Radical and Rational Equations Alıştırma Soruları — SAT — Sayfa 2

What is the only real solution to the equation $\sqrt{3x + 13} = x + 3$ ?

-4

Both

-4

and

1

1

-1

Cevabı ve açıklamayı göster

Cevap: $1$

Cevap

The correct answer is the value 1.

Squaring both sides of the equation

\sqrt{3x + 13} = x + 3

yields

3x + 13 = x^2 + 6x + 9

. Rearranging this into standard quadratic form gives

x^2 + 3x - 4 = 0

. Factoring the quadratic expression gives

(x + 4)(x - 1) = 0

, which yields the potential solutions

x = 1

and

x = -4

. Substituting

x = 1

back into the original equation gives

\sqrt{16} = 4

, which is true. Therefore, the value

1

is the only real solution.

Adım Adım Çözüm

1

Square both sides of the equation to eliminate the radical.

3x + 13 = (x + 3)^2

Squaring both sides eliminates the radical, allowing us to solve for

x

.

2

Expand the right side and rewrite the equation in standard quadratic form.

x^2 + 3x - 4 = 0

Expanding

(x+3)^2

yields

x^2 + 6x + 9

. Subtracting

3x

and

13

from both sides sets the quadratic equation to zero.

3

Factor the quadratic equation.

(x + 4)(x - 1) = 0

We find two numbers that multiply to

-4

and add to

3

, which are

4

and

-1

.

4

Solve for the potential solutions.

x = -4

or

x = 1

Setting each factor to zero yields the potential solutions.

5

Verify each potential solution in the original equation.

x = 1

is valid, and

x = -4

is extraneous.

Plugging in

x = 1

gives

\sqrt{16} = 4

, which is true. Plugging in

x = -4

gives

\sqrt{1} = -1

, which is false since the principal square root is non-negative.

Anahtar Kavram

Solving radical equations and checking for extraneous solutions.

Tahmini Süre:45s

Soru 22Soru →

If $\sqrt{4x + 9} = 5$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

Squaring both sides of the equation

\sqrt{4x + 9} = 5

yields

4x + 9 = 25

. Subtracting 9 from both sides gives

4x = 16

, and dividing by 4 yields

x = 4

. Checking this solution in the original equation yields

\sqrt{4(4) + 9} = \sqrt{25} = 5

, which confirms it is correct.

Adım Adım Çözüm

1

Square both sides of the equation

4x + 9 = 25

To eliminate the square root on the left side of the equation.

2

Subtract 9 from both sides of the equation

4x = 16

To isolate the variable term on the left side of the equation.

3

Divide both sides of the equation by 4

x = 4

To solve for

x

.

4

Substitute the solved value back into the original equation

\sqrt{4(4) + 9} = 5 \implies 5 = 5

To ensure the solution is valid and not extraneous.

Anahtar Kavram

Solving radical equations of the form

\sqrt{ax + b} = c

by squaring both sides.

Soru 23Soru →

If $x$ is a real number that satisfies the equation $\frac{x}{x-3} - \frac{6}{x^2-3x} = \frac{2}{x}$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

2

The correct answer is 2. Multiplying the entire equation by the lowest common denominator,

x(x-3)

, yields

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

. Distributing the right side gives

x^2 - 6 = 2x - 6

. Adding 6 to both sides and subtracting

2x

results in

x^2 - 2x = 0

. Factoring this equation gives

x(x-2) = 0

, which has solutions

x = 0

and

x = 2

. However,

x = 0

is extraneous because it results in division by zero in the original equation. Thus, the only valid real solution is 2.

Adım Adım Çözüm

1

Identify the common denominator and multiply the equation by it.

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

To clear the fractions and simplify the rational equation.

2

Distribute and simplify the equation.

x^2 - 2x = 0

To set up the quadratic equation in standard form.

3

Factor and solve for

x

.

x = 0

or

x = 2

To find potential solutions to the equation.

4

Check for extraneous solutions by substituting the values back into the original denominators.

x = 2

x = 0

makes the denominators equal to zero and is therefore extraneous.

Anahtar Kavram

Solving rational equations and identifying extraneous solutions.

Soru 24Soru →

If $\frac{x - 3}{4} = \frac{9}{x - 3}$ and $x > 3$ , what is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

9

Cross-multiplying the equation

\frac{x - 3}{4} = \frac{9}{x - 3}

gives

(x - 3)^2 = 36

. Taking the square root of both sides gives

x - 3 = 6

or

x - 3 = -6

. Solving for

x

yields

x = 9

or

x = -3

. Since the question specifies the constraint

x > 3

, the only valid solution is

9

.

Adım Adım Çözüm

1

Cross-multiply the equation to eliminate the denominators.

(x - 3)^2 = 36

Multiplying both sides of the equation by

4(x - 3)

simplifies the rational equation into a quadratic form.

2

Take the square root of both sides of the equation.

x - 3 = 6

or

x - 3 = -6

Taking the square root of a squared term yields both positive and negative root options.

