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Powers of i repeat in a cycle of length four: i^1 = i, i^2 = -1, i^3 = -i, and i^4 = 1. Dividing 26 by 4 leaves remainder 2, so i^26 equals i^2, which is -1.
Add real parts to real parts and imaginary parts to imaginary parts: (3 + 5) + (2 + (-7))i. Since 2 - 7 = -5, the sum is 8 - 5i.
Distribute i over both terms: i(4) - i(3i) = 4i - 3i^2. Because i^2 = -1, the term -3i^2 equals +3, so the product is 3 + 4i.
Isolating the square gives x^2 = -49. Taking square roots gives x = ±√(-49), and since √(-1) = i, the solutions are x = ±7i. Both 7i and -7i satisfy the equation.
For a rational exponent m/n, take the nth root of the base and then raise the result to the m power. The cube root of 27 is 3, and 3^2 = 9, so 27^(2/3) = 9.
A radical can be rewritten with a rational exponent using the rule that the nth root of x^m equals x^(m/n). Here the index is 3 and the power is 5, so the cube root of x^5 equals x^(5/3).
A rational number can be written as a ratio of integers, and its decimal either terminates or repeats. √16 = 4 and 0.75 = 3/4 are rational, and 22/7 is a ratio of integers. Since 7 is not a perfect square, √7 has a nonterminating, nonrepeating decimal and is irrational.
The complex conjugate of a + bi is a - bi: the real part stays the same and only the sign of the imaginary part changes. Applying this to 6 - 5i gives 6 + 5i.
Subtract matching parts: (7 - 2) + (-3 - (-8))i. Subtracting a negative adds, so -3 + 8 = 5, and the difference is 5 + 5i.
Using the distributive property gives 12 - 3i + 8i - 2i^2. Replacing i^2 with -1 turns -2i^2 into +2, so the expression becomes 12 + 5i + 2, which simplifies to 14 + 5i.
Multiply numerator and denominator by the conjugate 1 - i. The denominator becomes (1 + i)(1 - i) = 1 + 1 = 2, and the numerator becomes 4 - 4i, so the quotient is 2 - 2i.
For a complex number a + bi, the modulus is √(a^2 + b^2), the distance from the origin in the complex plane. Here √(9 + 16) = √25 = 5.
The discriminant is b^2 - 4ac = 4^2 - 4(1)(13) = 16 - 52 = -36. Because the discriminant is negative, the equation has two nonreal complex solutions.
Reduce each power using the four-term cycle of i. Since 15 leaves remainder 3, i^15 = i^3 = -i; since 8 is a multiple of 4, i^8 = 1. The sum is 1 - i.
A negative exponent produces a reciprocal, so 16^(-3/4) = 1/16^(3/4). The fourth root of 16 is 2, and 2^3 = 8, so the value is 1/8.
Simplify each radical first: √50 = √(25·2) = 5√2 and √18 = √(9·2) = 3√2. Adding like radicals gives 5√2 + 3√2 = 8√2.
Multiply numerator and denominator by the conjugate 2 + √3. The denominator becomes 4 - 3 = 1, and the numerator becomes 4(2 + √3) = 8 + 4√3.
The product fits the difference-of-squares pattern (a + b)(a - b) = a^2 - b^2, so it equals 4 - 5 = -1. Because -1 is an integer, the product of these two irrational factors is rational.
When a polynomial has real coefficients, any nonreal complex root occurs together with its complex conjugate. Since 3 - 2i is a root, its conjugate 3 + 2i must also be a root.
Using the quadratic formula gives x = (-2 ± √(4 - 20))/2 = (-2 ± √(-16))/2. Since √(-16) = 4i, this becomes (-2 ± 4i)/2, which simplifies to x = -1 ± 2i.
Multiply numerator and denominator by the conjugate 1 + i. The denominator becomes (1 - i)(1 + i) = 1 + 1 = 2, and the numerator becomes (1 + i)^2 = 1 + 2i - 1 = 2i, so the quotient is 2i/2 = i.
Multiplying (3 + 4i)(3 - 4i) gives 9 - 12i + 12i - 16i^2 = 9 + 16 = 25. In general, a complex number times its conjugate equals a^2 + b^2, the square of its modulus.
A quadratic has two nonreal complex solutions exactly when its discriminant is negative. Here b^2 - 4ac = k^2 - 100, so the condition k^2 - 100 < 0 gives k^2 < 100, which means -10 < k < 10.
The exponent 4n is a multiple of 4, so i^(4n) = (i^4)^n = 1. Multiplying by i^3 then gives i^(4n + 3) = 1 · i^3 = -i, and this holds for every integer n.
Rewrite both radicals with rational exponents: √x = x^(1/2) and the cube root of x^2 = x^(2/3). Multiplying powers of the same base adds exponents, and 1/2 + 2/3 = 3/6 + 4/6 = 7/6, so the product is x^(7/6).
