Rational and Radical Expressions and Equations Alıştırma Soruları — ACT — Sayfa 2

Determine the values of $x$ for which the given rational expression is undefined.

Aşağıdaki boşlukları doldurun

The rational expression

\frac{x + 5}{x^2 - 4x - 12}

is undefined for two real values of

x

. The smaller of these values is and the larger of these values is .

Cevabı ve açıklamayı göster

Cevap

The rational expression is undefined for

x = -2

and

x = 6

.

The expression is undefined when the denominator is zero. Solving

x^2 - 4x - 12 = 0

yields

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

, which gives

x = -2

and

x = 6

. The smaller value is

-2

and the larger value is

6

.

Adım Adım Çözüm

1

Identify the condition that makes a rational expression undefined.

The denominator must equal zero:

x^2 - 4x - 12 = 0

Division by zero is undefined in the set of real numbers.

2

Factor the quadratic equation

x^2 - 4x - 12 = 0

.

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

We need to find two numbers that multiply to

-12

and add to

-4

. These numbers are

-6

and

2

.

3

Solve for

x

by setting each factor to zero.

x = 6

or

x = -2

By the zero product property, if

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

, then

x - 6 = 0

or

x + 2 = 0

.

4

Determine the smaller and larger values to place in the blanks.

Smaller value is

-2

and larger value is

6

.

Comparing the two numbers,

-2

is less than

6

.

Anahtar Kavram

A rational expression is undefined when its denominator equals zero.

Soru 22Soru →

When solving the radical equation $\sqrt{3x+10} + \sqrt{x+2} = 2$ for all real values of $x$ , squaring both sides yields two potential solutions. What is the numerical value of the extraneous solution?

Aşağıdaki boşlukları doldurun

The numerical value of the extraneous solution is .

Cevabı ve açıklamayı göster

Cevap

The extraneous solution is

x = 2

.

The value

x = 2

is the extraneous solution because substituting it into the original equation yields

\sqrt{16} + \sqrt{4} = 6 \neq 2

, showing it does not satisfy the original equation despite being generated by the algebraic squaring process.

Adım Adım Çözüm

1

Isolate one of the radical terms in the equation.

\sqrt{3x+10} = 2 - \sqrt{x+2}

Isolating one radical simplifies the squaring process.

2

Square both sides of the equation.

3x+10 = 4 - 4\sqrt{x+2} + (x+2)

, which simplifies to

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

.

Squaring eliminates the isolated radical.

3

Divide both sides by 2 and square both sides again.

x+2 = -2\sqrt{x+2} \Rightarrow (x+2)^2 = 4(x+2)

, which simplifies to

x^2 - 4 = 0

.

This eliminates the remaining radical term.

4

Solve the quadratic equation for

x

.

x = 2

or

x = -2

Factoring

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

gives the potential solutions.

5

Check both potential solutions in the original equation.

For

x = -2

:

\sqrt{4} + \sqrt{0} = 2

, which is true. For

x = 2

:

\sqrt{16} + \sqrt{4} = 6 \neq 2

, which is false.

This identifies which solution is extraneous.

Anahtar Kavram

Solving radical equations and verifying solutions to identify extraneous roots

Tahmini Süre:2m 0s

Soru 23Soru →

For all real numbers $x$ such that $x \neq 0$ and $x \neq -1$ , which of the following is equivalent to the expression $\frac{1}{x} + \frac{1}{x+1}$ ?

\frac{2}{2x+1}

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

\frac{2x+1}{x^2+1}

\frac{2x+1}{x^2+x}

\frac{3}{x+1}

Cevabı ve açıklamayı göster

Cevap: $\frac{2x+1}{x^2+x}$

Cevap

The correct expression is

\frac{2x+1}{x^2+x}

.

The correct expression is obtained by finding a common denominator of

x(x+1)

, rewriting the numerators as

x+1

and

x

, and then adding them to get

2x+1

over the expanded denominator

x^2+x

.

Adım Adım Çözüm

1

Find a common denominator for the two rational terms.

The common denominator for

\frac{1}{x}

and

\frac{1}{x+1}

is

x(x+1)

, which expands to

x^2+x

.

To add fractions with different denominators, they must share a common denominator.

2

Rewrite each fraction with the common denominator by multiplying their numerators and denominators by the appropriate factor.