3

Solve each linear equation for

x

and apply the constraint

x > 3

.

x = 9

Adding

3

to both sides gives

x = 9

or

x = -3

. The constraint

x > 3

excludes the negative solution, leaving

9

as the only valid value.

Anahtar Kavram

Solving rational equations by cross-multiplication and factoring the resulting quadratic equation while adhering to domain constraints.

Soru 25Soru →

If

x

is a real solution to the equation

\frac{2x - 5}{x - 3} + \frac{x - 5}{x^2 - 4x + 3} = -\frac{1}{x - 1}

what is the value of

1 - 2x

?

-5

2

-2

0

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

2

The correct answer is 2. Multiplying the entire equation by the least common denominator

(x - 3)(x - 1)

simplifies the rational equation to the quadratic equation

2x^2 - 5x - 3 = 0

. This quadratic factors into

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

, yielding two potential solutions:

x = -1/2

and

x = 3

. However,

x = 3

is an extraneous solution because it results in a denominator of zero in the original equation. Thus, the only valid real solution is

x = -1/2

. Substituting this into the expression

1 - 2x

yields

1 - 2(-1/2) = 2

.

Adım Adım Çözüm

1

Factor the quadratic expression in the denominator to identify the least common denominator.

The quadratic in the denominator of the second term factors as

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

. Thus, the least common denominator for all terms in the equation is

(x - 3)(x - 1)

, with the restrictions that

x \neq 3

and

x \neq 1

.

Finding a common denominator allows us to multiply both sides of the equation to clear all rational expressions.

2

Multiply both sides of the equation by the least common denominator

(x - 3)(x - 1)

.

(2x - 5)(x - 1) + (x - 5) = -1(x - 3)

This clears the fractions and converts the rational equation into a polynomial equation.

3

Expand and simplify both sides of the equation.

(2x^2 - 7x + 5) + (x - 5) = -x + 3

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

Expanding the terms allows us to combine like terms and set up a quadratic equation.

4

Rearrange the terms to set the quadratic equation equal to zero.

2x^2 - 5x - 3 = 0

Quadratic equations must be set to zero to be solved by factoring or using the quadratic formula.

5

Solve the quadratic equation by factoring.

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

This gives two potential solutions:

x = -1/2

and

x = 3

.

Factoring is the most direct algebraic method to find the roots of the quadratic equation.

6

Check the potential solutions against the original domain restrictions to identify any extraneous solutions.

For

x = 3

, the denominators

x - 3

and

x^2 - 4x + 3

equal zero, so

x = 3

is an extraneous solution. For

x = -1/2

, the denominators are non-zero, so

x = -1/2

is the only valid solution.

Squaring or multiplying by variable expressions can introduce solutions that are undefined in the original equation.

7

Substitute the valid solution

x = -1/2

into the expression

1 - 2x

.

1 - 2\left(-\frac{1}{2}\right) = 1 + 1 = 2

The question asks for the value of the expression

1 - 2x

, not the value of

x

itself.

Anahtar Kavram

Solving rational equations by clearing denominators and checking for extraneous solutions.

Soru 26Soru →

If $x$ is the real solution to the equation below, what is the value of $x$ ?

\frac{2x^2 - 30x + 100}{x - 5} + \frac{140}{x + 2} = 10

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The correct answer is 8.

To solve the rational equation, we first state the domain restrictions as

x \neq 5

and

x \neq -2

because these values make the denominators zero. Next, we factor the numerator of the first term:

2x^2 - 30x + 100 = 2(x^2 - 15x + 50) = 2(x - 5)(x - 10)

. Since

x \neq 5

, we can cancel the common factor of

x - 5

from the numerator and denominator, which simplifies the first term to

2(x - 10) = 2x - 20

. The equation can then be rewritten as

2x - 20 + \frac{140}{x + 2} = 10

. Subtracting 10 from both sides yields

2x - 30 + \frac{140}{x + 2} = 0

. Dividing the entire equation by 2 simplifies the coefficients, giving

x - 15 + \frac{70}{x + 2} = 0

. To clear the fraction, we multiply the entire equation by

x + 2

, leading to

(x - 15)(x + 2) + 70 = 0

. Expanding the product gives

x^2 - 13x - 30 + 70 = 0

, which simplifies to

x^2 - 13x + 40 = 0

. Factoring this quadratic equation yields

(x - 5)(x - 8) = 0

, giving the potential solutions

x = 5

and

x = 8

. However,

x = 5

is extraneous because it is restricted from the domain of the original rational expression. Therefore, the only real solution is 8.

Adım Adım Çözüm

1

Find the values of

x

that make any denominator equal to zero to identify domain restrictions.

x \neq 5

and

x \neq -2

Dividing by zero is undefined in real numbers, so these values are excluded from the domain.

2

Factor the numerator of the first term,

2x^2 - 30x + 100

.

2(x - 5)(x - 10)

Factoring allows us to check for common factors in the numerator and denominator.

3

Simplify the first rational term by dividing the numerator and denominator by the common factor

x - 5

, under the assumption that

x \neq 5

.

