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Subtract 2x from both sides to get x - 7 = 5, then add 7 to both sides to obtain x = 12. The key step is collecting variable terms on one side and constants on the other before isolating x.
Multiplying each term of the first bracket by each term of the second gives x² - 3x + 4x - 12. Combining the like terms -3x and 4x yields x² + x - 12.
We need two numbers whose product is -24 and whose sum is -5. The pair -8 and 3 satisfies both conditions, giving (x - 8)(x + 3). Expanding confirms the middle term is -8x + 3x = -5x.
When multiplying powers with the same base, add the indices: 2³ × 2⁴ = 2⁷. When dividing, subtract the indices: 2⁷ ÷ 2⁵ = 2². The result equals 4.
Subtract 4 from both sides to get -2x > 6. Dividing both sides by -2 reverses the inequality sign, giving x < -3. Reversing the sign when dividing by a negative number is the crucial step.
The first term is a = 3 and the common difference is d = 4, so the 20th term is 3 + 19×4 = 79. Using S = n(a + l)/2 gives 20 × (3 + 79)/2 = 820.
Squaring a product means squaring each factor: 3² = 9, (x²)² = x⁴ and (y³)² = y⁶. Multiplying the indices by the outer power 2 gives the result 9x⁴y⁶.
A repeated root occurs when the discriminant b² - 4ac equals zero. Here k² - 4×1×9 = 0, so k² = 36 and k = ±6. Both values give a perfect square trinomial.
From the second equation x = y + 1. Substituting into the first gives 2(y + 1) + 3y = 12, so 5y = 10 and y = 2, hence x = 3. Checking: 2(3) + 3(2) = 12 and 3 - 2 = 1.
Multiply numerator and denominator by the conjugate √5 + 1. The denominator becomes 5 - 1 = 4, so the expression is 6(√5 + 1)/4 = (3/2)(√5 + 1).
Combining the logarithms gives log₂(x(x - 2)) = 3, so x² - 2x = 8. Solving x² - 2x - 8 = 0 gives x = 4 or x = -2; since logarithms require x > 0 and x - 2 > 0, only x = 4 is valid.
Let x = 0.4444... Then 10x = 4.4444..., and subtracting gives 9x = 4, so x = 4/9. This fraction is already in lowest terms.
Dividing P(x) by (x - 2) using synthetic division gives x² - 2x - 3. Factorising this quadratic gives (x - 3)(x + 1), so the full factorisation is (x - 2)(x - 3)(x + 1).
The absolute value equation splits into 2x - 5 = 7 or 2x - 5 = -7. Solving gives x = 6 or x = -1. Both values satisfy the original equation.
The first term is a = 5 and the common ratio is r = 3. Using Sₙ = a(rⁿ - 1)/(r - 1) gives 5(3⁶ - 1)/2 = 5(729 - 1)/2 = 1820.
Factorising gives (x - 3)(x + 3) in the numerator and (x + 4)(x - 3) in the denominator. Cancelling the common factor (x - 3) leaves (x + 3)/(x + 4).
The composite function f(g(x)) means substituting g(x) into f. Since g(x) = x², we compute f(x²) = 2x² + 1. The order of composition matters, so the inner function is evaluated first.
Using the binomial theorem, the term in x³ is C(5,3)·2²·x³ = 10 × 4 × x³ = 40x³. The coefficient is therefore 40.
The sum of the roots is 4 and the product is (2 + √3)(2 - √3) = 4 - 3 = 1. The equation is x² - (sum)x + product = 0, giving x² - 4x + 1 = 0.
The general term is C(6,r)·x^(6-r)·(-1/x)^r = C(6,r)·(-1)^r·x^(6-2r). The constant term requires 6 - 2r = 0, so r = 3, giving C(6,3)·(-1)³ = 20 × (-1) = -20.
For a cubic ax³ + bx² + cx + d, the sum of all three roots is -b/a = 3/2. Since one root is 3, the sum of the other two roots is 3/2 - 3 = -3/2.
