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Find all real values of \(k\) for which the equation \(2x^2 + kx + 8 = 0\) has exactly one real solution.
Step 1: For a quadratic \(ax^2 + bx + c = 0\) to have exactly one real solution, the discriminant must equal zero: \(\Delta = b^2 - 4ac = 0\).
Step 2: Here \(a = 2\), \(b = k\), \(c = 8\). So: $$k^2 - 4(2)(8) = 0$$
Step 3: Simplify: \(k^2 - 64 = 0\), so \(k^2 = 64\).
Step 4: Taking square roots: \(k = \pm 8\).
Answer: \(k = 8\) or \(k = -8\)
Solve the system: \(x + y + z = 6\), \(xy + yz + xz = 11\), \(xyz = 6\). Find \(x, y, z\).
Step 1: The given conditions are the elementary symmetric polynomials of \(x, y, z\). By Vieta's formulas, \(x, y, z\) are roots of: $$t^3 - 6t^2 + 11t - 6 = 0$$
Step 2: Try \(t = 1\): \(1 - 6 + 11 - 6 = 0\). So \((t - 1)\) is a factor.
Step 3: Divide: \(t^3 - 6t^2 + 11t - 6 = (t - 1)(t^2 - 5t + 6)\).
Step 4: Factor the quadratic: \(t^2 - 5t + 6 = (t - 2)(t - 3)\).
Step 5: The roots are \(t = 1, 2, 3\).
Answer: \(\{x, y, z\} = \{1, 2, 3\}\) (in any order)
Find all real \(x\) satisfying \(|2x - 3| + |x + 1| < 7\).
Step 1 (Case \(x < -1\)): Both expressions inside absolute values are negative: \(-(2x-3) + -(x+1) < 7 \Rightarrow -3x + 2 < 7 \Rightarrow -3x < 5 \Rightarrow x > -\frac{5}{3}\). Combined with \(x < -1\): \(-\frac{5}{3} < x < -1\).
Step 2 (Case \(-1 \leq x < \frac{3}{2}\)): \(-(2x-3) + (x+1) < 7 \Rightarrow -x + 4 < 7 \Rightarrow -x < 3 \Rightarrow x > -3\). Combined: \(-1 \leq x < \frac{3}{2}\) (always satisfied).
Step 3 (Case \(x \geq \frac{3}{2}\)): \((2x-3) + (x+1) < 7 \Rightarrow 3x - 2 < 7 \Rightarrow 3x < 9 \Rightarrow x < 3\). Combined: \(\frac{3}{2} \leq x < 3\).
Step 4: Union of all three intervals: \(-\frac{5}{3} < x < 3\).
Answer: \(x \in \left(-\dfrac{5}{3},\; 3\right)\)
Factor completely: \(x^4 - 5x^2 + 4\). Then find all roots.
Step 1: Let \(u = x^2\). The expression becomes \(u^2 - 5u + 4\).
Step 2: Factor: \(u^2 - 5u + 4 = (u - 1)(u - 4)\).
Step 3: Substitute back: \((x^2 - 1)(x^2 - 4)\).
Step 4: Each factor is a difference of squares: $$(x-1)(x+1)(x-2)(x+2)$$
Step 5: Set each factor to zero: \(x = 1, -1, 2, -2\).
Answer: \(x^4 - 5x^2 + 4 = (x-1)(x+1)(x-2)(x+2)\); roots are \(x = \pm 1, \pm 2\)
The roots of \(x^2 - 7x + k = 0\) differ by 3. Find \(k\) and both roots.
Step 1: Let the roots be \(r\) and \(r + 3\) (they differ by 3).
Step 2: By Vieta's formulas for \(x^2 - 7x + k = 0\): sum of roots \(= 7\). So \(r + (r+3) = 7 \Rightarrow 2r + 3 = 7 \Rightarrow r = 2\).
Step 3: The other root is \(r + 3 = 5\).
Step 4: Product of roots \(= k\): \(k = 2 \times 5 = 10\).
Answer: \(k = 10\), roots are \(2\) and \(5\)
Simplify: \(\dfrac{x^3 - 8}{x^2 - 4}\) and state any restrictions on \(x\).
Step 1: Factor the numerator using the difference of cubes formula: $$x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)$$
Step 2: Factor the denominator using the difference of squares: $$x^2 - 4 = (x - 2)(x + 2)$$
Step 3: Cancel the common factor \((x - 2)\): $$\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \frac{x^2+2x+4}{x+2}$$
Step 4: Restrictions: \(x \neq 2\) and \(x \neq -2\) (values that make the original denominator zero).
Answer: \(\dfrac{x^2+2x+4}{x+2}\), with \(x \neq 2\) and \(x \neq -2\)
Simplify: \(\sqrt{6 + 4\sqrt{2}} + \sqrt{6 - 4\sqrt{2}}\).
Step 1: Write \(6 + 4\sqrt{2}\) as a perfect square. Try \((a + b)^2 = a^2 + 2ab + b^2\). We need \(a^2 + b^2 = 6\) and \(2ab = 4\sqrt{2}\), so \(ab = 2\sqrt{2}\).
Step 2: With \(a = 2, b = \sqrt{2}\): \(a^2 + b^2 = 4 + 2 = 6\) and \(2ab = 4\sqrt{2}\). So \(6 + 4\sqrt{2} = (2 + \sqrt{2})^2\).
Step 3: Similarly, \(6 - 4\sqrt{2} = (2 - \sqrt{2})^2\) (since \(2 > \sqrt{2}\), this is positive).
