(k^2 + 4k + 1) = -8k

. For real roots,

\Delta \ge 0 \implies -8k \ge 0 \implies k \le 0

.

A quadratic equation has real roots if and only if its discriminant is non-negative.

5

Filter candidate values of

k

against the condition

k \le 0

.

k = -2\sqrt{2}

satisfies

k \le 0

(yielding

\Delta = 16\sqrt{2} > 0

), while

k = 2\sqrt{2}

fails

k \le 0

(yielding

\Delta = -16\sqrt{2} < 0

). Thus,

k = -2\sqrt{2}

is the unique valid value.

Values of

k

that make the discriminant negative produce non-real complex roots and must be discarded as extraneous.

Anahtar Kavram

Quadratic Vieta's formulas combined with discriminant non-negativity constraint for real roots

Tahmini Süre:2m 30s

Quadratic Equations and Factoring

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