The powers of i repeat in a cycle of four: i^0 = 1, i^1 = i, i^2 = -1, and i^3 = -i. Adding these four values gives 1 + i - 1 - i, which equals 0. Every complete block of four consecutive powers of i sums to zero.
When multiplying powers with the same base, add the exponents: 1/2 + 1/3 = 3/6 + 2/6 = 5/6. So the product equals x^(5/6). The base x stays unchanged.
The sum of a rational and an irrational number is always irrational. If r + s were rational, then subtracting the rational number r would make s = (r + s) - r rational, which contradicts the assumption that s is irrational. The rational numbers are closed under subtraction, so this reasoning is valid.
For nonnegative factors, √a · √b = √(ab), so √3 · √27 = √81. Since √81 = 9, the product is 9. Alternatively, √27 = 3√3, and √3 · 3√3 = 3 · 3 = 9.
Square the binomial using (a + b)^2 = a^2 + 2ab + b^2 with a = 3 and b = √2. This gives 9 + 2(3)(√2) + 2 = 11 + 6√2. The middle term is twice the product of the two terms, which is 6√2.
Multiply numerator and denominator by the conjugate 2 - i. The denominator becomes (2)^2 + (1)^2 = 5, and the numerator becomes 8 - 4i - 6i + 3i^2 = 5 - 10i. Dividing each part by 5 gives 1 - 2i.
Factor 24 as 8 · 3 so that one factor is a perfect cube. Then the cube root of 24 equals the cube root of 8 times the cube root of 3, which is 2 times the cube root of 3. Choosing 4 · 6 or 3 · 8 differently does not help because 6 and 3 are not perfect cubes.
A quadratic has exactly one real solution when its discriminant b^2 - 4ac equals zero. Here a = 1 and b = -6, so 36 - 4c = 0, giving c = 9. Equivalently, the trinomial is a perfect square: (x - 3)^2 = x^2 - 6x + 9.
Since √12 = 2√3, the expression becomes 6/(2√3) = 3/√3. Rationalizing by multiplying numerator and denominator by √3 gives (3√3)/3 = √3.
Split the fraction: 3/i + i/i. Since 1/i = 1/i · i/i = i/i^2 = -i, the first term is -3i, and the second term is 1. So the value is 1 - 3i.
The roots have sum 4 and product (2)^2 + (3)^2 = 13, so the equation is x^2 - 4x + 13 = 0. Equivalently, (x - 2)^2 = -9, which expands to x^2 - 4x + 4 + 9 = 0.
Write each radical with i: √(-9) = 3i and √(-4) = 2i. Their product is 6i^2, and since i^2 = -1, the value is -6. The rule √a · √b = √(ab) does not apply when both radicands are negative.
First compute (1 + i)^2 = 1 + 2i + i^2 = 2i. Squaring again gives (2i)^2 = 4i^2 = -4. Each squaring step reduces the work and avoids expanding all four factors at once.
Real solutions require the discriminant to be nonnegative: m^2 - 4(m + 3) = m^2 - 4m - 12 ≥ 0. Factoring gives (m - 6)(m + 2) ≥ 0, which holds when m ≤ -2 or m ≥ 6. The endpoints are included because a zero discriminant still yields one real solution.
Squaring the binomial gives x^(1/2 · 2) + 2·x^(1/2)·x^(-1/2) + x^(-1/2 · 2). Since the exponents add to zero in the middle term, that term is 2·x^0 = 2, so the result is x + 2 + 1/x.
Combine like terms by adding coefficients of matching powers: 2x^3 + 4x^3 = 6x^3, -5x + 3x = -2x, and 1 + (-7) = -6. The simplified polynomial is therefore 6x^3 - 2x - 6.
A difference of squares a^2 - b^2 factors as (a - b)(a + b). Here a = x and b = 3, so x^2 - 9 = (x - 3)(x + 3).
Add the real parts and the imaginary parts separately: (3 + 1) + (2i - 5i) = 4 - 3i. The real part is 4 and the imaginary coefficient is -3.
Rewrite 27 as 3^3 so that both sides have base 3: 3^(x + 1) = 3^3. Equating exponents gives x + 1 = 3, so x = 2.
Write 50 as 25 · 2, so √50 = √25 · √2 = 5√2. This uses the product property of square roots with the largest perfect-square factor.
Substitute -2 for every x: f(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10. Squaring -2 gives 4, and multiplying -3 by -2 gives 6.
An absolute value equation splits into two cases: x - 3 = 7 or x - 3 = -7. Solving gives x = 10 or x = -4, so the solution set is {-4, 10}.
A rational function is undefined where its denominator equals zero. Since x - 4 = 0 when x = 4, the domain is all real numbers except 4.
Factor the numerator as (x - 2)(x + 2) and the denominator as (x - 2)(x + 3). The common factor x - 2 cancels, leaving (x + 2)/(x + 3).
Multiply every term by the least common denominator 2x, assuming x ≠ 0: 2 + x = 6. Solving gives x = 4, which checks in the original equation.