2(x - 10) = 2x - 20

Reducing the expression simplifies the algebraic structure of the equation.

4

Substitute the simplified expression back into the original equation and subtract 10 from both sides.

2x - 30 + \frac{140}{x + 2} = 0

To consolidate the constant terms and prepare the equation for clearing the remaining fraction.

5

Divide the entire equation by 2.

x - 15 + \frac{70}{x + 2} = 0

Dividing by the common factor simplifies the coefficients before expanding.

6

Multiply the entire equation by

x + 2

to clear the denominator, then expand and combine like terms.

x^2 - 13x + 40 = 0

This transforms the rational equation into a standard quadratic equation.

7

Factor the quadratic equation.

(x - 5)(x - 8) = 0

Factoring is the most direct method to find the roots of the quadratic equation.

8

Verify the candidate solutions against the domain restrictions identified in Step 1.

The candidate solution

x = 5

is extraneous because it is excluded from the domain of the original equation. The candidate solution

x = 8

is valid.

Extraneous solutions introduced during algebraic steps must be excluded from the final solution set.

Anahtar Kavram

Solving rational equations by simplifying terms, finding common denominators, and identifying extraneous solutions.

Soru 27Soru →

If

x

is a real solution to the equation

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

what is the value of

5x - 4

?

6

-14

-5

5

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

-5

The correct answer is

-5

. Solving the equation by multiplying both sides by the least common denominator

(x-1)(x-2)

yields the quadratic equation

5x^2 - 9x - 2 = 0

. Factoring this equation gives the potential solutions

x = -1/5

and

x = 2

. However, substituting

x = 2

into the original equation results in division by zero, meaning

x = 2

is extraneous and must be discarded. The only valid solution is

x = -1/5

. Substituting this value into the expression

5x - 4

yields

5(-1/5) - 4 = -5

.

Adım Adım Çözüm

1

Identify the least common denominator (LCD) and multiply both sides of the equation to clear the denominators.

The LCD is

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

. Multiplying both sides by the LCD gives:

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

Multiplying by the LCD simplifies the rational equation into a polynomial equation.

2

Expand the terms and collect like terms to form a quadratic equation.

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

5x^2 - 8x = x + 2

5x^2 - 9x - 2 = 0

This puts the equation in the standard quadratic form

ax^2 + bx + c = 0

.

3

Factor the quadratic equation to find potential solutions.

The quadratic factors as:

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

This yields potential solutions

x = -\frac{1}{5}

and

x = 2

.

Factoring allows us to isolate the candidate values of

x

.

4

Check the potential solutions in the original equation to identify any extraneous solutions.

For

x = 2

, the denominators

x - 2

and

x^2 - 3x + 2

become zero, which is undefined. Therefore,

x = 2

is an extraneous solution. For

x = -\frac{1}{5}

, all denominators are non-zero, so it is the only valid solution.

Values that make any denominator in the original equation equal to zero must be excluded.

5

Substitute the valid solution into the target expression

5x - 4

.

5\left(-\frac{1}{5}\right) - 4 = -1 - 4 = -5

Evaluating the expression with the only valid solution gives the final answer.

Anahtar Kavram

Solving rational equations by clearing denominators and checking for extraneous solutions.

Tahmini Süre:2m 30s

Soru 28Soru →

An equation is given as follows:

\frac{x}{x - 3} - \frac{4}{x + 2} = \frac{20}{x^2 - x - 6}

What is the only value of $x$ for which this equation is true?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

To solve the equation, factor the denominator on the right side:

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

. The least common denominator is

(x - 3)(x + 2)

. Multiplying both sides by

(x - 3)(x + 2)

clears the fractions, resulting in

x(x + 2) - 4(x - 3) = 20

. Expanding the terms gives

x^2 + 2x - 4x + 12 = 20

. Simplifying and writing this in standard form yields

x^2 - 2x - 8 = 0

. Factoring the quadratic gives

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

, which gives the candidate solutions

x = 4

and

x = -2

. Substituting

x = -2

back into the original equation results in a denominator of zero, so

x = -2

is an extraneous solution. The only valid solution is

x = 4

.

Adım Adım Çözüm

1

Factor the quadratic trinomial in the denominator of the right side of the equation.

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

Identifying the factors of the quadratic trinomial helps find the least common denominator of the rational equation.

2

Multiply the entire equation by the least common denominator

(x - 3)(x + 2)

to clear the fractions.

x(x + 2) - 4(x - 3) = 20

, with the constraints that

x \neq 3

and

x \neq -2

.

This simplifies the rational equation into a standard polynomial equation.

3

Distribute and combine like terms to write the equation in standard quadratic form.

x^2 - 2x - 8 = 0

Rewriting the equation in the form

ax^2 + bx + c = 0

is necessary to solve it by factoring.

4

Factor the quadratic equation.

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

Factoring allows us to apply the zero product property to find candidate solutions.

5

Check the candidate solutions

x = 4

and

x = -2

against the original equation to identify extraneous solutions.

x = 4

is the only valid solution because

x = -2

makes the denominators in the original equation equal to zero.

