💡First, I need to convert the decimal number 93 to binary. I'll use the division-by-2 method, which involves repeatedly dividing the number by 2 and recording the remainders.Starting with 93, I divide by 2 to get a quotient of 46 and a remainder of 1. I'll write down the remainder.Next, I take the quotient 46 and divide it by 2, resulting in a quotient of 23 and a remainder of 0. I'll record this remainder.Then, I divide 23 by 2 to get a quotient of 11 and a remainder of 1. I'll note this remainder.Continuing, I divide 11 by 2, obtaining a quotient of 5 and a remainder of 1. I'll add this remainder to my list.Next, I divide 5 by 2, which gives a quotient of 2 and a remainder of 1. I'll record this remainder.I then divide 2 by 2, resulting in a quotient of 1 and a remainder of 0. I'll note this remainder.Finally, I divide 1 by 2, yielding a quotient of 0 and a remainder of 1. I'll record this last remainder.Now, I'll arrange all the remainders in reverse order, starting from the last division to the first. This gives me the binary representation of the decimal number 93.

💡First, I need to find the roots of the quadratic equation ( x^2 - 8x + 15 = 0 ) to determine the lengths of the two equal sides of the isosceles triangle.I'll start by factoring the quadratic equation. Looking for two numbers that multiply to 15 and add up to -8, I find that -3 and -5 satisfy these conditions. Therefore, the equation factors as:[(x - 3)(x - 5) = 0]Setting each factor equal to zero gives the roots:[x - 3 = 0 Rightarrow x = 3][x - 5 = 0 Rightarrow x = 5]So, the lengths of the two equal sides are 3 and 5.Next, I need to consider the possible configurations of the isosceles triangle. There are two scenarios:1. **Two sides of length 3 and one side of length 5:** [ text{Perimeter} = 3 + 3 + 5 = 11 ]2. **Two sides of length 5 and one side of length 3:** [ text{Perimeter} = 5 + 5 + 3 = 13 ]Both configurations are valid, so the perimeter of the isosceles triangle can be either 11 or 13.

💡Okay, so I have this quadratic equation: ( x^2 + bx + 16 ). The problem says that it has two non-real roots, and I need to find all the possible real values of ( b ). Hmm, okay. I remember that for a quadratic equation ( ax^2 + bx + c ), the nature of the roots is determined by the discriminant. The discriminant is ( b^2 - 4ac ). If the discriminant is positive, there are two distinct real roots. If it's zero, there's exactly one real root (a repeated root), and if it's negative, the roots are non-real and complex.So, in this case, since we want the roots to be non-real, the discriminant must be less than zero. Let me write that down:Discriminant ( D = b^2 - 4ac ).Given the quadratic ( x^2 + bx + 16 ), the coefficients are ( a = 1 ), ( b = b ), and ( c = 16 ). Plugging these into the discriminant formula:( D = b^2 - 4(1)(16) ).Simplifying that:( D = b^2 - 64 ).We want this discriminant to be negative for the roots to be non-real, so:( b^2 - 64 < 0 ).Let me solve this inequality for ( b ). Adding 64 to both sides:( b^2 < 64 ).This means that ( b ) must be a real number whose square is less than 64. To find the values of ( b ), I can take the square root of both sides. Remembering that when we take square roots in inequalities, we have to consider both the positive and negative roots:( |b| < 8 ).Which translates to:( -8 < b < 8 ).So, ( b ) must be greater than -8 and less than 8. That means all real numbers between -8 and 8, not including -8 and 8 themselves, will make the quadratic equation ( x^2 + bx + 16 ) have two non-real roots.Let me double-check my steps to make sure I didn't make a mistake. I started with the discriminant, plugged in the coefficients correctly, simplified the inequality, and solved for ( b ). Everything seems to check out. Just to visualize, if I think about the graph of the quadratic equation ( x^2 + bx + 16 ), it's a parabola opening upwards because the coefficient of ( x^2 ) is positive. If the discriminant is negative, the parabola doesn't intersect the x-axis, meaning it's entirely above the x-axis. So, the vertex of the parabola is above the x-axis, and the arms go upwards. This makes sense because if the roots are non-real, the graph doesn't touch the x-axis.Another way to think about it is that the quadratic doesn't cross the x-axis, so it's always positive. Therefore, the minimum value of the quadratic is positive. The vertex form of a quadratic is ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. For our equation, the vertex occurs at ( x = -frac{b}{2a} ), which is ( x = -frac{b}{2} ). Plugging this back into the equation gives the y-coordinate of the vertex:( y = left(-frac{b}{2}right)^2 + bleft(-frac{b}{2}right) + 16 ).Simplifying:( y = frac{b^2}{4} - frac{b^2}{2} + 16 ).Combining like terms:( y = -frac{b^2}{4} + 16 ).Since the parabola opens upwards, the minimum value is ( y = -frac{b^2}{4} + 16 ). For the quadratic to be always positive, this minimum value must be greater than zero:( -frac{b^2}{4} + 16 > 0 ).Multiplying both sides by 4:( -b^2 + 64 > 0 ).Which simplifies to:( b^2 < 64 ).Again, we get the same inequality ( -8 < b < 8 ). So, this method also confirms that ( b ) must lie between -8 and 8.I think I've covered this problem from different angles, and each time I arrive at the same conclusion. It seems solid. I don't see any mistakes in my reasoning, so I feel confident that the interval for ( b ) is indeed ( (-8, 8) ).**Final Answer**The set of all real values of ( b ) is boxed{(-8, 8)}.