Write 4^x as (2^x)² and let u = 2^x. Then u² - 5u + 4 = 0 factors as (u - 1)(u - 4) = 0, so u = 1 or u = 4. Thus 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2.
Use the identity x² + y² = (x + y)² - 2xy. Substituting the given values gives 25 - 6 = 19.
The derivative is f'(x) = 3x² - 6x = 3x(x - 2), and f''(x) = 6x - 6. At x = 2, f''(2) = 6 > 0 suggests a minimum, but the second derivative test is inconclusive here because f'(x) has the same sign on both sides; evaluating shows f'(x) = 3x(x - 2) changes sign at x = 0 but not at x = 2, so x = 2 is a stationary point of inflection.
Substituting y = mx + 2 into x² + y² = 1 gives (1 + m²)x² + 4mx + 3 = 0. For two distinct intersections the discriminant 16m² - 12(1 + m²) must be positive, so 4m² - 12 > 0 and m² > 3, giving m > √3 or m < -√3.
By the Remainder Theorem, the remainder is P(2). Substituting x = 2 gives 2(8) - 5(4) + 3(2) - 7 = 16 - 20 + 6 - 7 = -5. Thus the remainder is -5.
Using the binomial theorem, the term containing x² is C(4,2)(3x)²(-2)² = 6 * 9x² * 4 = 216x². Hence the coefficient is 216.
The first term is a = 8 and the common ratio is r = 1/2, with |r| < 1. The sum to infinity is S = a/(1 - r) = 8/(1 - 1/2) = 8/(1/2) = 16.
Factorising gives (x - 2)(x - 3) < 0. The product is negative when x lies between the roots 2 and 3, so the solution is 2 < x < 3.
Multiply numerator and denominator by the conjugate 1 + √2: [(1 + √2)²]/[(1 - √2)(1 + √2)] = (1 + 2√2 + 2)/(1 - 2) = (3 + 2√2)/(-1) = -3 - 2√2.
Combine logs: log₃[(x + 2)(x - 4)] = 2, so (x + 2)(x - 4) = 9. This gives x² - 2x - 17 = 0, whose roots are 1 ± 3√2. Only 1 + 3√2 ≈ 5.24 is greater than 4 and thus valid. However, none of the options match exactly; the closest is 5, but on re-evaluation, the equation yields x = 5? Let's check: log₃(7) + log₃(1) = log₃7 ≠ 2. So 5 is not correct. The correct solution is 1 + 3√2 ≈ 5.24, which is not listed. Therefore, the question is flawed. To fix, change the equation to log₃(x + 2) + log₃(x - 4) = 2? Actually, if the right side is 2, then (x+2)(x-4)=9, so x²-2x-8=9 => x²-2x-17=0, roots 1±3√2. None of the options. So we must rewrite the question. Let's choose a different equation: log₂(x + 1) + log₂(x - 1) = 3. Then (x+1)(x-1)=8 => x²-1=8 => x²=9 => x=3 (since x>1). So x=3. That works. So change the question to: The equation log₂(x + 1) + log₂(x - 1) = 3 has solution x equal to: options: 3, -3, 5, 2. Correct index 0. Explanation: Combine logs: log₂[(x+1)(x-1)] = 3, so (x+1)(x-1)=8, x²-1=8, x²=9, x=3 (since x>1). So x=3. Wrong explanations: -3 is invalid because log of negative; 5 gives log₂6+log₂4=log₂24≠3; 2 gives log₂3+log₂1=log₂3≠3. Good.
Expand: 2(1) + 2(-4i) + 3i(1) + 3i(-4i) = 2 - 8i + 3i - 12i². Since i² = -1, -12i² = 12, so the result is (2 + 12) + (-8i + 3i) = 14 - 5i.
From the second equation, y = 2x - 4. Substitute into the first: 3x + 2(2x - 4) = 7 → 3x + 4x - 8 = 7 → 7x = 15 → x = 15/7. Then y = 2(15/7) - 4 = 30/7 - 28/7 = 2/7. So the solution is (15/7, 2/7).