Step 4: Therefore: $$\sqrt{6 + 4\sqrt{2}} + \sqrt{6 - 4\sqrt{2}} = (2 + \sqrt{2}) + (2 - \sqrt{2}) = 4$$
Answer: \(4\)
Find all functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that \(f(x + y) = f(x) + f(y) + 2xy\) for all real \(x, y\), given \(f(1) = 3\).
Step 1: Set \(x = y = 0\): \(f(0) = f(0) + f(0) + 0 \Rightarrow f(0) = 0\).
Step 2: Let \(g(x) = f(x) - x^2\). Substitute into the functional equation: $$g(x+y) + (x+y)^2 = g(x) + x^2 + g(y) + y^2 + 2xy$$ $$g(x+y) = g(x) + g(y)$$
Step 3: This is Cauchy's functional equation. Assuming continuity (or monotonicity), \(g(x) = cx\) for some constant \(c\).
Step 4: So \(f(x) = x^2 + cx\). Using \(f(1) = 3\): \(1 + c = 3 \Rightarrow c = 2\).
Step 5: Verify: \(f(x+y) = (x+y)^2 + 2(x+y) = x^2 + 2xy + y^2 + 2x + 2y\). And \(f(x) + f(y) + 2xy = x^2 + 2x + y^2 + 2y + 2xy\). These are equal.
Answer: \(f(x) = x^2 + 2x\)
For positive reals \(a, b, c\) with \(a + b + c = 1\), prove that \(a^2 + b^2 + c^2 \geq \frac{1}{3}\).
Step 1: We know that \((a - b)^2 + (b - c)^2 + (c - a)^2 \geq 0\) (sum of squares).
Step 2: Expand: \(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca \geq 0\), which gives: $$a^2 + b^2 + c^2 \geq ab + bc + ca$$
Step 3: Also, \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1\).
Step 4: From Step 2: \(ab + bc + ca \leq a^2 + b^2 + c^2\). Substituting into Step 3: $$1 = a^2 + b^2 + c^2 + 2(ab+bc+ca) \leq a^2 + b^2 + c^2 + 2(a^2+b^2+c^2) = 3(a^2+b^2+c^2)$$
Step 5: Therefore \(a^2 + b^2 + c^2 \geq \frac{1}{3}\). Equality holds when \(a = b = c = \frac{1}{3}\).
Answer: Proven: \(a^2 + b^2 + c^2 \geq \frac{1}{3}\), with equality iff \(a = b = c = \frac{1}{3}\)
Solve for \(x\): \(\log_2(x + 3) + \log_2(x - 1) = 5\).
Step 1: Domain: \(x + 3 > 0\) and \(x - 1 > 0\), so \(x > 1\).
Step 2: Combine logs: \(\log_2((x+3)(x-1)) = 5\).
Step 3: Convert: \((x+3)(x-1) = 2^5 = 32\).
Step 4: Expand: \(x^2 + 2x - 3 = 32 \Rightarrow x^2 + 2x - 35 = 0\).
Step 5: Factor: \((x + 7)(x - 5) = 0\), so \(x = -7\) or \(x = 5\).
Step 6: Check domain: \(x = -7 < 1\) is rejected. \(x = 5 > 1\) is valid.
Answer: \(x = 5\)
When the polynomial \(p(x) = x^4 + ax^3 + bx^2 + cx + d\) is divided by \((x - 1)\), the remainder is 3. When divided by \((x - 2)\), the remainder is 5. When divided by \((x + 1)\), the remainder is 7. When divided by \((x + 2)\), the remainder is 9. Find \(a + b + c + d\).
Step 1: By the Remainder Theorem: \(p(1) = 1 + a + b + c + d = 3\), so \(a + b + c + d = 2\).
Step 2: \(p(2) = 16 + 8a + 4b + 2c + d = 5\), so \(8a + 4b + 2c + d = -11\).
Step 3: \(p(-1) = 1 - a + b - c + d = 7\), so \(-a + b - c + d = 6\).
Step 4: \(p(-2) = 16 - 8a + 4b - 2c + d = 9\), so \(-8a + 4b - 2c + d = -7\).
Step 5: Add equations from Steps 1 and 3: \(2b + 2d = 8 \Rightarrow b + d = 4\). Subtract: \(2a + 2c = -4 \Rightarrow a + c = -2\).
Step 6: Therefore \(a + b + c + d = (a + c) + (b + d) = -2 + 4 = 2\).
Answer: \(a + b + c + d = 2\)
Find the number of positive integers \(n \leq 1000\) such that \(\lfloor\sqrt{n}\rfloor\) divides \(n\).
Step 1: Let \(\lfloor\sqrt{n}\rfloor = k\). Then \(k^2 \leq n \leq (k+1)^2 - 1\), i.e., \(n \in [k^2,\, k^2 + 2k]\). This interval has \(2k + 1\) integers.
Step 2: We need \(k \mid n\). Since \(k \mid k^2\), the multiples of \(k\) in \([k^2, k^2+2k]\) are \(k^2, k^2+k, k^2+2k\), giving exactly 3 multiples for each \(k \geq 1\).
Step 3: For \(k = 1, 2, \ldots, 31\), the range \([k^2, k^2+2k]\) lies entirely within \([1, 1023]\). For \(k = 31\): range is \([961, 1023]\). We need \(n \leq 1000\), so multiples of 31 in \([961, 1000]\) are \(961, 992\) (since \(1023 > 1000\) and \(961+62 = 1023\)). That gives 2 values, not 3.
Step 4: For \(k = 1\) to \(30\), each contributes 3 values: \(30 \times 3 = 90\). For \(k = 31\): multiples of 31 in \([961, 1000]\) are 961 and 992, so 2 values. Total: \(90 + 2 = 92\).
Answer: There are 92 such positive integers