Intermediate Algebra

272 soru

Soru 21Soru

For the imaginary unit ii, if the complex number zz is defined by z=(1+2i)22iz = \frac{(1 + 2i)^2}{2 - i}, what is the real part of zz?

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Cevap: 2-2

Cevap

The real part of the complex number zz is 2-2.
To find the real part of the complex number, we first simplify the expression by expanding the squared binomial in the numerator, which yields 3+4i-3 + 4i. Next, we rationalize the fraction by multiplying both the numerator and the denominator by the complex conjugate of the denominator, 2+i2 + i. This multiplication yields 10+5i5\frac{-10 + 5i}{5}. Dividing both the real and imaginary terms by 55 results in the standard form 2+i-2 + i. Thus, the real part of this complex number is 2-2.

Adım Adım Çözüm

1
Expand the squared binomial in the numerator of the expression for zz.
(1+2i)2=12+2(1)(2i)+(2i)2=1+4i+4i2(1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2 = 1 + 4i + 4i^2
Apply the algebraic identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.
2
Simplify the expanded numerator using the definition of the imaginary unit.
1+4i+4(1)=3+4i1 + 4i + 4(-1) = -3 + 4i
Since i2=1i^2 = -1, the term 4i24i^2 simplifies to 4-4. Combining this with 11 gives the real part 3-3.
3
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator to rationalize it.
z=3+4i2i2+i2+i=(3+4i)(2+i)(2i)(2+i)z = \frac{-3 + 4i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{(-3 + 4i)(2 + i)}{(2 - i)(2 + i)}
Multiplying the denominator by its complex conjugate, 2+i2 + i, eliminates the imaginary unit from the denominator.
4
Expand and simplify the numerator and denominator.
z=63i+8i+4i24i2=6+5i44(1)=10+5i5z = \frac{-6 - 3i + 8i + 4i^2}{4 - i^2} = \frac{-6 + 5i - 4}{4 - (-1)} = \frac{-10 + 5i}{5}
Using the distributive property in the numerator gives 6+5i+4i2-6 + 5i + 4i^2. Since i2=1i^2 = -1, this simplifies to 10+5i-10 + 5i. In the denominator, (2i)(2+i)=4i2=5(2-i)(2+i) = 4 - i^2 = 5.
5
Divide both terms of the simplified numerator by the denominator to express zz in standard form a+bia + bi.
z=2+iz = -2 + i
Dividing the real part 10-10 by 55 yields the real part 2-2, and dividing the imaginary part 5i5i by 55 yields the imaginary part ii.

Anahtar Kavram

Division of complex numbers using the complex conjugate
Tahmini Süre:2m 0s
Soru 22Soru

For the imaginary unit ii, where i2=1i^2 = -1, the complex number zz is defined by z=11+3i3iz = \frac{11 + 3i}{3 - i}. What is the real part of zz?

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Cevap: 3

Cevap

The real part of the complex number is 3.
By multiplying both the numerator and the denominator of 11+3i3i\frac{11 + 3i}{3 - i} by the conjugate of the denominator, 3+i3 + i, we obtain (11+3i)(3+i)(3i)(3+i)=33+11i+9i+3i29i2=30+20i10=3+2i\frac{(11+3i)(3+i)}{(3-i)(3+i)} = \frac{33 + 11i + 9i + 3i^2}{9 - i^2} = \frac{30 + 20i}{10} = 3 + 2i. The real part of this complex number is the term without ii, which is 3.

Adım Adım Çözüm

1
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator, which is 3+i3 + i.
z=(11+3i)(3+i)(3i)(3+i)z = \frac{(11 + 3i)(3 + i)}{(3 - i)(3 + i)}
To eliminate the imaginary unit from the denominator.
2
Expand and simplify the numerator using the distributive property and substituting 1-1 for i2i^2.
(11+3i)(3+i)=33+11i+9i+3i2=33+20i+3(1)=30+20i(11 + 3i)(3 + i) = 33 + 11i + 9i + 3i^2 = 33 + 20i + 3(-1) = 30 + 20i
To combine the real and imaginary terms of the numerator.
3
Expand and simplify the denominator using the difference of squares property and substituting 1-1 for i2i^2.
(3i)(3+i)=9i2=9(1)=10(3 - i)(3 + i) = 9 - i^2 = 9 - (-1) = 10
To find the real number denominator.
4
Divide each term in the simplified numerator by the denominator.
z=30+20i10=3+2iz = \frac{30 + 20i}{10} = 3 + 2i
To express the complex number in the standard form a+bia + bi.
5
Extract the real part of the resulting complex number 3+2i3 + 2i.
3
The real part of a complex number in the form a+bia + bi is aa.

Anahtar Kavram

Division of complex numbers using the complex conjugate

Alternatif Yöntem

Instead of simplifying the fraction directly, assume the resulting complex number is x+yix + yi, where xx represents the real part and yy represents the imaginary part. We can set up the equation x+yi=11+3i3ix + yi = \frac{11 + 3i}{3 - i} and multiply both sides by 3i3 - i to get (x+yi)(3i)=11+3i(x + yi)(3 - i) = 11 + 3i. Expanding the left side gives (3x+y)+(3yx)i=11+3i(3x + y) + (3y - x)i = 11 + 3i. Equating the real and imaginary parts yields a system of linear equations: 3x+y=113x + y = 11 and x+3y=3-x + 3y = 3. Multiplying the second equation by 3 and adding it to the first equation gives 10y=2010y = 20, which means y=2y = 2. Substituting y=2y = 2 back into the first equation yields 3x+2=113x + 2 = 11, which simplifies to 3x=93x = 9, or x=3x = 3. The real part is therefore 3.
Tahmini Süre:1m 30s
Soru 23Soru

For all real numbers xx such that x2x \neq -2 and x3x \neq 3, the expression 2xx35x+2\frac{2x}{x - 3} - \frac{5}{x + 2} is equivalent to which of the following?