Factor the numerator as a difference of cubes: x³ - 8 = (x - 2)(x² + 2x + 4). The denominator is a difference of squares: x² - 4 = (x - 2)(x + 2). Cancelling the common factor (x - 2) gives (x² + 2x + 4)/(x + 2).
The nth term is Sₙ - Sₙ₋₁. Compute S₁₀ = 3(100) + 20 = 320 and S₉ = 3(81) + 18 = 261. Thus the 10th term is 320 - 261 = 59.
Let y = 2^x. Then y² - 5y + 4 = 0, so (y - 1)(y - 4) = 0, giving y = 1 or y = 4. Thus 2^x = 1 → x = 0, and 2^x = 4 → x = 2. The solutions are x = 0 and x = 2.
By Vieta's formulas, α + β = 3/2 and αβ = 2. Then α² + β² = (α + β)² - 2αβ = (9/4) - 4 = (9/4) - (16/4) = -7/4.
The term containing x³ is C(5,3)(1)²(2x)³ = 10 * 1 * 8x³ = 80x³. Hence the coefficient is 80.
For two distinct real roots, the discriminant must be positive: Δ = (k - 2)² - 4(1)(k + 1) = k² - 4k + 4 - 4k - 4 = k² - 8k = k(k - 8). So k(k - 8) > 0, which holds when k < 0 or k > 8.
Multiply numerator and denominator by the conjugate 1 + i: [(1 + i)²]/[(1 - i)(1 + i)] = (1 + 2i + i²)/(1 - i²) = (1 + 2i - 1)/(1 + 1) = (2i)/2 = i.
For a quadratic ax² + bx + c, the vertex x-coordinate is −b/(2a) = 8/4 = 2. Substituting gives f(2) = 8 − 16 + 5 = −3, so the vertex is (2, −3). Completing the square confirms f(x) = 2(x − 2)² − 3.
A horizontal shift left by 3 replaces x with x + 3 inside the function, and a vertical shift down by 2 subtracts 2 outside. Combining these gives f(x) = |x + 3| − 2. The vertex moves from (0, 0) to (−3, −2), which matches.
The radicand must be non-negative, so 5 − x ≥ 0, giving x ≤ 5. The denominator must be nonzero, so x ≠ 1. Intersecting these conditions yields x ≤ 5 with x ≠ 1.
For an odd function, h(−x) = −h(x), so h(−3) = −h(3) = 4. An odd function defined at the origin satisfies h(0) = 0, since h(0) = −h(0). Therefore the sum is 4 + 0 = 4.
A point (a, b) on y = f(x) corresponds to the point (a + 2, b + 3) on y = f(x − 2) + 3. Applying this to (1, 4) gives (3, 7). The transformation shifts the graph 2 units right and 3 units up.
The leading term is −2x⁴, of even degree with a negative coefficient. For large |x|, this term dominates and is negative in both directions. Hence p(x) tends to −∞ as x → +∞ and as x → −∞.
Solve 3x − 7 = 5, giving 3x = 12 and x = 4. Equivalently, f⁻¹(x) = (x + 7)/3, so f⁻¹(5) = 12/3 = 4.
Reflecting in the y-axis replaces each x-coordinate by its negative while keeping y unchanged, so the new equation is y = f(−x). A reflection in the x-axis would instead give y = −f(x).
Completing the square gives f(x) = (x + 2)² + 3, whose minimum on the restricted domain occurs at x = −2 and equals 3. Since the parabola opens upward and the domain extends rightward from the vertex, the range is y ≥ 3.
For y = f(2x), an intercept at x = a of f occurs where 2x = a, that is x = a/2. Thus −3 and 2 become −1.5 and 1, which is not the target; instead consider that scaling the input by 1/2 gives x = 2a. Check y = f(x/2): intercepts occur where x/2 = a, so x = 2a, giving −6 and 4. Among the options, only y = f(x/2) doubles the intercept positions.