\frac{1}{x} = \frac{x+1}{x(x+1)}

and

\frac{1}{x+1} = \frac{x}{x(x+1)}

Multiplying the numerator and denominator of each fraction by the missing factor of the common denominator preserves the value of the fraction.

3

Add the numerators together over the common denominator and simplify.

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

Combining the numerators and expanding the denominator gives the simplified equivalent expression.

Anahtar Kavram

Adding rational expressions by finding a common denominator

Soru 24Soru →

What is the positive real solution to the equation $\frac{3}{x-2} - \frac{2}{x} = 1$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The positive real solution to the equation is 4.

By multiplying the entire equation by the least common denominator,

x(x-2)

, the rational equation is cleared of fractions, resulting in

3x - 2(x-2) = x(x-2)

. Simplifying this yields the quadratic equation

x^2 - 3x - 4 = 0

. Factoring the quadratic equation gives

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

, which has roots of

x = 4

and

x = -1

. The positive solution is 4.

Adım Adım Çözüm

1

Multiply both sides of the equation by the least common denominator,

x(x-2)

, to eliminate the fractions.

3x - 2(x-2) = x(x-2)

Eliminating the denominators simplifies the rational equation into a polynomial equation.

2

Distribute and combine like terms to simplify both sides of the equation.

x + 4 = x^2 - 2x

Simplification is necessary before rearranging the equation into a solvable form.

3

Rearrange the terms into standard quadratic form:

ax^2 + bx + c = 0

.

x^2 - 3x - 4 = 0

Setting the quadratic expression equal to zero allows for factoring and solving.

4

Factor the quadratic equation.

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

Finding the factors reveals the potential values for

x

.

5

Solve for the roots and select the positive real solution.

x = 4

The equation yields two solutions,

x = 4

and

x = -1

, and the question explicitly requests the positive solution.

Anahtar Kavram

Solving rational equations by converting them to quadratic equations using the least common denominator.

Soru 25Soru →

For all real numbers $x$ such that $x \neq 1$ and $x \neq -1$ , which of the following is equivalent to the expression $\left(\frac{1}{x-1} + \frac{1}{x+1}\right)^2$ ?

\frac{1}{x^2}

\frac{2x^2}{(x^2-1)^2}

\frac{2x^2+2}{(x^2-1)^2}

\frac{4x^2}{x^4-1}

\frac{4x^2}{(x^2-1)^2}

Cevabı ve açıklamayı göster

Cevap: $\frac{4x^2}{(x^2-1)^2}$

Cevap

The expression is equivalent to the fraction with numerator four times x squared and denominator the square of the difference x squared minus one.

The correct answer is found by first rewriting the terms inside the parentheses with the common denominator

(x^2-1)

, which simplifies the sum to

\frac{2x}{x^2-1}

. Squaring the numerator

(2x)^2

yields

4x^2

, and squaring the denominator yields

(x^2-1)^2

.

Adım Adım Çözüm

1

Find a common denominator for the two fractions inside the parentheses.

The common denominator is

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

. The expression inside the parentheses becomes

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

.

To add rational expressions, they must have a common denominator.

2

Combine the numerators over the common denominator.

The sum is

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

.

Combine the like terms in the numerator to simplify the expression before squaring.

3

Square the simplified rational expression by squaring the numerator and the denominator separately.

\left(\frac{2x}{x^2-1}\right)^2 = \frac{(2x)^2}{(x^2-1)^2} = \frac{4x^2}{(x^2-1)^2}.

Apply the power of a quotient property to obtain the final simplified expression.

Anahtar Kavram

Simplifying rational expressions and applying the power of a quotient rule.

Tahmini Süre:1m 30s

Soru 26Soru →

If $x$ is a real number, what is the sum of all distinct real solutions to the rational equation $\frac{3}{(x-2)^2} - \frac{2}{x-2} = 1$ ?

3

-3

-2

2

4

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The sum of all distinct real solutions is 2.

The correct answer is obtained by multiplying the equation by the least common denominator,

(x-2)^2

, which yields the quadratic equation

x^2 - 2x - 3 = 0

. Factoring this equation gives two solutions:

3

and

-1

. Since neither value makes the original denominator equal to zero, both are valid solutions. Summing these values gives a result of

2

.