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Cevap: \frac{2x^2 - x + 15}{x^2 - x - 6}

Cevap

The expression is equivalent to 2x2x+15x2x6\frac{2x^2 - x + 15}{x^2 - x - 6}.
To subtract the rational expressions, we find the common denominator, which is (x3)(x+2)=x2x6(x - 3)(x + 2) = x^2 - x - 6. We rewrite each fraction with this denominator, which gives 2x(x+2)x2x6=2x2+4xx2x6\frac{2x(x + 2)}{x^2 - x - 6} = \frac{2x^2 + 4x}{x^2 - x - 6} and 5(x3)x2x6=5x15x2x6\frac{5(x - 3)}{x^2 - x - 6} = \frac{5x - 15}{x^2 - x - 6}. Subtracting the second numerator from the first gives (2x2+4x)(5x15)=2x2x+15(2x^2 + 4x) - (5x - 15) = 2x^2 - x + 15. Placing this over the common denominator gives the simplified expression 2x2x+15x2x6\frac{2x^2 - x + 15}{x^2 - x - 6}.

Adım Adım Çözüm

1
Identify the common denominator.
The common denominator is (x3)(x+2)=x2x6(x - 3)(x + 2) = x^2 - x - 6.
To subtract rational expressions, we need a common denominator.
2
Rewrite each rational expression with the common denominator.
The first term becomes 2x(x+2)(x3)(x+2)=2x2+4xx2x6\frac{2x(x + 2)}{(x - 3)(x + 2)} = \frac{2x^2 + 4x}{x^2 - x - 6}, and the second term becomes 5(x3)(x3)(x+2)=5x15x2x6\frac{5(x - 3)}{(x - 3)(x + 2)} = \frac{5x - 15}{x^2 - x - 6}.
Multiplying the numerator and denominator of each term by the missing factor keeps the values of the expressions unchanged.
3
Subtract the numerators.
(2x2+4x)(5x15)=2x2+4x5x+15=2x2x+15(2x^2 + 4x) - (5x - 15) = 2x^2 + 4x - 5x + 15 = 2x^2 - x + 15.
Subtracting the second numerator requires distributing the negative sign to both terms of the expression (5x15)(5x - 15).
4
Combine the result over the common denominator.
\frac{2x^2 - x + 15}{x^2 - x - 6}
Write the simplified numerator over the common denominator.

Anahtar Kavram

Subtraction of rational expressions involves finding a common denominator, expanding the numerators, and distributing negative signs carefully.
Soru 24Soru

If 3x2=4\sqrt{3x - 2} = 4, what is the value of xx?

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Cevap: 6

Cevap

The value of xx is 6.
Squaring both sides of the equation 3x2=4\sqrt{3x - 2} = 4 eliminates the radical, leading to 3x2=163x - 2 = 16. Adding 2 to both sides results in 3x=183x = 18. Dividing both sides by 3 gives x=6x = 6. Substituting 6 back into the original equation yields 3(6)2=182=16=4\sqrt{3(6) - 2} = \sqrt{18 - 2} = \sqrt{16} = 4, which verifies that the solution is correct.

Adım Adım Çözüm

1
Square both sides of the equation to remove the radical.
3x2=163x - 2 = 16
Squaring a square root removes the radical sign since (a)2=a(\sqrt{a})^2 = a for non-negative values.
2
Add 2 to both sides of the equation.
3x=183x = 18
Adding the constant term moves it to the other side to isolate the term containing the variable.
3
Divide both sides by 3.
x=6x = 6
Dividing by the coefficient of xx yields the final solution.

Anahtar Kavram

Solving basic radical equations by isolating the radical and squaring both sides.
Soru 25Soru

For the imaginary unit ii, where i2=1i^2 = -1, which of the following is equivalent to the expression 52i\frac{5}{2 - i}?

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Cevap: 2+i2 + i

Cevap

The simplified expression is 2+i2 + i.
To simplify the expression, multiply both the numerator and denominator by the complex conjugate of the denominator, which is 2+i2 + i. This results in 5(2+i)(2i)(2+i)=10+5i4i2\frac{5(2+i)}{(2-i)(2+i)} = \frac{10+5i}{4-i^2}. Since i2=1i^2 = -1, the denominator becomes 4(1)=54 - (-1) = 5. Dividing both terms in the numerator by 55 gives 2+i2 + i.

Adım Adım Çözüm

1
Multiply the numerator and the denominator by the complex conjugate of the denominator, 2+i2 + i.
5(2+i)(2i)(2+i)\frac{5(2 + i)}{(2 - i)(2 + i)}
Multiplying by the conjugate rationalizes the denominator, converting it into a real number.
2
Expand the numerator and the denominator, substituting 1-1 for i2i^2.
10+5i4(1)=10+5i5\frac{10 + 5i}{4 - (-1)} = \frac{10 + 5i}{5}
Using the distributive property for the numerator and the difference of squares identity for the denominator, along with the definition i2=1i^2 = -1.
3
Divide each term in the numerator by the denominator.
2+i2 + i
Distributing the division by 55 to both the real and imaginary parts of the numerator simplifies the expression to standard form.

Anahtar Kavram

Rationalizing the denominator of a complex fraction by multiplying by the complex conjugate of the denominator.
Tahmini Süre:45s
Soru 26Soru

For the imaginary unit ii, where i2=1i^2 = -1, let zz be the complex number defined by z=10+ki2iz = \frac{10 + ki}{2 - i}, where kk is a real constant. If the imaginary part of zz is 44, what is the value of kk?

