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For ax² + bx + c = 0 the discriminant is b² − 4ac. Here a = 2, b = −5, c = 3, so the discriminant equals 25 − 24 = 1. A positive value confirms two distinct real roots.
The modulus of a + bi is √(a² + b²). With a = 3 and b = −4 we get √(9 + 16) = √25 = 5. The modulus is always a non-negative real number.
Factorising gives (x − 1)(x − 3) < 0. The parabola opens upwards, so it is negative strictly between its roots 1 and 3. Hence the solution set is the open interval ]1, 3[.
In a geometric sequence each term is obtained by multiplying the previous one by a fixed ratio r. Here −6 ÷ 3 = −2 and 12 ÷ (−6) = −2, so the common ratio is −2. The alternating signs confirm a negative ratio.
Using log₂ 8 = 3 and log₂ 4 = 2, the sum is 5. Equivalently, log₂ 8 + log₂ 4 = log₂ 32 = 5 by the product rule for logarithms.
Repeated multiplication by a constant factor models exponential decay. Starting from 200 g and multiplying by 0.8 once per year gives m(t) = 200 · 0.8^t, an exponential function of t.
C(n, k) = n! / (k!(n − k)!). Here C(6, 2) = 6! / (2! · 4!) = (6 · 5) / 2 = 15. Order does not matter, so permutations would overcount.
Since 27 = 3³, the equation becomes 3^(x−1) = 3³. The exponential function with base 3 is injective, so x − 1 = 3 and therefore x = 4.
Substituting x = 2 gives 8 − 8 + 2 − 2 = 0. Since P(2) = 0, the factor theorem tells us that (x − 2) is a factor of P(x).
The factor condition gives P(1) = 1 + a + b + 6 = 0, so a + b = −7. The remainder condition gives P(2) = 8 + 4a + 2b + 6 = 12, so 2a + b = −1. Subtracting yields a = 6, but checking the system: from a + b = −7, b = −7 − a; substituting gives 2a − 7 − a = −1, so a = 6. Recomputing the first equation with a = 6 gives b = −13, and P(2) = 8 + 24 − 26 + 6 = 12, consistent.
Write 4^x = 2^(2x) and 0.5 = 2^(−1). The inequality becomes 2^x > 2^(2x−1), and since the base 2 exceeds 1 the exponent order is preserved: x > 2x − 1, hence x < 1.
The sum of an arithmetic sequence is S_n = (n/2)(2u₁ + (n − 1)d). Here S₂₀ = 10 · (14 − 57) = 10 · (−43) = −430. The terms decrease steadily, so a negative sum is expected.
The first term is 1 and the ratio is 1/3, whose absolute value is less than 1, so the series converges. Its sum is a/(1 − r) = 1/(1 − 1/3) = 3/2.
The general term is C(5, k) · 2^(5−k) · x^k. For x³ we take k = 3, giving C(5, 3) · 2² = 10 · 4 = 40. Only the constant part contributes the factor 4.
Expanding (2 + i)(1 − 3i) = 2 − 6i + i − 3i². Since i² = −1, the last term becomes +3, so the product is 5 − 5i.
The modulus is √(1 + 3) = 2. The point lies in the second quadrant with equal reference angle π/3, so the argument is π − π/3 = 2π/3.
The product rule gives log₃(x(x − 2)) = 1, so x² − 2x = 3, i.e. x² − 2x − 3 = 0, giving x = 3 or x = −1. The logarithm requires x > 2, so only x = 3 is valid.
Let y = 2^x. Then y² − 5y + 4 = 0, so (y − 1)(y − 4) = 0, giving y = 1 or y = 4. Therefore 2^x = 1 gives x = 0 and 2^x = 4 gives x = 2, and both are valid.
Substituting x = −1 gives −1 − 3 + 4 = 0, so (x + 1) is a factor. The factor theorem guarantees that a root corresponds to a linear factor, and x = −1 is the only listed value that makes the polynomial vanish.
Factorising the numerator gives (x − 3)(x + 3), so for x ≠ 3 the function equals x + 3. The limit as x tends to 3 is therefore 6, even though f(3) is undefined.
The arithmetic series 1 + 2 + … + n has first term 1, last term n and n terms, so its sum is n(n + 1)/2. This formula can be proved by induction on n.
The term in x² is C(6, 2) · (−2x)² = 15 · 4x² = 60x². The negative sign of −2 is squared, so the coefficient is positive.
A double root at x = 1 means (x − 1)² divides P. Writing P(x) = (x − 1)²(x − r) and comparing the x² coefficient, which is zero, gives r = −2. Expanding yields x³ − 3x + 2, so p = −3 and q = 2.
Two distinct real roots require the discriminant to be strictly positive: m² − 4(m + 3) > 0, i.e. m² − 4m − 12 > 0. Factorising gives (m − 6)(m + 2) > 0, which holds for m < −2 or m > 6.