Adım Adım Çözüm

1

Identify the domain constraints of the rational equation.

The denominator

x-2

cannot be zero, so the domain constraint is

x \neq 2

.

Division by zero is undefined.

2

Eliminate the denominators by multiplying the entire equation by the least common denominator, which is

(x-2)^2

.

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

This transforms the rational equation into a polynomial equation.

3

Expand both sides of the equation and combine like terms.

3 - 2x + 4 = x^2 - 4x + 4 \implies 7 - 2x = x^2 - 4x + 4

Standard algebraic expansion simplifies the equation into a quadratic form.

4

Rearrange the quadratic equation into standard form

ax^2 + bx + c = 0

.

x^2 - 2x - 3 = 0

Setting the equation to zero allows us to solve for

x

by factoring.

5

Factor the quadratic equation.

(x-3)(x+1) = 0 \implies x = 3

or

x = -1

Factoring finds the values of

x

that satisfy the quadratic equation.

6

Verify if the potential solutions satisfy the domain constraint and calculate their sum.

Both

3

and

-1

are within the domain (

x \neq 2

). The sum is

3 + (-1) = 2

.

Extraneous solutions must be discarded before calculating the final sum.

Anahtar Kavram

Solving rational equations by clearing denominators and verifying domain constraints

Soru 27Soru →

For all real values of $x$ , what is the real solution to the equation $\sqrt{x+13} - 1 = x$ ?

-4

-3

3

4

11

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The only real solution to the equation is 3.

Isolating the radical term gives

\sqrt{x+13} = x+1

. Squaring both sides yields

x+13 = x^2 + 2x + 1

. Rearranging into standard quadratic form gives

x^2 + x - 12 = 0

, which factors as

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

. This yields candidate solutions

x = 3

and

x = -4

. Checking these values in the original equation shows that

x = 3

is a valid solution because

\sqrt{3+13} - 1 = 3

, whereas

x = -4

is extraneous because

\sqrt{-4+13} - 1 = 2 \neq -4

.

Adım Adım Çözüm

1

Isolate the radical expression on one side of the equation.

\sqrt{x+13} = x+1

This prepares the equation for squaring both sides to eliminate the radical.

2

Square both sides of the equation.

x+13 = (x+1)^2 \implies x+13 = x^2 + 2x + 1

Squaring both sides eliminates the radical and yields a polynomial equation.

3

Move all terms to one side to set the quadratic equation equal to zero.

x^2 + x - 12 = 0

This puts the equation in the standard quadratic form

ax^2 + bx + c = 0

so it can be solved by factoring.

4

Factor the quadratic equation.

(x+4)(x-3) = 0 \implies x = 3

or

x = -4

Factoring allows us to find the potential roots of the quadratic equation.

5

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

For

x = 3

:

\sqrt{3+13} - 1 = 4 - 1 = 3

(valid). For

x = -4

:

\sqrt{-4+13} - 1 = 3 - 1 = 2 \neq -4

(extraneous).

Squaring both sides of an equation can introduce extraneous roots that must be checked and discarded.

Anahtar Kavram

Solving radical equations and checking for extraneous solutions

Tahmini Süre:1m 30s

Soru 28Soru →

The rational expression $\frac{x^2 - 4}{2x^2 + 5x - 3}$ is undefined for two real values of $x$ . If the smaller value is $a$ and the larger value is $b$ , what are the values of $a$ and $b$ ?

Aşağıdaki boşlukları doldurun

a =

b =

Cevabı ve açıklamayı göster

Cevap

The smaller value is

-3

and the larger value is

1/2

(or

0.5

).

A rational expression is undefined when the denominator is equal to zero. To find these values, set the denominator

2x^2 + 5x - 3

equal to

0

and solve for

x

. Factoring the quadratic yields

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

. Setting each factor to zero gives

x = 1/2

and

x = -3

. Since

-3

is less than

1/2

, the smaller value is

-3

and the larger value is

1/2

(or

0.5

).

Adım Adım Çözüm

1

Identify the condition that makes a rational expression undefined.

The rational expression is undefined when its denominator is equal to zero:

2x^2 + 5x - 3 = 0

.

Division by zero is undefined in the set of real numbers.

2

Factor the quadratic expression in the denominator.