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Cevap: 5

Cevap

The value of kk is 55.
Multiplying the numerator and denominator of z=10+ki2iz = \frac{10 + ki}{2 - i} by the complex conjugate 2+i2 + i gives z=(20k)+(10+2k)i5z = \frac{(20 - k) + (10 + 2k)i}{5}. The imaginary part is 10+2k5\frac{10 + 2k}{5}. Setting this expression equal to 44 and solving for kk yields k=5k = 5.

Adım Adım Çözüm

1
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator, which is 2+i2 + i.
z=(10+ki)(2+i)(2i)(2+i)z = \frac{(10 + ki)(2 + i)}{(2 - i)(2 + i)}
To eliminate the imaginary unit from the denominator and express the complex number in standard form.
2
Expand both the numerator and the denominator, using the property i2=1i^2 = -1.
z=20+10i+2ki+ki24i2=(20k)+(10+2k)i5z = \frac{20 + 10i + 2ki + ki^2}{4 - i^2} = \frac{(20 - k) + (10 + 2k)i}{5}
To separate the real terms and imaginary terms in the numerator and simplify the denominator to a real number.
3
Express the complex number in standard form a+bia + bi to identify the imaginary part.
z=20k5+(10+2k5)iz = \frac{20 - k}{5} + \left(\frac{10 + 2k}{5}\right)i
The imaginary part of a complex number is the coefficient of ii, which is 10+2k5\frac{10 + 2k}{5}.
4
Set the imaginary part equal to 44 and solve the linear equation for kk.
10+2k5=4    10+2k=20    2k=10    k=5\frac{10 + 2k}{5} = 4 \implies 10 + 2k = 20 \implies 2k = 10 \implies k = 5
To find the specific value of the constant kk that makes the imaginary part of zz equal to 44.

Anahtar Kavram

Rationalizing complex numbers and identifying real and imaginary components
Soru 27Soru
For all real values of xx where the expression is defined, consider the equation:
2xx+3=x+31\frac{2x}{\sqrt{x + 3}} = \sqrt{x + 3} - 1
Which of the following represents the complete set of real solutions to this equation?
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Cevap: {1}\{1\}

Cevap

The set containing only 1
The correct answer is the set containing only 1. To solve the equation, we first multiply both sides by the denominator x+3\sqrt{x+3}, which yields 2x=x+3x+32x = x + 3 - \sqrt{x+3}. Isolating the radical gives x+3=3x\sqrt{x+3} = 3-x. Squaring both sides produces the quadratic equation x+3=x26x+9x+3 = x^2-6x+9, which simplifies to x27x+6=0x^2-7x+6=0. Solving this quadratic gives candidate solutions of 1 and 6. Substituting 6 back into the original equation results in an invalid statement (4=24 = 2), making it extraneous. Substituting 1 yields a valid statement (1=11 = 1), meaning the only real solution is 1.

Adım Adım Çözüm

1
Determine the domain of the equation.
x>3x > -3
The expression inside the square root must be non-negative (x+30x+3 \ge 0), and since it is in the denominator, it cannot be zero (x+30x+3 \neq 0).
2
Clear the denominator by multiplying both sides by x+3\sqrt{x + 3}.
2x=x+3x+32x = x + 3 - \sqrt{x + 3}
Multiplying both sides by the denominator simplifies the rational expression into a form where we can isolate the radical.
3
Isolate the radical term.
x+3=3x\sqrt{x + 3} = 3 - x
Grouping all non-radical terms on one side prepares the equation for squaring to eliminate the radical.
4
Analyze constraints on the variable.
x3x \le 3
Since the principal square root on the left side is non-negative, the right side 3x3 - x must also be non-negative, which restricts any valid solutions to x3x \le 3.
5
Square both sides and simplify to form a quadratic equation.
x27x+6=0x^2 - 7x + 6 = 0
Squaring both sides eliminates the radical: x+3=(3x)2    x+3=x26x+9    x27x+6=0x + 3 = (3 - x)^2 \implies x + 3 = x^2 - 6x + 9 \implies x^2 - 7x + 6 = 0.
6
Solve the quadratic equation by factoring.
x=1x = 1 or x=6x = 6
Factoring (x1)(x6)=0(x-1)(x-6) = 0 gives the potential solutions.
7
Verify the solutions in the original equation.
x=1x = 1 is valid; x=6x = 6 is extraneous.
Substituting x=6x = 6 yields 123=4\frac{12}{3} = 4 on the left and 91=2\sqrt{9}-1 = 2 on the right, which are not equal. Substituting x=1x = 1 yields 22=1\frac{2}{2} = 1 on the left and 41=1\sqrt{4}-1 = 1 on the right, which are equal.

Anahtar Kavram

Solving equations containing both rational and radical expressions requires clearing denominators, isolating the radical, squaring both sides, and verifying candidate solutions to eliminate extraneous solutions.
Soru 28Soru

Which of the following is the complete set of real solutions to the equation 5x9=x3\sqrt{5x - 9} = x - 3?

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Cevap: {9}\{9\}

Cevap

{9}\{9\}
The correct option is the set containing only the value 9. To solve the equation, we square both sides to get 5x9=x26x+95x - 9 = x^2 - 6x + 9, which simplifies to the quadratic equation x211x+18=0x^2 - 11x + 18 = 0. Factoring this yields the potential solutions 9 and 2. Substituting these back into the original equation, we find that 9 satisfies the equation while 2 results in a contradiction, making it extraneous.