Any solution that makes a denominator in the original rational expression equal to zero is extraneous and must be excluded.

Anahtar Kavram

Solving rational equations by finding a common denominator, clearing fractions, and checking for extraneous solutions.

Soru 29Soru →

What is the sum of all positive real solutions to the equation

\frac{x^2 - 3x}{x^2 - 3x - 2} + \frac{x^2 - 3x - 2}{x^2 - 3x} = \frac{5}{2}

?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The sum of all positive real solutions is 7.

By substituting

u = x^2 - 3x

, the original rational equation simplifies to

\frac{u}{u - 2} + \frac{u - 2}{u} = \frac{5}{2}

. Multiplying both sides by the common denominator

2u(u-2)

and simplifying results in the quadratic equation

u^2 - 2u - 8 = 0

. Factoring gives

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

, so

u = 4

or

u = -2

. Substituting back

x^2 - 3x

for

u

leads to two quadratic equations:

x^2 - 3x = 4

(which has solutions

x = 4

and

x = -1

) and

x^2 - 3x = -2

(which has solutions

x = 2

and

x = 1

). Checking the denominators, none of these solutions make the original denominators zero, so they are all valid. The positive solutions are

1

,

2

, and

4

, and their sum is

1 + 2 + 4 = 7

.

Adım Adım Çözüm

1

Introduce a substitution variable to simplify the rational equation.

Letting

u = x^2 - 3x

transforms the equation into

\frac{u}{u - 2} + \frac{u - 2}{u} = \frac{5}{2}

.

This reduces the degree of the rational expression and simplifies the algebraic manipulation required to solve it.

2

Eliminate the denominators by multiplying by the least common denominator.

Multiplying by

2u(u-2)

gives

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

, which simplifies to

u^2 - 2u - 8 = 0

.

This converts the rational equation into a standard quadratic equation in terms of

u

.

3

Solve the quadratic equation for

u

by factoring.

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

, which gives

u = 4

or

u = -2

.

Finding the values of

u

allows us to set up equations to solve for the original variable

x

.

4

Substitute back

x^2 - 3x

for

u

and solve the resulting quadratic equations for

x

.

From

x^2 - 3x = 4

, we get

(x-4)(x+1) = 0 \implies x = 4, -1

. From

x^2 - 3x = -2

, we get

(x-2)(x-1) = 0 \implies x = 2, 1

.

This yields all real values of

x

that satisfy the original algebraic structure.

5

Filter for positive real solutions and calculate their sum.

The positive solutions are

1

,

2

, and

4

. Their sum is

1 + 2 + 4 = 7

.

The question specifically asks for the sum of only the positive real solutions.

Anahtar Kavram

Solving rational equations using algebraic substitution and factoring quadratic equations.

Alternatif Yöntem

Instead of using substitution directly, the equation can be solved by multiplying by the common denominator

(x^2 - 3x - 2)(x^2 - 3x)

to get a fourth-degree polynomial:

2(x^2 - 3x)^2 + 2(x^2 - 3x - 2)^2 = 5(x^2 - 3x)(x^2 - 3x - 2)

. Letting

z = x^2 - 3x

at this stage simplifies this expression to

2z^2 + 2(z-2)^2 = 5z(z-2)

, which avoids full expansion into a fourth-degree polynomial and leads to the same quadratic in

z

.

Tahmini Süre:3m 0s

Soru 30Soru →

In the equation below,

x

is a real number.

\sqrt{7 - 6x} = x + 8

What is the value of

x + 5

?

8

-14

2

-3

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The correct answer is the value 2.

Substituting the only valid solution of the radical equation, which is

-3

, into the expression

x + 5

yields

2

.

Adım Adım Çözüm

1

Square both sides of the equation to eliminate the radical sign.

7 - 6x = (x + 8)^2

, which expands to

7 - 6x = x^2 + 16x + 64

.

Squaring both sides of a radical equation allows it to be rewritten as a standard polynomial equation.

2

Rearrange the terms to set the quadratic equation equal to zero.

x^2 + 22x + 57 = 0

Grouping all terms on one side of the equation puts it in standard quadratic form.

3

Factor the quadratic equation to find potential solutions for

x

.

(x + 19)(x + 3) = 0

, which gives the potential solutions

x = -19

and

x = -3

.

Factoring allows us to solve the quadratic equation easily.

4

Test the potential solutions in the original equation to identify any extraneous solutions.

Substituting

x = -3

yields

\sqrt{7 - 6(-3)} = -3 + 8 \Rightarrow \sqrt{25} = 5

, which is true. Substituting

x = -19

yields

\sqrt{7 - 6(-19)} = -19 + 8 \Rightarrow \sqrt{121} = -11

, which is false because a principal square root must be non-negative. Therefore,

x = -3

is the only valid solution.

Squaring both sides can introduce extraneous solutions that must be discarded.

5

Substitute the valid solution

x = -3

into the target expression

x + 5

.

-3 + 5 = 2

The question asks for the value of the expression

x + 5

.

Anahtar Kavram

Solving radical equations by squaring and checking for extraneous solutions.