2x^2 + 5x - 3 = (2x - 1)(x + 3) = 0

.

Factoring allows us to find the roots of the quadratic equation using the zero product property.

3

Solve for the roots of the factored equation.

2x - 1 = 0 \Rightarrow x = 1/2

, and

x + 3 = 0 \Rightarrow x = -3

.

Setting each linear factor to zero determines the values of

x

that make the denominator zero.

4

Assign the values to the variables based on the inequality constraint.

Since

-3 < 1/2

, the smaller value is

a = -3

and the larger value is

b = 1/2

(or

0.5

).

The question specifies that

a

is the smaller value and

b

is the larger value.

Anahtar Kavram

Identifying domain restrictions of rational expressions by finding where the denominator is equal to zero.

Tahmini Süre:1m 30s

Soru 29Soru →

Pipe A can fill a water tank in $x$ hours, where $x > 0$ . Pipe B can fill the same water tank in $x + 3$ hours. If both pipes are opened at the same time, they can fill the tank together in exactly 2 hours. What is the value of $x$ ?

0.5

1

2

3

6

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of

x

is 3.

The correct value is 3. The work rate equation for the two pipes is

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

. Combining the fractions on the left gives

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

. Cross-multiplying yields

4x + 6 = x^2 + 3x

, which simplifies to the quadratic equation

x^2 - x - 6 = 0

. Factoring this quadratic equation gives

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

, which has the solutions

x = 3

and

x = -2

. Since

x

represents a positive number of hours, the only valid solution is 3.

Adım Adım Çözüm

1

Set up the work rate equation based on the individual and combined rates.

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

The sum of the individual rates (in tanks per hour) equals the combined rate of the two pipes.

2

Find a common denominator to combine the fractions on the left side of the equation.

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

To add fractions, they must share a common denominator, which is

x(x+3)

.

3

Cross-multiply to eliminate the denominators and form a quadratic equation.

2(2x+3) = x^2+3x \implies 4x + 6 = x^2 + 3x \implies x^2 - x - 6 = 0

Cross-multiplication converts the rational equation into a polynomial equation.

4

Factor the quadratic equation to find the solutions for

x

.

(x-3)(x+2) = 0 \implies x = 3

or

x = -2

Since

x

represents time, it must be positive (

x > 0

), so the only valid solution is

x = 3

.

Anahtar Kavram

Solving rational equations in the context of work-rate problems

Soru 30Soru →

For all real numbers $x$ such that $x \neq 0$ and $x \neq 3$ , what is the real solution to the equation $\frac{x}{x-3} + \frac{2}{x} = \frac{3}{x-3}$ ?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The correct answer is -2.

Subtracting

\frac{x}{x-3}

from both sides yields

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

. Since

3-x = -(x-3)

, the right side simplifies to

-1

for all

x \neq 3

. The equation becomes

\frac{2}{x} = -1

, which gives

x = -2

. Since

-2

does not violate the domain constraints, it is the correct solution.

Adım Adım Çözüm

1

Subtract

\frac{x}{x-3}

from both sides of the equation.

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

To group terms with common denominators on the same side.

2

Simplify the fraction on the right side of the equation.

\frac{2}{x} = -1

The numerator

3-x

is the negative of the denominator

x-3

, so their quotient is

-1

for all

x \neq 3

.

3

Solve the simplified equation

\frac{2}{x} = -1

for

x

.

x = -2

Multiplying both sides by

x

yields

2 = -x

, and dividing by

-1

gives

x = -2

.

4

Check the solution against the domain restrictions

x \neq 0

and

x \neq 3

.

The solution

x = -2

is valid.

Since

-2

is neither

0

nor

3

, it does not cause any denominator in the original equation to equal zero.

Anahtar Kavram

Solving rational equations by isolating terms with common denominators and checking for extraneous solutions.

Soru 31Soru →

What is the real solution to the equation $\sqrt{3x + 1} = x - 3$ ?

1

1 and 8

8

-2 and 5

-2

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The only real solution is 8.