Adım Adım Çözüm

1
Isolate the radical and square both sides of the equation.
5x9=(x3)25x - 9 = (x - 3)^2
Squaring both sides eliminates the square root to allow solving for the variable.
2
Expand the right-hand side using the binomial squaring rule.
5x9=x26x+95x - 9 = x^2 - 6x + 9
The square of a binomial (ab)2(a - b)^2 is a22ab+b2a^2 - 2ab + b^2.
3
Move all terms to one side to set the quadratic equation to zero, then factor.
x211x+18=0(x9)(x2)=0x^2 - 11x + 18 = 0 \Rightarrow (x - 9)(x - 2) = 0
Rearranging terms simplifies the equation into a standard quadratic form that can be factored.
4
Solve for the potential roots and check for extraneous solutions in the original equation.
x=9x = 9 (valid) and x=2x = 2 (extraneous)
Checking x=9x = 9 gives 5(9)9=936=6\sqrt{5(9) - 9} = 9 - 3 \Rightarrow 6 = 6 (true). Checking x=2x = 2 gives 5(2)9=231=1\sqrt{5(2) - 9} = 2 - 3 \Rightarrow 1 = -1 (false).

Anahtar Kavram

Solving radical equations and verifying for extraneous solutions
Soru 29Soru

For the imaginary unit ii, where i2=1i^2 = -1, the complex number ww is defined as w=6+4i2iw = \frac{6 + 4i}{2i}. What is the imaginary part of ww?

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Cevap: -3

Cevap

The imaginary part of ww is 3-3.
Dividing each term in the numerator of 6+4i2i\frac{6 + 4i}{2i} by the denominator 2i2i yields 62i+4i2i\frac{6}{2i} + \frac{4i}{2i}, which simplifies to 3i+2\frac{3}{i} + 2. Since i2=1i^2 = -1, the term 3i\frac{3}{i} can be rationalized to 3i-3i. Thus, the complex number in standard form is 23i2 - 3i. The imaginary part is the real coefficient of ii, which is 3-3.

Adım Adım Çözüm

1
Divide each term in the numerator by the denominator.
w=62i+4i2iw = \frac{6}{2i} + \frac{4i}{2i}
This separates the quotient into two simpler terms that can be simplified individually.
2
Simplify both terms.
w=3i+2w = \frac{3}{i} + 2
Reduce the fractions by dividing out common factors in both the numerators and the denominators.
3
Rationalize the denominator of the imaginary term.
3iii=3ii2=3i1=3i\frac{3}{i} \cdot \frac{i}{i} = \frac{3i}{i^2} = \frac{3i}{-1} = -3i
Multiply the numerator and denominator by ii to eliminate the imaginary unit from the denominator, using the property i2=1i^2 = -1.
4
Combine the real and imaginary parts into standard form a+bia + bi.
w=23iw = 2 - 3i
Group the real constant and the simplified imaginary term together.
5
Identify the imaginary part of the complex number.
3-3
The imaginary part of a complex number a+bia + bi is the real coefficient bb of the imaginary unit ii.

Anahtar Kavram

Simplifying a quotient of complex numbers by dividing by a pure imaginary number.
Soru 30Soru

If xx is a real number that satisfies the equation 2x+7+x+3=1\sqrt{2x + 7} + \sqrt{x + 3} = 1, what is the value of xx?

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Cevap: -3

Cevap

The only real solution to the equation is 3-3.
The value 3-3 is the only real number that satisfies the original equation. Substituting 3-3 back into the original equation yields 2(3)+7+3+3=1+0=1\sqrt{2(-3) + 7} + \sqrt{-3 + 3} = \sqrt{1} + 0 = 1, which is true.

Adım Adım Çözüm

1
Isolate the first radical term.
2x+7=1x+3\sqrt{2x + 7} = 1 - \sqrt{x + 3}
This allows for squaring both sides to eliminate one radical.
2
Square both sides and simplify.
2x+7=x+42x+32x + 7 = x + 4 - 2\sqrt{x + 3}
Squaring removes the radical on the left side, though it creates a middle term on the right side.
3
Isolate the remaining radical term.
x+3=2x+3x + 3 = -2\sqrt{x + 3}
Grouping the non-radical terms on one side prepares the equation for a second squaring step.
4
Square both sides again to eliminate the remaining radical.
x2+6x+9=4(x+3)x^2 + 6x + 9 = 4(x + 3)
Squaring both sides eliminates the radical completely, converting the expression into a polynomial equation.
5
Solve the quadratic equation.
x=3x = -3 and x=1x = 1
Rearranging to x2+2x3=0x^2 + 2x - 3 = 0 and factoring as (x+3)(x1)=0(x + 3)(x - 1) = 0 gives the candidate solutions.
6
Substitute candidates back into the original equation to check for extraneous solutions.
The only valid solution is x=3x = -3.
Substituting x=1x = 1 yields 5=15 = 1 (invalid), while substituting x=3x = -3 yields 1=11 = 1 (valid).

Anahtar Kavram

Solving radical equations by isolating radicals and squaring, then testing for extraneous solutions.
Soru 31Soru
What is the sum of all real values of xx that satisfy the equation x23xx23x+2+x23x1x23x3=113\frac{x^2 - 3x}{x^2 - 3x + 2} + \frac{x^2 - 3x - 1}{x^2 - 3x - 3} = \frac{11}{3}?
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Cevap: 3

Cevap

The sum of all real values of xx that satisfy the equation is 33.
Substituting y=x23xy = x^2 - 3x simplifies the original rational equation into the quadratic form y2y12=0y^2 - y - 12 = 0. Solving for yy yields the values 44 and 3-3. Substituting back x23xx^2 - 3x for yy produces two quadratic equations. The first, x23x4=0x^2 - 3x - 4 = 0, has real solutions of 44 and 1-1. The second, x23x+3=0x^2 - 3x + 3 = 0, has a negative discriminant and produces no real solutions. Summing the valid real solutions gives 4+(1)=34 + (-1) = 3.