Soru 31Soru →

What is the larger solution to the equation below?

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

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The larger solution to the equation is 6.

To solve the rational equation, we clear the denominators by multiplying both sides by

x(x - 2)

, which yields the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation gives

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

, leading to the solutions

x = 6

and

x = 1

. Both values are valid because they do not make any denominator of the original expression equal to zero. The larger of the two values is 6.

Adım Adım Çözüm

1

Multiply both sides of the equation by the common denominator

x(x - 2)

to clear the fractions.

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

Clearing denominators simplifies the rational equation into a polynomial equation.

2

Expand both sides of the equation.

x^2 + 3x - 6 = 2x^2 - 4x

Distributing the multiplication allows us to combine like terms.

3

Move all terms to one side of the equation to write it in the standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 7x + 6 = 0

Setting the quadratic expression to zero prepares it for factoring.

4

Factor the quadratic equation.

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

Finding two numbers that multiply to

6

and sum to

-7

gives

-6

and

-1

.

5

Identify the values of

x

that satisfy the factored equation and check for extraneous solutions.

x = 6

or

x = 1

Neither solution makes the denominators in the original equation,

x-2

or

x

, equal to zero, so both are valid. The larger of these two solutions is 6.

Anahtar Kavram

Solving rational equations by clearing denominators to form a quadratic equation, and verifying solutions against the domain constraints.

Soru 32Soru →

For all real values of

x

that satisfy the equation below, what is the value of

x + 1

?

\sqrt{5x + 39} - 3 = x

-5

-4

6

7

Cevabı ve açıklamayı göster

Cevap: $6$

Cevap

The correct answer is

6

.

By adding

3

to both sides, we get

\sqrt{5x + 39} = x + 3

. Squaring both sides yields

5x + 39 = x^2 + 6x + 9

. Setting this quadratic equation to zero gives

x^2 + x - 30 = 0

, which factors as

(x + 6)(x - 5) = 0

. The potential solutions are

x = 5

and

x = -6

. Checking these values in the original equation shows that only

x = 5

is valid:

\sqrt{5(5) + 39} - 3 = 8 - 3 = 5

, which satisfies the equation. The value

x = -6

is extraneous since

\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6

. Therefore, the value of

x + 1

is

5 + 1 = 6

.

Adım Adım Çözüm

1

Isolate the radical expression on one side of the equation.

\sqrt{5x + 39} = x + 3

Isolating the radical allows us to square both sides to eliminate the square root.

2

Square both sides of the equation.

5x + 39 = (x + 3)^2 \implies 5x + 39 = x^2 + 6x + 9

Squaring a square root removes the radical, allowing us to solve the resulting quadratic equation.

3

Rearrange the terms to form a standard quadratic equation of the form

ax^2 + bx + c = 0

.

x^2 + x - 30 = 0

Subtracting

5x

and

39

from both sides sets the quadratic expression equal to zero.

4

Factor the quadratic equation to find potential solutions.

(x + 6)(x - 5) = 0 \implies x = -6 \text{ or } x = 5

Finding two numbers that multiply to

-30

and add up to

1

gives

6

and

-5

.

5

Substitute both candidate solutions back into the original equation to check for extraneous solutions.

For

x = 5

:

\sqrt{5(5) + 39} - 3 = 8 - 3 = 5

(Valid). For

x = -6

:

\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6

(Extraneous).

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original relation.

6

Use the valid solution to calculate the requested expression.

x + 1 = 5 + 1 = 6

The question asks for the value of

x + 1

.

Anahtar Kavram

Solving radical equations and identifying extraneous solutions

Soru 33Soru →

If $x$ satisfies the equation below, what is the value of $x - 3$ ?

x - \sqrt{2x + 7} = 4

-2

6

9

-4

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

The correct answer is the value

6

. Isolating the radical in

x - \sqrt{2x + 7} = 4

gives

x - 4 = \sqrt{2x + 7}

. Squaring both sides yields

x^2 - 8x + 16 = 2x + 7

, which simplifies to the quadratic equation

x^2 - 10x + 9 = 0

. Factoring this equation gives

(x - 9)(x - 1) = 0

, yielding potential solutions of

x = 9

and

x = 1

. Substituting these potential solutions back into the original equation reveals that

x = 9

is a valid solution because

9 - \sqrt{25} = 4

, whereas

x = 1

is extraneous because

1 - \sqrt{9} = -2 \neq 4

. The value of the expression

x - 3

is

9 - 3 = 6

.

Adım Adım Çözüm

1

Isolate the radical term in the equation.

x - 4 = \sqrt{2x + 7}

Before squaring both sides of a radical equation, it is best to isolate the radical term on one side.

2

Square both sides of the equation to eliminate the radical.

(x - 4)^2 = 2x + 7

Squaring a square root removes the radical, allowing us to solve the resulting algebraic equation.

3

Expand the squared binomial and move all terms to one side to set the quadratic equation to zero.

x^2 - 10x + 9 = 0

(x - 4)^2

expands to

x^2 - 8x + 16

. Subtracting

2x

and

7

from both sides gives the standard quadratic form

ax^2 + bx + c = 0

.