The correct answer is the single value 8. Squaring both sides of the equation

\sqrt{3x+1} = x-3

leads to the quadratic equation

x^2-9x+8=0

, which factors as

(x-8)(x-1)=0

. This yields potential solutions of 8 and 1. Checking 8 in the original equation gives

\sqrt{25} = 5

, which is correct. Checking 1 in the original equation gives

\sqrt{4} = -2

, which is incorrect because the principal square root must be non-negative. Therefore, 1 is an extraneous solution, and 8 is the only valid solution.

Adım Adım Çözüm

1

Square both sides of the equation to eliminate the radical.

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

To clear the square root and obtain a polynomial equation.

2

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

x^2 - 9x + 8 = 0

Expanding

(x - 3)^2

gives

x^2 - 6x + 9

. Subtracting

3x

and

1

from both sides gives the standard form of the quadratic equation.

3

Factor the quadratic equation.

(x - 8)(x - 1) = 0

Finding two numbers that multiply to 8 and add to -9, which are -8 and -1, allows us to find potential solutions

x = 8

and

x = 1

.

4

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

Checking

x = 8

:

\sqrt{3(8) + 1} = \sqrt{25} = 5

and

8 - 3 = 5

(True). Checking

x = 1

:

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

and

1 - 3 = -2

(False).

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

Anahtar Kavram

Solving radical equations and identifying extraneous solutions

Tahmini Süre:1m 30s

Soru 32Soru →

For what positive integer value of $x$ is the rational expression $\frac{2x - 5}{3x^2 - 10x - 8}$ undefined?

Aşağıdaki boşlukları doldurun

The expression is undefined when

x

is equal to .

Cevabı ve açıklamayı göster

Cevap

The expression is undefined when the positive integer x is equal to 4.

The rational expression is undefined when its denominator is 0. Setting the denominator

3x^2 - 10x - 8

equal to 0 and factoring gives

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

. The roots of this equation are

x = -\frac{2}{3}

and

x = 4

. Since the question restricts the target value to a positive integer, the correct value is 4.

Adım Adım Çözüm

1

Set the denominator of the rational expression equal to 0.

3x^2 - 10x - 8 = 0

A rational expression is undefined when its denominator is equal to 0 because division by zero is undefined.

2

Factor the quadratic equation

3x^2 - 10x - 8 = 0

.

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

Factoring the quadratic trinomial helps identify the values of x that satisfy the equation.

3

Solve for the roots by setting each factor to 0.

x = -\frac{2}{3}

or

x = 4

According to the zero product property, if the product of two factors is zero, then at least one of the factors must be zero.

4

Filter the solutions based on the given constraint.

x = 4

The question specifies that x must be a positive integer. Since

-\frac{2}{3}

is a negative fraction, we reject it and select 4.

Anahtar Kavram

Finding values that make a rational expression undefined by setting the denominator to zero and solving the resulting quadratic equation.

Tahmini Süre:1m 30s

Soru 33Soru →

If $\sqrt{3x + 19} - x = 3$ , what is the value of $x$ ?

-10

-5

-2

2

5

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

x

is 2.

To solve the equation, isolate the radical to obtain

\sqrt{3x + 19} = x + 3

. Squaring both sides yields

3x + 19 = x^2 + 6x + 9

. Rearranging terms results in the quadratic equation

x^2 + 3x - 10 = 0

, which factors into

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

. The potential solutions are

x = 2

and

x = -5

. Testing these in the original equation shows that

2

is a valid solution because

\sqrt{3(2) + 19} - 2 = 3

. Testing

-5

results in

\sqrt{3(-5) + 19} - (-5) = 7

, which does not equal

3

. Thus, the only real solution is

2

.

Adım Adım Çözüm

1

Isolate the radical expression on one side of the equation.

\sqrt{3x + 19} = x + 3

Isolating the radical allows us to eliminate it by squaring both sides.

2

Square both sides of the equation.

3x + 19 = x^2 + 6x + 9

Squaring a square root removes the radical. Remember to expand

(x+3)^2

correctly as

x^2 + 6x + 9

.

3

Rearrange the terms into standard quadratic form,

ax^2 + bx + c = 0

.

x^2 + 3x - 10 = 0

Moving all terms to one side sets up the quadratic equation for factoring.

4

Factor the quadratic expression.

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

Factoring allows us to find the potential values of

x

by setting each factor to zero.

5

Solve for the potential values of

x

.

x = -5

or

x = 2

Setting

x + 5 = 0

gives

x = -5

, and setting

x - 2 = 0

gives

x = 2

.