Adım Adım Çözüm

1
Substitute y=x23xy = x^2 - 3x into the equation to simplify the rational terms.
yy+2+y1y3=113\frac{y}{y+2} + \frac{y-1}{y-3} = \frac{11}{3}
Using a temporary variable simplifies the algebraic manipulation of the rational expressions.
2
Multiply the entire equation by the least common denominator, 3(y+2)(y3)3(y+2)(y-3), to eliminate all fractions.
3y(y3)+3(y1)(y+2)=11(y+2)(y3)3y(y-3) + 3(y-1)(y+2) = 11(y+2)(y-3), where y2y \neq -2 and y3y \neq 3
This clears the denominators so the equation can be solved as a polynomial.
3
Expand the terms on both sides of the equation.
(3y29y)+(3y2+3y6)=11y211y66(3y^2 - 9y) + (3y^2 + 3y - 6) = 11y^2 - 11y - 66
Expanding allows for combining like terms.
4
Combine like terms and move all terms to one side of the equation to set it equal to zero.
5y25y60=05y^2 - 5y - 60 = 0
This sets up the expression in the standard quadratic form ay2+by+c=0ay^2 + by + c = 0.
5
Divide the entire quadratic equation by its greatest common factor, 55, and factor the resulting expression.
y2y12=0    (y4)(y+3)=0    y=4 or y=3y^2 - y - 12 = 0 \implies (y-4)(y+3) = 0 \implies y = 4 \text{ or } y = -3
Factoring solves for the possible values of the substituted variable yy.
6
Substitute back y=x23xy = x^2 - 3x for each case and solve the resulting quadratic equations for xx.
For y=4y = 4: x23x4=0    (x4)(x+1)=0    x=4 or x=1x^2 - 3x - 4 = 0 \implies (x-4)(x+1) = 0 \implies x = 4 \text{ or } x = -1.
For y=3y = -3: x23x+3=0x^2 - 3x + 3 = 0. The discriminant is (3)24(1)(3)=3<0(-3)^2 - 4(1)(3) = -3 < 0, which means there are no real solutions.
This determines the real values of xx that solve the original equation.
7
Verify that neither solution makes the original denominators zero, and add the valid real solutions.
4+(1)=34 + (-1) = 3
Since both x=4x = 4 and x=1x = -1 result in non-zero denominators, both are valid real solutions. Their sum is 33.

Anahtar Kavram

Solving rational equations by utilizing algebraic substitution to reduce complexity and analyzing quadratic equations for real solutions.
Soru 32Soru

For the imaginary unit ii, where i2=1i^2 = -1, the complex number zz is defined by:

z=13i451+i95z = \frac{1 - 3i^{45}}{1 + i^{95}}

Which of the following is equivalent to z2z^2?

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Cevap: 3 - 4i

Cevap

3 - 4i
The correct answer is found by first simplifying the powers of ii in the expression for zz: i45=ii^{45} = i and i95=ii^{95} = -i, which yields z=13i1iz = \frac{1 - 3i}{1 - i}. Rationalizing the fraction by multiplying both the numerator and the denominator by the conjugate of the denominator (1+i1 + i) simplifies the expression to z=2iz = 2 - i. Finally, squaring this result using the binomial expansion formula gives (2i)2=44i+i2=34i(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i.

Adım Adım Çözüm

1
Simplify the high integer powers of the imaginary unit ii by using the fact that powers of ii repeat in a cycle of four: i1=ii^1 = i, i2=1i^2 = -1, i3=ii^3 = -i, and i4=1i^4 = 1.
i45=ii^{45} = i and i95=ii^{95} = -i
Since 45=4(11)+145 = 4(11) + 1, the remainder is 11, so i45=i1=ii^{45} = i^1 = i. Since 95=4(23)+395 = 4(23) + 3, the remainder is 33, so i95=i3=ii^{95} = i^3 = -i.
2
Substitute these simplified values back into the expression for zz.
z=13i1iz = \frac{1 - 3i}{1 - i}
This sets up the fraction with simplified imaginary terms in both the numerator and the denominator.
3
Rationalize the denominator by multiplying the numerator and denominator of the fraction by the complex conjugate of the denominator, which is 1+i1 + i.
z=2iz = 2 - i
Multiplying by the conjugate eliminates the imaginary unit from the denominator: (13i)(1+i)(1i)(1+i)=1+i3i3i21i2=42i2=2i\frac{(1 - 3i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + i - 3i - 3i^2}{1 - i^2} = \frac{4 - 2i}{2} = 2 - i.
4
Calculate the value of z2z^2 by squaring the simplified complex number 2i2 - i.
z2=34iz^2 = 3 - 4i
Using the binomial squaring formula (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2, we expand (2i)2(2 - i)^2 to get 222(2)(i)+i2=44i1=34i2^2 - 2(2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i.

Anahtar Kavram

Simplifying complex numbers by evaluating powers of ii, rationalizing fractions with complex conjugates, and expanding complex binomials.
Tahmini Süre:1m 30s
Soru 33Soru

Let ii be the imaginary unit such that i2=1i^2 = -1. What is the simplified form of the expression (1+2i)2(3i)(1 + 2i)^2(3 - i)?