4

Factor the quadratic equation.

(x - 9)(x - 1) = 0

Finding two numbers that multiply to

9

and add to

-10

gives

-9

and

-1

. Thus, the potential solutions are

x = 9

and

x = 1

.

5

Substitute both potential solutions back into the original equation to check for extraneous solutions.

x = 9

is valid, while

x = 1

is extraneous.

Checking

x = 9

:

9 - \sqrt{2(9)+7} = 9 - 5 = 4

(true). Checking

x = 1

:

1 - \sqrt{2(1)+7} = 1 - 3 = -2 \neq 4

(false). Thus,

x = 9

is the only real solution.

6

Calculate the value of the requested expression

x - 3

.

9 - 3 = 6

Since the only valid solution is

x = 9

, we substitute

9

into the expression

x - 3

to find the final answer.

Anahtar Kavram

Radical equations require isolating the radical, squaring both sides, solving the resulting equation, and checking for extraneous solutions that do not satisfy the original equation.

Tahmini Süre:1m 30s

Soru 34Soru →

An equation is shown below.

\frac{x}{x - 1} - \frac{2}{x + 2} = \frac{6}{x^2 + x - 2}

What is the value of the real solution to the equation?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The correct answer is 2.

The correct answer is 2. Multiplying both sides by the least common denominator

(x - 1)(x + 2)

results in the quadratic equation

x(x + 2) - 2(x - 1) = 6

. Simplifying this equation yields

x^2 + 2 = 6

, which has solutions

x = 2

and

x = -2

. However, substituting

x = -2

into the original equation results in division by zero, making it an extraneous solution. Therefore,

2

is the only valid real solution.

Adım Adım Çözüm

1

Multiply the entire equation by the least common denominator,

(x - 1)(x + 2) = x^2 + x - 2

, to clear the denominators.

x(x + 2) - 2(x - 1) = 6

This simplifies the rational equation into a polynomial equation.

2

Expand the terms and simplify the equation.

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

, which simplifies to

x^2 + 2 = 6

.

Distributing the terms allows us to group like terms and solve for the variable.

3

Solve the quadratic equation for

x

.

x^2 = 4

, which gives

x = 2

or

x = -2

.

Subtracting 2 from both sides isolates the squared variable.

4

Check for extraneous solutions by substituting the potential solutions back into the original denominators.

For

x = -2

, the denominator

x + 2

becomes 0, which is undefined. For

x = 2

, all denominators are non-zero.

Solutions that make any denominator in the original equation equal to zero are extraneous and must be excluded.

Anahtar Kavram

Solving rational equations by finding a common denominator and checking for extraneous solutions.

Soru 35Soru →

If

x

is a real number that satisfies the equation below, what is the value of

x

?

\frac{x + 2}{x - 4} - \frac{3}{x} = \frac{12}{x^2 - 4x}

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The only valid real solution to the equation is

1

.

Multiplying the equation by the common denominator

x(x - 4)

simplifies the equation to

x^2 - x = 0

. Solving this quadratic equation yields

x = 0

and

x = 1

. Because

x = 0

leads to a division by zero in the original equation, it is extraneous. Therefore, the only valid real solution is

1

.

Adım Adım Çözüm

1

Determine the common denominator and restrictions for the rational equation.

The common denominator is

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

. The restrictions are

x \neq 0

and

x \neq 4

.

Finding the common denominator allows us to eliminate fractions, while the restrictions help us identify potential extraneous solutions.

2

Multiply the entire equation by the common denominator

x(x - 4)

.

x(x + 2) - 3(x - 4) = 12

This step clears the rational expressions, leaving a polynomial equation.

3

Expand and simplify the polynomial equation.

x^2 + 2x - 3x + 12 = 12

, which simplifies to

x^2 - x = 0

.

Expanding the terms allows us to combine like terms and set the quadratic equation to zero.

4

Factor the quadratic equation to solve for

x

.

x(x - 1) = 0

, giving candidate solutions

x = 0

or

x = 1

.

Applying the zero-product property identifies the roots of the quadratic equation.

5

Verify the candidate solutions against the initial restrictions.

Since

x = 0

makes the denominators in the original equation equal to zero, it is extraneous. The candidate solution

x = 1

is valid.

We must verify solutions because multiplying by a variable expression can introduce extraneous roots that make the original expressions undefined.

Anahtar Kavram

Solving rational equations by clearing denominators and checking for extraneous solutions.

Tahmini Süre:1m 30s

Soru 36Soru →

If

x

satisfies the equation below, what is the value of

x - 1

?

\sqrt{30 - 2x} = x - 3

-4

2

6

8

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

The correct answer is

6

. Squaring both sides of the equation

\sqrt{30 - 2x} = x - 3

yields

30 - 2x = x^2 - 6x + 9

. Rearranging terms into standard quadratic form gives

x^2 - 4x - 21 = 0

, which factors as

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

. This gives potential solutions of

x = 7

and

x = -3

. Checking these in the original equation shows that

x = 7

is valid since

\sqrt{30 - 2(7)} = 7 - 3 \implies 4 = 4

, while

x = -3

is extraneous since

\sqrt{30 - 2(-3)} = -3 - 3 \implies 6 \neq -6

. Therefore, the value of the requested expression

x - 1

is

7 - 1 = 6

.

