Algebra Alıştırma Soruları — GRE General Test — Sayfa 18

For all real numbers

x

and

y

such that

x \neq y

, which of the following expressions is equivalent to

\frac{x^3 - x^2y - xy^2 + y^3}{x - y} - (x - y)^2

?

2xy - 2y^2

2xy

-2y^2

2x^2 - 2xy

0

Cevabı ve açıklamayı göster

Cevap: $2xy - 2y^2$

Cevap

The equivalent expression is

2xy - 2y^2

.

Factoring the numerator by grouping gives

x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y)

. Dividing by

(x - y)

yields

x^2 - y^2

. Expanding

(x - y)^2

gives

x^2 - 2xy + y^2

. Subtracting this expansion from

x^2 - y^2

results in

(x^2 - y^2) - (x^2 - 2xy + y^2) = 2xy - 2y^2

.

Adım Adım Çözüm

1

Factor the numerator of the rational fraction by grouping terms

x^3 - x^2y - xy^2 + y^3 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x + y)(x - y)^2

Grouping terms allows factoring out common binomial components.

2

Simplify the fraction by dividing by

(x - y)

\frac{(x + y)(x - y)^2}{x - y} = (x + y)(x - y) = x^2 - y^2

Since

x \neq y

,

x - y \neq 0

, so common factors can be canceled.

3

Expand the subtracted term

(x - y)^2

(x - y)^2 = x^2 - 2xy + y^2

Applying the standard perfect square trinomial expansion formula.

4

Subtract the expanded expression from the simplified fraction result

(x^2 - y^2) - (x^2 - 2xy + y^2) = x^2 - y^2 - x^2 + 2xy - y^2 = 2xy - 2y^2

Distribute the negative sign across all terms inside the parentheses and combine like terms.

Anahtar Kavram

Simplifying algebraic expressions using factoring by grouping, difference of squares, and expanding binomial squares.

Soru 342Soru →

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

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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

Soru 343Soru →

For all real numbers

x

such that

x \neq -3

,

x \neq 0

, and

x \neq 3

, which of the following expressions is equivalent to

\frac{x^4 - 81}{x^3 + 3x^2} \div \frac{x^2 + 9}{x^2}

?

x - 3

x + 3

x^2 - 3

\frac{x - 3}{x}

x - 9

Cevabı ve açıklamayı göster

Cevap: $x - 3$

Cevap

x - 3

Factoring the numerator

x^4 - 81

gives

(x - 3)(x + 3)(x^2 + 9)

and factoring the denominator

x^3 + 3x^2

gives

x^2(x + 3)

. Multiplying by the reciprocal

\frac{x^2}{x^2 + 9}

allows the factors

(x + 3)

,

(x^2 + 9)

, and

x^2

to cancel completely, leaving

x - 3

.

Adım Adım Çözüm

1

Rewrite division of rational expressions as multiplication by the reciprocal

\frac{x^4 - 81}{x^3 + 3x^2} \cdot \frac{x^2}{x^2 + 9}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

2

Factor the numerator

x^4 - 81

using the difference of squares formula twice

x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)

Recognize

x^4 - 81

as

(x^2)^2 - 9^2

and then factor

x^2 - 9

further into

(x - 3)(x + 3)

.

3

Factor out the greatest common factor from the denominator

x^3 + 3x^2

x^3 + 3x^2 = x^2(x + 3)

Extract

x^2

from both terms of the binomial.

4

Substitute the factored expressions into the product and cancel common factors

\frac{(x - 3)(x + 3)(x^2 + 9)}{x^2(x + 3)} \cdot \frac{x^2}{x^2 + 9} = x - 3

The common terms

(x + 3)

,

(x^2 + 9)

, and

x^2

cancel out completely from the numerator and denominator for all

x \neq -3, 0, 3

.

Anahtar Kavram

Factoring polynomial expressions using difference of squares and common terms to simplify rational expressions.

Tahmini Süre:1m 30s

Soru 344Soru →

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

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Quadratic Discriminant and Factoring

Tahmini Süre:1m 30s

Soru 345Soru →

For all real numbers $x$ such that $x \neq -2$ , which of the following expressions are equivalent to $\frac{x^6 - 64}{x^3 + 8}$ ? Select all such expressions.