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Cevap: 5+15i-5 + 15i

Cevap

5+15i-5 + 15i
The correct answer is 5+15i-5 + 15i. We first expand the squared binomial (1+2i)2(1 + 2i)^2, which results in 1+4i+4i21 + 4i + 4i^2. Substituting i2=1i^2 = -1 yields 3+4i-3 + 4i. We then multiply this by the second binomial (3i)(3 - i) to get 9+3i+12i4i2-9 + 3i + 12i - 4i^2. Substituting i2=1i^2 = -1 one more time and combining like terms leads to the final simplified result of 5+15i-5 + 15i.

Adım Adım Çözüm

1
Expand the squared binomial (1+2i)2(1 + 2i)^2
1+4i+4i21 + 4i + 4i^2
Before multiplying by the second binomial, we must apply the exponent to the first binomial according to the order of operations.
2
Substitute i2=1i^2 = -1 to simplify the expression from Step 1
3+4i-3 + 4i
Since i2=1i^2 = -1, the term 4i24i^2 becomes 4(1)=44(-1) = -4, and combining the real parts gives 14=31 - 4 = -3.
3
Multiply the simplified term by (3i)(3 - i) using the FOIL method
9+3i+12i4i2-9 + 3i + 12i - 4i^2
We distribute each term of the first binomial into the second binomial: (3)(3)=9(-3)(3) = -9, (3)(i)=3i(-3)(-i) = 3i, (4i)(3)=12i(4i)(3) = 12i, and (4i)(i)=4i2(4i)(-i) = -4i^2.
4
Simplify the resulting expression by combining like terms and substituting i2=1i^2 = -1
5+15i-5 + 15i
Combining the imaginary parts gives 3i+12i=15i3i + 12i = 15i. Substituting i2=1i^2 = -1 into 4i2-4i^2 gives 4(1)=+4-4(-1) = +4. Finally, combining the real parts yields 9+4=5-9 + 4 = -5.

Anahtar Kavram

Simplification of complex expressions involving binomial squaring and multiplication under the definition i2=1i^2 = -1.
Tahmini Süre:1m 30s
Soru 34Soru

If kk is a positive real number such that k+4k=15\sqrt{k} + \sqrt{4k} = 15, what is the value of kk?

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Cevap: 25

Cevap

25
The correct answer is 2525. To find this, we simplify 4k\sqrt{4k} to 2k2\sqrt{k}. Substituting this into the equation gives k+2k=15\sqrt{k} + 2\sqrt{k} = 15. Combining like terms yields 3k=153\sqrt{k} = 15. Dividing both sides by 33 gives k=5\sqrt{k} = 5. Squaring both sides of the equation results in k=25k = 25. Checking the solution: 25+4(25)=5+10=15\sqrt{25} + \sqrt{4(25)} = 5 + 10 = 15, which is true.

Adım Adım Çözüm

1
Simplify the radical term 4k\sqrt{4k} using the product property of radicals.
4k=4k=2k\sqrt{4k} = \sqrt{4} \cdot \sqrt{k} = 2\sqrt{k}
This allows us to express both terms in the equation using the same radical base, k\sqrt{k}.
2
Substitute the simplified term back into the equation and combine like terms.
k+2k=15    3k=15\sqrt{k} + 2\sqrt{k} = 15 \implies 3\sqrt{k} = 15
Combining like radical terms simplifies the equation to a single radical term.
3
Isolate the radical by dividing both sides of the equation by 33.
k=5\sqrt{k} = 5
Isolating the radical term is necessary before squaring both sides to solve for the variable.
4
Square both sides of the equation to solve for kk.
k=25k = 25
Squaring is the inverse operation of taking the square root, which isolates kk.

Anahtar Kavram

Solving radical equations by simplifying and combining like radical terms
Soru 35Soru

For the imaginary unit ii, where i2=1i^2 = -1, what is the real part of the complex number z=(2i)3+9iz = (2 - i)^3 + 9i?

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Cevap: 2

Cevap

The real part of the complex number is 2.
Expanding the expression (2i)3(2 - i)^3 yields 211i2 - 11i. Adding 9i9i gives 22i2 - 2i. The real part of this complex number is the term without ii, which is 2.

Adım Adım Çözüm

1
Expand the squared binomial (2i)2(2 - i)^2.
34i3 - 4i
Apply the identity (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2 and substitute i2=1i^2 = -1.
2
Multiply the result of the square by (2i)(2 - i) to calculate (2i)3(2 - i)^3.
211i2 - 11i
Distribute the terms (34i)(2i)=63i8i+4i2(3 - 4i)(2 - i) = 6 - 3i - 8i + 4i^2 and substitute i2=1i^2 = -1.
3
Add 9i9i to the simplified cube to find the complex number zz.
22i2 - 2i
Combine the imaginary components: 11i+9i=2i-11i + 9i = -2i.
4
Identify the real part of zz.
2
The real part of a complex number a+bia + bi is aa.

Anahtar Kavram

Expanding complex binomials and simplifying powers of the imaginary unit
Soru 36Soru

For all real numbers yy that satisfy the inequality y362||y - 3| - 6| \leq 2, what is the sum of all possible integer values of yy?

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Cevap: 3030

Cevap

The sum of all possible integer values of yy is 3030.
The compound inequality 4y384 \leq |y - 3| \leq 8 splits into two parts: y38|y - 3| \leq 8 (which gives 5y11-5 \leq y \leq 11) and y34|y - 3| \geq 4 (which gives y1y \leq -1 or y7y \geq 7). The intersection of these intervals is [5,1][7,11][-5, -1] \cup [7, 11]. Summing all the integers in these intervals gives (5+4+3+2+1)+(7+8+9+10+11)=15+45=30(-5 + -4 + -3 + -2 + -1) + (7 + 8 + 9 + 10 + 11) = -15 + 45 = 30.