Adım Adım Çözüm

1

Square both sides of the equation to eliminate the radical.

30 - 2x = (x - 3)^2

To solve a radical equation, squaring both sides isolates the terms under the square root.

2

Expand the right side and move all terms to one side to set the quadratic equation to zero.

x^2 - 4x - 21 = 0

Expanding

(x - 3)^2

yields

x^2 - 6x + 9

. Subtracting

30

and adding

2x

to both sides results in a standard quadratic form.

3

Factor the quadratic equation.

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

Factoring the quadratic helps find the potential solutions for

x

.

4

Identify potential solutions and substitute them back into the original equation to check for extraneous solutions.

x = 7

is the only valid solution;

x = -3

is extraneous.

Substituting

x = 7

gives

\sqrt{30 - 14} = 7 - 3

, which simplifies to

4 = 4

(true). Substituting

x = -3

gives

\sqrt{30 - 2(-3)} = -3 - 3

, which simplifies to

6 = -6

(false).

5

Calculate the value of the expression

x - 1

using the valid solution

x = 7

.

7 - 1 = 6

The question asks for the value of

x - 1

, so we substitute

7

for

x

.

Anahtar Kavram

Solving radical equations and identifying extraneous solutions.

Tahmini Süre:1m 35s

Soru 37Soru →

The equation $\sqrt{2x + 12} - x = -6$ has one real solution. What is this solution?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

Substituting 12 into the original equation

\sqrt{2x + 12} - x = -6

yields

\sqrt{2(12) + 12} - 12 = \sqrt{36} - 12 = 6 - 12 = -6

. Since this creates a true statement, 12 is the unique real solution to the equation.

Adım Adım Çözüm

1

Isolate the radical term by adding

x

to both sides of the equation.

\sqrt{2x + 12} = x - 6

Isolating the radical on one side allows us to eliminate it by squaring both sides in the next step.

2

Square both sides of the equation to clear the square root.

2x + 12 = (x - 6)^2

Squaring a square root cancels the radical, allowing us to solve the equation algebraically.

3

Expand the squared binomial on the right-hand side.

2x + 12 = x^2 - 12x + 36

Expanding

(x - 6)^2

to

x^2 - 12x + 36

helps set up a standard quadratic form.

4

Subtract

2x

and

12

from both sides to form a quadratic equation equal to zero.

x^2 - 14x + 24 = 0

A quadratic equation must be in the form

ax^2 + bx + c = 0

to solve by factoring or the quadratic formula.

5

Factor the quadratic equation.

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

Factoring finds two numbers that multiply to 24 and add to -14, which are -12 and -2.

6

Test the potential solutions

x = 12

and

x = 2

in the original equation to identify any extraneous solutions.

For

x = 2

:

\sqrt{2(2) + 12} - 2 = \sqrt{16} - 2 = 4 - 2 = 2 \neq -6

. For

x = 12

:

\sqrt{2(12) + 12} - 12 = \sqrt{36} - 12 = 6 - 12 = -6

. Thus,

x = 12

is the only valid solution.

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.

Anahtar Kavram

Solving radical equations by isolating the radical, squaring both sides, solving the resulting quadratic equation, and verifying all potential solutions to check for extraneous values.

Tahmini Süre:1m 30s

Soru 38Soru →

An equation is shown below.

\sqrt{2x + 7} = x - 4

If

x

satisfies the equation above, what is the value of

x + 3

?

4

2

12

6

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12

To solve the equation, square both sides to obtain

2x + 7 = x^2 - 8x + 16

. Rearranging terms gives the quadratic equation

x^2 - 10x + 9 = 0

, which factors as

(x - 9)(x - 1) = 0

. This yields potential solutions of

x = 9

and

x = 1

. Testing these in the original equation shows that

x = 9

is a valid solution because

\sqrt{25} = 5

, but

x = 1

is extraneous because

\sqrt{9} \neq -3

. Thus, the only real solution is

x = 9

. Evaluating

x + 3

for this solution gives

9 + 3 = 12

.

Adım Adım Çözüm

1

Square both sides of the equation to eliminate the radical.

2x + 7 = (x - 4)^2

, which expands to

2x + 7 = x^2 - 8x + 16

.

Squaring both sides is the standard method to solve radical equations.

2

Rearrange the terms to set the quadratic equation to zero.

x^2 - 10x + 9 = 0

.

This allows us to solve the quadratic equation by factoring.

3

Factor the quadratic equation.

(x - 9)(x - 1) = 0

, giving potential solutions of

x = 9

and

x = 1

.

Factoring helps find the roots of the quadratic equation.