Geçerli olan tümünü seçin

x^3 - 8

(x - 2)(x^2 + 2x + 4)

(x - 2)^3

x^3 + 8

x^2 - 8

Cevabı ve açıklamayı göster

Cevap: $x^3 - 8$ ; $(x - 2)(x^2 + 2x + 4)$

Cevap

The expressions equivalent to the given rational expression are

x^3 - 8

and

(x - 2)(x^2 + 2x + 4)

.

Factoring the numerator

x^6 - 64

as a difference of squares yields

(x^3 - 8)(x^3 + 8)

. Dividing by the denominator

(x^3 + 8)

leaves

x^3 - 8

. Further factoring

x^3 - 8

as a difference of cubes gives

(x - 2)(x^2 + 2x + 4)

. Both

x^3 - 8

and

(x - 2)(x^2 + 2x + 4)

are valid equivalent expressions.

Adım Adım Çözüm

1

Factor the numerator using the difference of squares identity

x^6 - 64 = (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8)

The expression

x^6 - 64

is a difference of two squares.

2

Simplify the fraction by canceling the common non-zero term

(x^3 + 8)

\frac{(x^3 - 8)(x^3 + 8)}{x^3 + 8} = x^3 - 8

Since

x \neq -2

,

x^3 + 8 \neq 0

, so

(x^3 + 8)

can be canceled from numerator and denominator.

3

Factor

x^3 - 8

using the difference of cubes identity

x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)

Applying

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

with

a = x

and

b = 2

.

Anahtar Kavram

Simplifying rational expressions by factoring polynomial numerators and denominators using algebraic identities such as difference of squares and difference of cubes.

Soru 346Soru →

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.

Geçerli olan tümünü seçin

-3

-2

3

4

9

Cevabı ve açıklamayı göster

Cevap: $-3$ ; $-2$ ; $3$

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

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

Soru 347Soru →

For all real numbers

x

such that

x \neq -3

,

x \neq 0

, and

x \neq 3

, which of the following expressions is equivalent to

\frac{x^{-2} - 9^{-1}}{x^{-1} + 3^{-1}} \cdot \frac{3x^2 + 9x}{x - 3}?

-x - 3

x + 3

x - 3

\frac{x + 3}{3x}

-x + 3

Cevabı ve açıklamayı göster

Cevap: $-x - 3$

Cevap

-x - 3

The numerator of the first factor is a difference of squares

(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})

. Dividing by

x^{-1} + 3^{-1}

leaves

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

. Factoring

3x

out of the numerator of the second expression gives

3x(x + 3)

. Multiplying these terms yields

\frac{3 - x}{3x} \cdot \frac{3x(x + 3)}{x - 3}

. Since

3 - x = -(x - 3)

, the binomials

(3 - x)

and

(x - 3)

cancel to

-1

, and

3x

cancels out completely, resulting in

-(x + 3) = -x - 3

.

Adım Adım Çözüm

1

Factor the numerator of the first expression as a difference of squares.

x^{-2} - 9^{-1} = (x^{-1})^2 - (3^{-1})^2 = (x^{-1} - 3^{-1})(x^{-1} + 3^{-1})

Expressing negative exponents as squares allows cancellation with the denominator.

2

Divide by

(x^{-1} + 3^{-1})

and convert negative exponents into a single rational term.

\frac{(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})}{x^{-1} + 3^{-1}} = x^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}

Finding a common denominator simplifies the first factor.

3

Factor the numerator of the second expression

3x^2 + 9x

.

3x^2 + 9x = 3x(x + 3)

Factoring out the greatest common factor

3x

enables further simplification.

4

Multiply the simplified expressions and reduce.

\left(\frac{3 - x}{3x}\right) \cdot \left(\frac{3x(x + 3)}{x - 3}\right) = \frac{-(x - 3)}{3x} \cdot \frac{3x(x + 3)}{x - 3} = -(x + 3) = -x - 3

Canceling

3x

and noting that

(3 - x)/(x - 3) = -1

leaves the linear expression

-x - 3

.