Adım Adım Çözüm

1
Set up the compound inequality representing the outer absolute value.
2y362-2 \leq |y - 3| - 6 \leq 2
By definition, ua|u| \leq a is equivalent to aua-a \leq u \leq a for a0a \geq 0.
2
Isolate the inner absolute value term by adding 66 to all parts of the inequality.
4y384 \leq |y - 3| \leq 8
Isolating the absolute value allows us to split the compound inequality into two separate cases.
3
Split the compound inequality into two separate inequalities and solve each one.
y38|y - 3| \leq 8 and y34|y - 3| \geq 4
The expression y3|y - 3| must be simultaneously less than or equal to 88 and greater than or equal to 44.
4
Solve the first inequality, y38|y - 3| \leq 8.
5y11-5 \leq y \leq 11
Rewriting the inequality gives 8y38-8 \leq y - 3 \leq 8, and adding 33 to all parts yields the interval [5,11][-5, 11].
5
Solve the second inequality, y34|y - 3| \geq 4.
y1y \leq -1 or y7y \geq 7
By definition, ua|u| \geq a is equivalent to uau \geq a or uau \leq -a. Thus, y34    y7y - 3 \geq 4 \implies y \geq 7, and y34    y1y - 3 \leq -4 \implies y \leq -1.
6
Determine the intersection of the two solution sets.
y[5,1][7,11]y \in [-5, -1] \cup [7, 11]
The values of yy must lie within [5,11][-5, 11] and also satisfy y1y \leq -1 or y7y \geq 7.
7
Identify the integers within the final intervals and calculate their sum.
Sum = 3030
The integers in [5,1][-5, -1] are 5,4,3,2,1-5, -4, -3, -2, -1 (sum = 15-15). The integers in [7,11][7, 11] are 7,8,9,10,117, 8, 9, 10, 11 (sum = 4545). The total sum is 15+45=30-15 + 45 = 30.

Anahtar Kavram

Solving compound and nested absolute value inequalities

Alternatif Yöntem

Instead of solving the inequality algebraically, one can test the integer values around the critical points. Since yy must satisfy 4y384 \leq |y - 3| \leq 8, we can see that the distance of yy from 33 must be between 44 and 88 units. The integers at distance 4,5,6,7,84, 5, 6, 7, 8 to the right of 33 are 7,8,9,10,117, 8, 9, 10, 11. The integers at distance 4,5,6,7,84, 5, 6, 7, 8 to the left of 33 are 1,2,3,4,5-1, -2, -3, -4, -5. Summing these ten integers yields 3030.
Tahmini Süre:3m 0s
Soru 37Soru

When solving the radical equation 2x24x6=x3\sqrt{2x^2 - 4x - 6} = x - 3 for all real values of xx, one of the solutions obtained from the squared equation is extraneous. What is the value of this extraneous solution?

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Cevap: -5

Cevap

The value of the extraneous solution is 5-5.
The correct answer is 5-5. Squaring both sides of the equation 2x24x6=x3\sqrt{2x^2 - 4x - 6} = x - 3 yields 2x24x6=x26x+92x^2 - 4x - 6 = x^2 - 6x + 9, which simplifies to x2+2x15=0x^2 + 2x - 15 = 0. Factoring this quadratic gives (x+5)(x3)=0(x + 5)(x - 3) = 0, leading to potential solutions of 33 and 5-5. Substituting 5-5 back into the original equation results in the left side simplifying to 88 and the right side simplifying to 8-8. Since 888 \neq -8, the value 5-5 is an extraneous solution.

Adım Adım Çözüm

1
Square both sides of the original radical equation to eliminate the square root.
2x24x6=(x3)22x^2 - 4x - 6 = (x - 3)^2
Squaring both sides removes the radical on the left side.
2
Expand the right side and move all terms to the left side to set the quadratic equation to zero.
x2+2x15=0x^2 + 2x - 15 = 0
Expanding (x3)2(x - 3)^2 gives x26x+9x^2 - 6x + 9. Subtracting this expression from both sides yields the simplified quadratic equation.
3
Factor the quadratic equation to find the potential solutions.
(x+5)(x3)=0(x + 5)(x - 3) = 0, which gives x=5x = -5 and x=3x = 3.
The factors of 15-15 that add up to 22 are 55 and 3-3.
4
Substitute each potential solution back into the original equation to check for extraneous roots.
For x=3x = 3, 2(3)24(3)6=33\sqrt{2(3)^2 - 4(3) - 6} = 3 - 3 simplifies to 0=00 = 0 (valid). For x=5x = -5, 2(5)24(5)6=53\sqrt{2(-5)^2 - 4(-5) - 6} = -5 - 3 simplifies to 8=88 = -8 (invalid).
Extraneous solutions satisfy the squared equation but do not satisfy the original radical equation due to the sign difference introduced by squaring.

Anahtar Kavram

Solving radical equations and identifying extraneous solutions
Tahmini Süre:2m 0s
Soru 38Soru

Given that i=1i = \sqrt{-1}, what is the value of the expression i83i^{83}?

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Cevap: i-i

Cevap

The value of the expression is i-i.
The expression i83i^{83} can be simplified by dividing the exponent 83 by 4. Because the remainder is 3, the expression is equivalent to i3i^3. Since i3=i2i=(1)i=ii^3 = i^2 \cdot i = (-1) \cdot i = -i, the correct value is i-i.