De Moivre's theorem gives modulus 2³ = 8 and argument 3 · π/6 = π/2. Thus z³ = 8(cos π/2 + i sin π/2) = 8i in Cartesian form.
The general term is C(4,k)(3x)^k(−2)^(4−k); for x² we take k = 2, giving C(4,2)·9·4 = 6·36 = 216. The sign is positive because (−2)² is positive.
Multiplying by the conjugate gives z·z̄ = (3 − 4i)(3 + 4i) = 9 + 16 = 25, which equals |z|². The imaginary parts cancel, so the product is always real and non-negative.
Using Sₙ = a(1 − rⁿ)/(1 − r) with a = 5, r = −2, n = 6 gives S₆ = 5(1 − 64)/3 = 5(−63)/3 = −105. The even power r⁶ = 64 is positive, so the numerator is negative.
By the remainder theorem the remainder equals P(2) = 8 − 16 + 2 + 6 = 0. Hence (x − 2) is a factor of P, and the division is exact.
Two distinct real roots require a positive discriminant: Δ = k² − 36 > 0. This holds exactly when |k| > 6, that is k < −6 or k > 6.
Using the quotient rule, log₂(48) − log₂(3) = log₂(48/3) = log₂(16) = 4. The difference of logarithms with the same base is the logarithm of the quotient.
Iterating the recurrence: u₂ = 3·2 − 1 = 5, u₃ = 3·5 − 1 = 14, u₄ = 3·14 − 1 = 41. Each term depends on the immediately preceding one.
First (1 + i)² = 2i, so (1 + i)⁴ = (2i)² = −4 and (1 + i)⁸ = (−4)² = 16. Alternatively, the modulus is √2 and the argument is π/4, so the eighth power has modulus (√2)⁸ = 16 and argument 2π.
Dividing P by (x − 3) gives the quadratic 2x² + 3x − 2, whose discriminant is 9 + 16 = 25 and whose roots are 1/2 and −2. Hence P(x) = (x − 3)(2x − 1)(x + 2).
The common difference is 3, and 100 = 1 + 3(n − 1) gives n = 34 terms. Then S = (n/2)(first + last) = 17·101 = 1717.
Adding the same real number to both sides of an inequality preserves its direction, an order axiom of the reals. The other statements fail for suitable signs, for example multiplying by a negative c or squaring across zero.
The point (1, −1) lies in the fourth quadrant, so its argument is arctan(−1/1) = −π/4, which lies in the principal range (−π, π]. The modulus is √2.
Writing t = 3ˣ > 0 gives t² − 4t + 3 = 0, so t = 1 or t = 3. Hence x = 0 or x = 1, and both values are admissible since t is positive.
C(9,3) = 9!/(3!·6!) = (9·8·7)/(3·2·1) = 504/6 = 84. The coefficient counts unordered selections of 3 objects from 9.
By the factor theorem P(x) = (x − 1)(x + 2)(x − 4), and expanding gives x³ − 3x² − 6x + 8. The coefficient of x² is the negative of the sum of the roots: −(1 − 2 + 4) = −3.
For a quadratic ax² + bx + c the vertex lies at x = −b/(2a). Here a = 3 and b = −12, so x = 12/6 = 2. Substituting gives the minimum value f(2) = −5.
Replacing x by x − 4 shifts the graph 4 units in the positive x-direction (right), because the input that produced a given output now occurs 4 larger. Adding 3 outside the function shifts every y-value up by 3.
A square root is real only when its argument is non-negative, so 5 − x ≥ 0. Rearranging gives x ≤ 5, and the endpoint x = 5 is included because √0 = 0 is defined.
Compute f(−x) = (−x)³ − (−x) = −x³ + x = −(x³ − x) = −f(x). Since f(−x) = −f(x) for every x, the function is odd, and its graph has rotational symmetry about the origin.
Setting the expressions equal gives x² − 4x + 4 = 0, i.e. (x − 2)² = 0, whose discriminant is zero. A zero discriminant means the line is tangent to the parabola, so there is exactly one intersection point, at x = 2.
The denominator vanishes at x = 3 while the numerator does not, giving a vertical asymptote there. For large |x| the ratio (x + 1)/(x − 3) tends to the ratio of leading coefficients, 1/1 = 1, so y = 1 is the horizontal asymptote.
By definition (f ∘ g)(x) = f(g(x)), so substitute g(x) = x² into f: f(x²) = 2x² + 5. Order matters, and here the squaring is applied first.
Complete the square: f(x) = (x − 3)² − 4, so y + 4 = (x − 3)² and x = 3 ± √(y + 4). The restriction x ≥ 3 selects the positive root, giving f⁻¹(x) = 3 + √(x + 4).
A linear function has constant slope, here (12 − 4)/(5 − 1) = 2. Since x = 3 is the midpoint of x = 1 and x = 5, f(3) is the midpoint of the outputs: (4 + 12)/2 = 8.
A rational function is undefined where its denominator equals zero. Solving x² − 9 = 0 gives x² = 9, hence x = 3 or x = −3, and both must be excluded from the domain.