Anahtar Kavram

Simplifying algebraic expressions containing negative exponents, rational fractions, and opposite-sign binomial factors

Soru 348Soru →

When the algebraic expression $\frac{4x^3 - 16x}{2x^2 + 4x}$ is simplified for all $x \neq 0$ and $x \neq -2$ , it reduces to the linear polynomial $ax + b$ , where $a$ and $b$ are constants. What is the value of $a + b$ ?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The correct numerical answer is -2.

Factoring the numerator yields

4x(x - 2)(x + 2)

and factoring the denominator yields

2x(x + 2)

. Canceling the common factors

2x

and

(x + 2)

leaves

2(x - 2) = 2x - 4

. Comparing

2x - 4

to

ax + b

gives

a = 2

and

b = -4

. Summing these values gives

a + b = 2 + (-4) = -2

.

Adım Adım Çözüm

1

Factor out the greatest common factor and apply the difference of squares formula to the numerator.

4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)

Fully factoring the numerator allows identification of all linear factors.

2

Factor out the greatest common factor from the denominator.

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

Extracting

2x

reveals the common terms shared with the numerator.

3

Divide the numerator by the denominator by canceling identical non-zero factors

2x

and

(x + 2)

.

\frac{4x(x - 2)(x + 2)}{2x(x + 2)} = \frac{4}{2}(x - 2) = 2(x - 2) = 2x - 4

Simplifying rational expressions requires canceling common factors present in both numerator and denominator.

4

Compare the simplified expression

2x - 4

to the form

ax + b

to find

a

and

b

, then compute their sum.

a = 2

and

b = -4 \implies a + b = 2 + (-4) = -2

Matching corresponding terms identifies the values of the target constants.

Anahtar Kavram

Simplifying rational expressions by factoring out common terms and applying the difference of squares identity.

Soru 349Soru →

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

Cevabı ve açıklamayı göster

Cevap: $15$

Cevap

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

.

Adım Adım Çözüm

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}

.

Anahtar Kavram

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

Soru 350Soru →

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 $a - 2b$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Factoring Quadratic Equations

Soru 351Soru →

For all real numbers $x$ and $y$ such that $y \neq 0$ and $3x^2 + y^2 \neq 0$ , which of the following expressions is equivalent to $\frac{(x + y)^3 - (x - y)^3}{2y(3x^2 + y^2)}$ ?

1

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

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

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

\frac{1}{3}

Cevabı ve açıklamayı göster

Cevap: $1$

Cevap

1

Expanding the numerator yields

(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3 = 2y(3x^2 + y^2)

. Since the numerator and denominator are identical non-zero expressions, the quotient simplifies to 1.

Adım Adım Çözüm

1

Expand

(x + y)^3

and

(x - y)^3

using the binomial theorem.

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

and

(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

Expanding the binomial expressions allows combination of like terms in the numerator.

2

Subtract

(x - y)^3

from

(x + y)^3

.

(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3

The terms

x^3

and

3xy^2

subtract to zero, while the remaining terms double.

3

Factor out common terms from the simplified numerator.

6x^2y + 2y^3 = 2y(3x^2 + y^2)

Factoring out

2y

reveals a factor identical to the denominator.

4

Divide the numerator by the denominator

2y(3x^2 + y^2)

.

\frac{2y(3x^2 + y^2)}{2y(3x^2 + y^2)} = 1

Since

y \neq 0

and

3x^2 + y^2 \neq 0

, identical non-zero factors cancel out.

Anahtar Kavram

Binomial expansion of cubic expressions and factoring algebraic expressions.

Tahmini Süre:1m 30s

Soru 352Soru →

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

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Quadratic discriminant and root relationships (Vieta's Formulas)

Tahmini Süre:1m 30s

Soru 353Soru →

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

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The value of

x^2 + y^2

is 80.

Simplifying the first equation by factoring the difference of squares

\frac{(x-y)(x+y)}{x-y} = 12

gives

x + y = 12

. Simplifying the second equation by factoring out

xy

yields

\frac{xy(x-y)}{xy} = 4

, which gives

x - y = 4

. Solving this system yields

x = 8

and

y = 4

. Squaring and summing these values gives

8^2 + 4^2 = 64 + 16 = 80

. Alternatively, using the algebraic identity

\frac{(x+y)^2 + (x-y)^2}{2} = \frac{12^2 + 4^2}{2} = \frac{144 + 16}{2} = 80

directly yields the correct answer.