Adım Adım Çözüm

1
Divide the exponent 83 by 4 to determine the remainder.
The quotient is 20 with a remainder of 3, which means 83=4×20+383 = 4 \times 20 + 3.
Powers of the imaginary unit ii repeat in a cycle of four: i1=ii^1 = i, i2=1i^2 = -1, i3=ii^3 = -i, and i4=1i^4 = 1.
2
Rewrite the expression using the rules of exponents and substitute the values of i4i^4 and i3i^3.
i83=i4(20)+3=(i4)20i3=(1)20(i)=ii^{83} = i^{4(20) + 3} = (i^4)^{20} \cdot i^3 = (1)^{20} \cdot (-i) = -i.
Since i4=1i^4 = 1, any power of ii that is a multiple of 4 simplifies to 1, leaving only the remainder power to determine the final value.

Anahtar Kavram

Simplifying powers of the imaginary unit ii
Soru 39Soru

Determine the value of xx for which the given rational expression is undefined.

Aşağıdaki boşlukları doldurun

The rational expression x+72x10\frac{x + 7}{2x - 10} is undefined when x=x = .
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Cevap

5
The rational expression is undefined when the denominator, 2x102x - 10, equals 00. Solving 2x10=02x - 10 = 0 gives 2x=102x = 10, which simplifies to x=5x = 5.

Adım Adım Çözüm

1
Identify the denominator of the rational expression and set it equal to zero.
2x10=02x - 10 = 0
A rational expression is undefined when its denominator is equal to zero, as division by zero is not defined in the real number system.
2
Solve the linear equation for xx by adding 10 to both sides and then dividing by 2.
2x=10x=52x = 10 \Rightarrow x = 5
Isolating xx yields the value that makes the denominator zero.

Anahtar Kavram

Rational expressions are undefined when the denominator is equal to zero.
Tahmini Süre:45s
Soru 40Soru

Let the functions ff and gg be defined by f(x)=2x5f(x) = |2x - 5| and g(x)=x+7g(x) = \sqrt{x + 7} for all real numbers in their respective domains. If aa is a real number such that (fg)(a)=3(f \circ g)(a) = 3, what is the sum of all possible values of aa?

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Cevap: 3

Cevap

The sum of all possible values of aa is 33.
The correct answer is 33. The composition (fg)(a)=3(f \circ g)(a) = 3 translates to f(g(a))=2a+75=3f(g(a)) = |2\sqrt{a + 7} - 5| = 3. This absolute value relation splits into two equations: 2a+75=32\sqrt{a + 7} - 5 = 3 and 2a+75=32\sqrt{a + 7} - 5 = -3. Solving the first equation yields a+7=4a=9\sqrt{a + 7} = 4 \Rightarrow a = 9. Solving the second equation yields a+7=1a=6\sqrt{a + 7} = 1 \Rightarrow a = -6. Since both 99 and 6-6 are greater than or equal to 7-7, they are within the domain of the radical function. The sum of these values is 9+(6)=39 + (-6) = 3.

Adım Adım Çözüm

1
Express the composition (fg)(a)(f \circ g)(a) in terms of aa.
2a+75=3|2\sqrt{a + 7} - 5| = 3
By definition of function composition, (fg)(a)=f(g(a))(f \circ g)(a) = f(g(a)). Substituting g(a)=a+7g(a) = \sqrt{a + 7} into the expression for f(x)f(x) yields f(g(a))=2g(a)5=2a+75f(g(a)) = |2g(a) - 5| = |2\sqrt{a + 7} - 5|.
2
Set up the two algebraic cases to eliminate the absolute value.
2a+75=32\sqrt{a + 7} - 5 = 3 or 2a+75=32\sqrt{a + 7} - 5 = -3
An absolute value equation of the form u=c|u| = c (where c0c \geq 0) has two possible cases: u=cu = c or u=cu = -c.
3
Solve the first equation case for aa.
a=9a = 9
Adding 55 to both sides of 2a+75=32\sqrt{a + 7} - 5 = 3 gives 2a+7=82\sqrt{a + 7} = 8. Dividing by 22 gives a+7=4\sqrt{a + 7} = 4. Squaring both sides yields a+7=16a + 7 = 16, which gives a=9a = 9.
4
Solve the second equation case for aa.
a=6a = -6
Adding 55 to both sides of 2a+75=32\sqrt{a + 7} - 5 = -3 gives 2a+7=22\sqrt{a + 7} = 2. Dividing by 22 gives a+7=1\sqrt{a + 7} = 1. Squaring both sides yields a+7=1a + 7 = 1, which gives a=6a = -6.
5
Verify domain constraints and sum the valid solutions.
33
Both a=9a = 9 and a=6a = -6 satisfy the domain requirement for g(x)=x+7g(x) = \sqrt{x+7}, which is x7x \geq -7. The sum of these two valid values is 9+(6)=39 + (-6) = 3.

Anahtar Kavram

Evaluating and solving equations containing composite functions, absolute values, and radical functions.

Alternatif Yöntem

Instead of expanding the composition immediately, substitute a temporary variable u=g(a)=a+7u = g(a) = \sqrt{a+7}. The equation becomes f(u)=32u5=3f(u) = 3 \Rightarrow |2u - 5| = 3. Solve this simplified absolute value equation to get 2u5=3u=42u - 5 = 3 \Rightarrow u = 4, and 2u5=3u=12u - 5 = -3 \Rightarrow u = 1. Next, substitute back a+7\sqrt{a+7} for uu: solving a+7=4\sqrt{a+7} = 4 yields a=9a = 9, and solving a+7=1\sqrt{a+7} = 1 yields a=6a = -6. Summing these two solutions gives 9+(6)=39 + (-6) = 3.
Tahmini Süre:2m 0s
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