6

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

For

x = 2

:

\sqrt{3(2) + 19} - 2 = \sqrt{25} - 2 = 5 - 2 = 3

(valid). For

x = -5

:

\sqrt{3(-5) + 19} - (-5) = \sqrt{4} + 5 = 2 + 5 = 7 \neq 3

(extraneous).

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

Anahtar Kavram

Solving radical equations and checking for extraneous solutions.

Soru 34Soru →

What is the value of the larger real solution to the equation $\frac{10}{x} - \frac{3}{x-1} = 2$ ?

Cevabı ve açıklamayı göster

Cevap: 2.5

Cevap

The larger real solution is

2.5

.

Multiplying the equation by the least common denominator

x(x-1)

yields

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

. Simplifying this leads to

2x^2 - 9x + 10 = 0

. Factoring the quadratic yields

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

, which gives the solutions

x = 2.5

and

x = 2

. The larger of these two solutions is

2.5

.

Adım Adım Çözüm

1

Multiply the entire equation by the least common denominator, which is

x(x-1)

, for

x \neq 0

and

x \neq 1

.

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

This clears the fractions from the rational equation.

2

Expand both sides of the equation.

10x - 10 - 3x = 2x^2 - 2x

Expanding the terms allows us to combine like terms.

3

Combine like terms on the left side and move all terms to one side to set the quadratic equation equal to zero.

2x^2 - 9x + 10 = 0

Setting the quadratic equation to zero is required to solve it by factoring.

4

Factor the quadratic equation by grouping.

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

Factoring allows us to find the roots of the quadratic equation.

5

Solve for

x

by setting each factor equal to zero.

x = 2.5

or

x = 2

By the zero product property, at least one of the factors must be zero.

6

Check for extraneous solutions and select the larger real value.

Both

2

and

2.5

are valid because they do not make the original denominators zero. The larger value is

2.5

.

The question specifically asks for the larger of the two real solutions.

Anahtar Kavram

Solving rational equations by clearing denominators

Tahmini Süre:1m 30s

Soru 35Soru →

Consider the system of equations below:

y = \sqrt{x + 37}

y = x - 5

If

(x, y)

is a real solution to this system, what is the value of

xy

?

-12

-4

6

24

84

Cevabı ve açıklamayı göster

Cevap: 84

Cevap

84

The correct answer is 84 because solving the system yields the quadratic equation

x^2 - 11x - 12 = 0

, which has roots

x = 12

and

x = -1

. Substituting these back into the original equations shows that only

x = 12

is valid (giving

y = 7

), while

x = -1

is extraneous (since

\sqrt{36} = 6 \neq -6

). The product of the coordinates of the valid solution is

12 \times 7 = 84

.

Adım Adım Çözüm

1

Set the two expressions for

y

equal to each other to solve for

x

.

\sqrt{x + 37} = x - 5

Since both equations are solved for

y

, their right-hand sides must be equal at the point of intersection.

2

Square both sides of the equation to eliminate the radical.

x + 37 = (x - 5)^2 \Rightarrow x + 37 = x^2 - 10x + 25

Squaring is the inverse operation of the square root. The binomial on the right must be expanded fully using

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

.

3

Rearrange the quadratic equation into standard form.

x^2 - 11x - 12 = 0

Subtracting

x

and

37

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

4

Factor the quadratic equation to find the algebraic solutions.

(x - 12)(x + 1) = 0 \Rightarrow x = 12 \text{ or } x = -1

Factoring the trinomial allows us to find the roots by setting each linear factor to zero.

5

Substitute each solution back into the original system to check for extraneous solutions.

For

x = 12

:

y = 12 - 5 = 7

and

y = \sqrt{12 + 37} = 7

(Valid). For

x = -1

:

y = -1 - 5 = -6

, but

y = \sqrt{-1 + 37} = 6 \neq -6

(Extraneous).

Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original radical relationship because the principal square root must be non-negative.

6

Calculate the product

xy

of the coordinates of the valid solution.

xy = 12 \times 7 = 84

The question asks for the product of

x

and

y

for the real solution

(12, 7)

.

Anahtar Kavram

Solving systems containing radical equations and verifying for extraneous solutions.

Rational and Radical Expressions and Equations

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