Adım Adım Çözüm

1

Simplify the first given algebraic expression using the difference of squares identity.

x + y = 12

Factoring

x^2 - y^2

gives

(x - y)(x + y)

. Since

x \neq y

, we can cancel the non-zero common factor

(x - y)

from the numerator and denominator.

2

Simplify the second given algebraic expression by factoring out the greatest common factor.

x - y = 4

Factoring

xy

from

x^2y - xy^2

yields

xy(x - y)

. Since

x, y \neq 0

, we cancel

xy

from the numerator and denominator.

3

Solve for the individual values of

x

and

y

.

x = 8

and

y = 4

Adding

(x + y = 12)

and

(x - y = 4)

gives

2x = 16

, so

x = 8

. Subtracting the equations gives

2y = 8

, so

y = 4

.

4

Calculate the target expression

x^2 + y^2

.

80

x^2 + y^2 = 8^2 + 4^2 = 64 + 16 = 80

.

Anahtar Kavram

Factoring Algebraic Expressions (Difference of Squares and GCF Extraction)

Soru 354Soru →

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

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Vieta's Formulas and Algebraic Identities for Quadratic Equations

Alternatif Yöntem

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

.

Tahmini Süre:1m 30s

Soru 355Soru →

For all real numbers $x$ and $y$ such that $x \neq 2y$ and $x \neq -2y$ , which of the following expressions is equivalent to $\frac{x^3 + 2x^2y - 4xy^2 - 8y^3}{x^2 - 4y^2}$ ?

x - 2y

x + 4y

x + 2y

\frac{x + 2y}{x - 2y}

x^2 + 2y

Cevabı ve açıklamayı göster

Cevap: $x + 2y$

Cevap

The simplified expression is

x + 2y

.

Grouping terms in the numerator gives

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

. Dividing this by the denominator

(x^2 - 4y^2)

cancels out the identical non-zero factor

(x^2 - 4y^2)

, leaving the linear binomial

x + 2y

.

Adım Adım Çözüm

1

Group the four terms in the numerator in pairs to factor by grouping

x^3 + 2x^2y - 4xy^2 - 8y^3 = x^2(x + 2y) - 4y^2(x + 2y)

Grouping the first two terms and the last two terms allows factoring out

x^2

and

-4y^2

respectively.

2

Factor out the common binomial factor

(x + 2y)

from the numerator

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

Both terms share the common factor

(x + 2y)

.

3

Divide the factored numerator by the denominator

(x^2 - 4y^2)

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

Since

x \neq \pm 2y

,

x^2 - 4y^2 \neq 0

, allowing the common polynomial factor

(x^2 - 4y^2)

to be canceled.

Anahtar Kavram

Factoring Four-Term Polynomials by Grouping and Rational Expression Simplification

Soru 356Soru →

If $x$ is a negative real number such that $2x^2 + 5x - 12 = 0$ , what is the value of $(2x - 3)^2$ ?

-121

0

25

55

121

Cevabı ve açıklamayı göster

Cevap: 121

Cevap

121

Factoring

2x^2 + 5x - 12 = 0

gives

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

, which yields the solutions

x = \frac{3}{2}

and

x = -4

. Because the problem specifies that

x

is negative,

x

must equal

-4

. Substituting

-4

for

x

in the target expression yields

(2(-4) - 3)^2 = (-11)^2 = 121

.

Adım Adım Çözüm

1

Factor the given quadratic equation to find its roots.

2x^2 + 5x - 12 = (2x - 3)(x + 4) = 0

, giving solutions

x = \frac{3}{2}

and

x = -4

.

Factoring isolates the linear terms to solve for the values of

x

.

2

Apply the given constraint to select the appropriate root.

Since

x

is specified as a negative real number,

x = -4

.

The root

x = \frac{3}{2}

is positive and violates the problem constraint.

3

Substitute the valid root into the target expression

(2x - 3)^2

.

(2(-4) - 3)^2 = (-8 - 3)^2 = (-11)^2 = 121

.

Evaluating the expression with

x = -4

gives the final requested numerical value.

Anahtar Kavram

Solving quadratic equations by factoring and evaluating algebraic expressions under given sign constraints.

Tahmini Süre:1m 30s

Algebra

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