4

Verify each potential solution in the original equation to check for extraneous roots.

x = 9

is valid because

\sqrt{2(9)+7} = 9-4

is true (

5=5

).

x = 1

is extraneous because

\sqrt{2(1)+7} = 1-4

is false (

3 \neq -3

).

Squaring both sides can introduce extraneous solutions that do not satisfy the original equation.

5

Substitute the valid solution

x = 9

into the expression

x + 3

.

9 + 3 = 12

.

The question asks for the value of the expression

x + 3

rather than just

x

.

Anahtar Kavram

Solving radical equations and checking for extraneous solutions

Soru 39Soru →

An equation is shown below.

\frac{x+1}{x-2} + \frac{1}{x} = \frac{6}{x^2-2x}

If

x

satisfies the equation above, what is the value of

x

?

-2

2

-4

4

Cevabı ve açıklamayı göster

Cevap: -4

Cevap

-4

The correct answer is

-4

. To solve the rational equation, we first identify the least common denominator as

x(x-2)

. Multiplying both sides by this expression eliminates the denominators, yielding

x(x+1) + (x-2) = 6

. Expanding and simplifying gives the quadratic equation

x^2 + 2x - 8 = 0

, which factors as

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

. This yields potential solutions of

x = -4

and

x = 2

. However, substituting

x = 2

back into the original equation results in division by zero, making

x = 2

extraneous. The only valid solution is

-4

.

Adım Adım Çözüm

1

Find the least common denominator (LCD) of the rational expressions in the equation.

The denominators are

x-2

,

x

, and

x^2-2x

. Since

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

, the LCD is

x(x-2)

, with the restriction that

x \neq 0

and

x \neq 2

.

Finding a common denominator allows us to clear the fractions by multiplying both sides of the equation.

2

Multiply each term of the equation by the LCD to eliminate the denominators.

x(x+1) + 1(x-2) = 6

This clears the denominators and converts the rational equation into a polynomial equation.

3

Expand and simplify the resulting equation.

x^2 + x + x - 2 = 6

, which simplifies to

x^2 + 2x - 2 = 6

.

Combining like terms prepares the equation to be written in standard quadratic form.

4

Rearrange the terms into standard quadratic form:

ax^2 + bx + c = 0

.

x^2 + 2x - 8 = 0

Subtracting

6

from both sides sets the quadratic equation to zero so it can be solved by factoring.

5

Factor the quadratic equation.

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

Finding two numbers that multiply to

-8

and add to

2

(

4

and

-2

) allows us to solve for

x

.

6

Find the potential solutions by setting each factor to zero.

x = -4

or

x = 2

Applying the zero product property gives the potential roots of the quadratic equation.

7

Check the potential solutions in the original equation to identify any extraneous solutions.

For

x = -4

, the equation is defined and valid. For

x = 2

, the denominators

x-2

and

x^2-2x

become zero, which is undefined.

Rational equations can produce extraneous solutions that make the original denominators zero, so they must be checked and discarded.

Anahtar Kavram

Solving rational equations by finding a common denominator and checking for extraneous solutions.

Soru 40Soru →

If $x$ satisfies the equation below, what is the value of $x + 5$ ?

x - 4 = \sqrt{4x + 5}

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The correct answer is 16.

Squaring both sides of the equation

x - 4 = \sqrt{4x + 5}

results in

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

, which expands to

x^2 - 8x + 16 = 4x + 5

. Subtracting

4x + 5

from both sides gives the quadratic equation

x^2 - 12x + 11 = 0

. Factoring this quadratic equation yields

(x - 11)(x - 1) = 0

, giving candidate solutions of

x = 11

and

x = 1

. Checking these solutions in the original equation shows that

x = 11

is valid (

7 = 7

), whereas

x = 1

is extraneous (

-3 = 3

is false). Therefore, the only real solution is

x = 11

, and the value of the expression

x + 5

is

11 + 5 = 16

.

Adım Adım Çözüm

1

Square both sides of the equation to remove the radical.

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

Squaring a square root isolates the expression under the radical.

2

Expand the squared binomial on the left side.

x^2 - 8x + 16 = 4x + 5

Applying the distributive property to

(x - 4)(x - 4)

yields

x^2 - 8x + 16

.

3

Rearrange the equation to set it equal to zero.

x^2 - 12x + 11 = 0

Subtracting

4x + 5

from both sides simplifies the equation into standard quadratic form.

4

Factor the quadratic equation.

(x - 11)(x - 1) = 0

Finding two numbers that multiply to 11 and add to -12 gives -11 and -1.

5

Solve for the candidate values of x.

x = 11

and

x = 1

Setting each factor equal to zero yields the possible solutions.

6

Check the candidate values in the original equation to identify any extraneous solutions.

x = 11

is valid;

x = 1

is extraneous.

Substituting

x = 1

results in

-3 = 3

, which is false because the principal square root is always non-negative.

7

Evaluate the target expression using the valid solution.

11 + 5 = 16

The question asks for the value of

x + 5

, so we substitute the only valid solution,

x = 11

.

Anahtar Kavram

Solving radical equations by squaring both sides and checking for extraneous solutions.

Radical and Rational Equations

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