Question: Find all functions $f : \mathbb{R} o \mathbb{R}$ such that $f(a + b) = f(a) + f(b) + ab$ for all real numbers $a, b$.

February 22, 2026

$Question: Find all functions $f : \mathbb{R} o \mathbb{R}$ such that $f(a + b) = f(a) + f(b) + ab$ for all real numbers $a, b$.$

Solution: Use the Cauchy-Schwarz inequality: $(2^2 + 3^2 + 4^2)(x^2 + y^2 + z^2) \geq (2x + 3y + 4z)^2$. This gives $29(x^2 + y^2 + z^2) \geq 144$, so $x^2 + y^2 + z^2 \geq rac{144}{29}$. Equality holds when $rac{x}{2} = rac{y}{3} = rac{z}{4} = k$, leading to $x = 2k$, $y = 3k$, $z = 4k$. Substituting into $2x + 3y + 4z = 12$ gives $4k + 9k + 16k = 29k = 12$, so $k = rac{12}{29}$. Thus, the minimum value is $oxed{\dfrac{144}{29}}$.
Question: Find the center of the hyperbola $9x^2 - 18x - 16y^2 + 64y = 144$.
Solution: Complete the square for $x$ and $y$. For $x$: $9(x^2 - 2x) = 9[(x - 1)^2 - 1] = 9(x - 1)^2 - 9$. For $y$: $-16(y^2 - 4y) = -16[(y - 2)^2 - 4] = -16(y - 2)^2 + 64$. Substitute back: $9(x - 1)^2 - 9 - 16(y - 2)^2 + 64 = 144$. Simplify: $9(x - 1)^2 - 16(y - 2)^2 = 89$. The center is at $(1, 2)$. Thus, the center is $oxed{(1, 2)}$.
Solution: Assume $f$ is quadratic. Let $f(x) = px^2 + qx + r$. Substitute into the equation: $p(a + b)^2 + q(a + b) + r = pa^2 + qa + r + pb^2 + qb + r + ab$. Expand and equate coefficients: $p(a^2 + 2ab + b^2) + q(a + b) + r = pa^2 + pb^2 + q(a + b) + 2r + ab$. Simplify: $2pab = ab + 2r$. For this to hold for all $a, b$, we require $2p = 1$ and $2r = 0$, so $p = rac{1}{2}$, $r = 0$. The linear term $q$ cancels out, so $f(x) = rac{1}{2}x^2 + qx$. Verifying, $f(a + b) = rac{1}{2}(a + b)^2 + q(
Solution: We compute $ \sum_{n=1}^{10} P(n) = \sum_{n=1}^{10} (n^2 + 3n + 5) = \sum_{n=1}^{10} n^2 + 3\sum_{n=1}^{10} n + \sum_{n=1}^{10} 5 $.
Using formulas:

Question: Find all functions $f : \mathbb{R} o \mathbb{R}$ such that $f(a + b) = f(a) + f(b) + ab$ for all real numbers $a, b$.

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