💡Okay, so I have this problem here with two parts. Let me try to figure it out step by step. First, part (I) says: Given the function f(x) = lg( (2x² + bx + c)/(x² + 1) ), and the solution set of the inequality bx² + 3x + c ≥ 0 is A = {x | -1/2 ≤ x ≤ 2}. I need to find the values of b and c.Alright, so the inequality bx² + 3x + c ≥ 0 has the solution set between -1/2 and 2. That means that the quadratic equation bx² + 3x + c = 0 has roots at x = -1/2 and x = 2. So, these are the solutions where the quadratic equals zero, and the inequality holds between these roots because the quadratic opens upwards or downwards.Wait, actually, if the solution set is between -1/2 and 2, that means the quadratic must open upwards if the coefficient b is positive, or downwards if b is negative. But since the inequality is ≥ 0, and the solution is between the roots, that suggests that the quadratic opens downward because it's positive between the roots. So, that would mean b is negative.But let me think again. If the quadratic is positive between its roots, it must open downward. So, yes, b is negative.So, using Vieta's formulas, which relate the sum and product of the roots to the coefficients of the quadratic. For a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a.In our case, the quadratic is bx² + 3x + c = 0, so the sum of the roots is -3/b and the product is c/b.Given the roots are -1/2 and 2, their sum is (-1/2) + 2 = 3/2, and their product is (-1/2)*2 = -1.So, according to Vieta's formulas:Sum of roots = 3/2 = -3/bProduct of roots = -1 = c/bSo, from the sum:3/2 = -3/bMultiply both sides by b:(3/2)b = -3Multiply both sides by 2:3b = -6Divide by 3:b = -2Okay, so b is -2.Now, from the product:-1 = c/bWe know b is -2, so:-1 = c/(-2)Multiply both sides by -2:c = 2So, c is 2.Alright, that seems straightforward. So, part (I) is done: b = -2, c = 2.Now, moving on to part (II). It says: Let M = {m | there exists a t in A such that f(t) = m holds true}, find the set M.So, M is the set of all m such that there's some t in A (which is [-1/2, 2]) where f(t) = m. So, essentially, M is the range of the function f(x) over the interval A.Given that f(x) = lg( (2x² + bx + c)/(x² + 1) ). From part (I), we know b = -2 and c = 2, so plugging those in:f(x) = lg( (2x² - 2x + 2)/(x² + 1) )So, f(x) is the logarithm (base 10, I assume) of the rational function (2x² - 2x + 2)/(x² + 1). To find the set M, I need to find all possible values of f(x) as x ranges over A = [-1/2, 2]. So, I need to find the range of f(x) over this interval.First, let me analyze the function inside the logarithm: g(x) = (2x² - 2x + 2)/(x² + 1). Since the logarithm is a continuous function where its argument is positive, and since g(x) is a rational function, I need to ensure that g(x) is positive for all x in A. Looking at the numerator: 2x² - 2x + 2. Let me check its discriminant to see if it ever becomes zero or negative.Discriminant D = (-2)² - 4*2*2 = 4 - 16 = -12 < 0. So, the quadratic in the numerator has no real roots and since the coefficient of x² is positive, it is always positive. Similarly, the denominator x² + 1 is always positive. So, g(x) is always positive, so f(x) is defined for all real x, including over A.Therefore, f(x) is continuous over A, so its range will be an interval. To find this interval, I need to find the minimum and maximum values of g(x) over A, and then take the logarithm of those to get the range of f(x).So, first, let me find the extrema of g(x) over A.To find the extrema, I can take the derivative of g(x) and find its critical points.g(x) = (2x² - 2x + 2)/(x² + 1)Let me compute g'(x):Using the quotient rule: (numerator derivative * denominator - numerator * denominator derivative) / denominator²Numerator: 2x² - 2x + 2, derivative is 4x - 2Denominator: x² + 1, derivative is 2xSo,g'(x) = [ (4x - 2)(x² + 1) - (2x² - 2x + 2)(2x) ] / (x² + 1)²Let me compute the numerator:First term: (4x - 2)(x² + 1) = 4x*(x² + 1) - 2*(x² + 1) = 4x³ + 4x - 2x² - 2Second term: (2x² - 2x + 2)(2x) = 4x³ - 4x² + 4xSo, subtracting the second term from the first term:[4x³ + 4x - 2x² - 2] - [4x³ - 4x² + 4x] = 4x³ + 4x - 2x² - 2 - 4x³ + 4x² - 4xSimplify:4x³ - 4x³ = 04x - 4x = 0-2x² + 4x² = 2x²-2 remainsSo, numerator is 2x² - 2Therefore, g'(x) = (2x² - 2)/(x² + 1)² = 2(x² - 1)/(x² + 1)²So, critical points occur where g'(x) = 0 or undefined. Since denominator is always positive, g'(x) is defined for all x. So, set numerator equal to zero:2(x² - 1) = 0 => x² - 1 = 0 => x = ±1So, critical points at x = 1 and x = -1.Now, we need to check these critical points within the interval A = [-1/2, 2].x = -1 is not in A, since A starts at -1/2. x = 1 is in A.So, the critical point within A is x = 1.Therefore, to find the extrema of g(x) over A, we need to evaluate g(x) at the endpoints x = -1/2 and x = 2, and at the critical point x = 1.Let me compute g(-1/2):g(-1/2) = [2*(-1/2)^2 - 2*(-1/2) + 2] / [(-1/2)^2 + 1] = [2*(1/4) + 1 + 2] / [1/4 + 1] = [ (1/2) + 1 + 2 ] / (5/4) = (3.5) / (1.25) = 2.8Wait, 3.5 divided by 1.25 is 2.8, which is 14/5.So, g(-1/2) = 14/5.g(1):g(1) = [2*(1)^2 - 2*(1) + 2] / [1 + 1] = [2 - 2 + 2]/2 = 2/2 = 1g(2):g(2) = [2*(4) - 2*(2) + 2] / [4 + 1] = [8 - 4 + 2]/5 = 6/5 = 1.2So, g(-1/2) = 14/5 = 2.8, g(1) = 1, g(2) = 6/5 = 1.2.So, over the interval A, g(x) attains its maximum at x = -1/2 with value 14/5 and its minimum at x = 1 with value 1.Therefore, the range of g(x) over A is [1, 14/5].Since f(x) = lg(g(x)), and the logarithm is a monotonically increasing function, the range of f(x) will be [lg(1), lg(14/5)].But lg(1) = 0, so the range of f(x) is [0, lg(14/5)].Therefore, the set M is all real numbers m such that 0 ≤ m ≤ lg(14/5).So, M = [0, lg(14/5)].Let me just double-check my calculations.First, for g(-1/2):2*(1/4) = 0.5, -2*(-1/2) = 1, so numerator is 0.5 + 1 + 2 = 3.5, which is 7/2. Denominator is (1/4) + 1 = 5/4. So, 7/2 divided by 5/4 is (7/2)*(4/5) = 28/10 = 14/5. Correct.g(1): 2 - 2 + 2 = 2, denominator 2, so 2/2 = 1. Correct.g(2): 8 - 4 + 2 = 6, denominator 5, so 6/5. Correct.Derivative: I think I did that correctly. Let me just recheck:g'(x) = [ (4x - 2)(x² + 1) - (2x² - 2x + 2)(2x) ] / (x² + 1)^2Expanding numerator:(4x - 2)(x² + 1) = 4x³ + 4x - 2x² - 2(2x² - 2x + 2)(2x) = 4x³ - 4x² + 4xSubtracting: 4x³ + 4x - 2x² - 2 - 4x³ + 4x² - 4x = (0)x³ + (0)x + (2x²) - 2So, numerator is 2x² - 2. Correct.Therefore, critical points at x = ±1. Since x = -1 is outside A, only x = 1 is considered.Thus, the maximum is at x = -1/2, which is 14/5, and the minimum at x = 1, which is 1.Therefore, f(x) ranges from lg(1) = 0 to lg(14/5).So, M is [0, lg(14/5)].I think that's correct.**Final Answer**(I) The values of ( b ) and ( c ) are ( boxed{-2} ) and ( boxed{2} ) respectively.(II) The set ( M ) is ( boxed{left[0, lg frac{14}{5}right]} ).

💡Okay, so I have this problem about a company's pension plan. It says that the pension each year is directly proportional to the square root of the number of years worked. Hmm, that means if someone works more years, their pension increases, but not linearly—it's based on the square root. Interesting.Let me try to break this down. Let's say the pension amount is y dollars, and the number of years worked is x. Since it's directly proportional, I can write this relationship as y = k√x, where k is the constant of proportionality. That makes sense.Now, the problem states that if an employee works a more years, their pension increases by p dollars. So, if they work a more years, their new pension would be y + p. Using the same formula, that would be y + p = k√(x + a). Similarly, if they work b more years instead, their pension increases by q dollars, so y + q = k√(x + b).Alright, so now I have two equations:1. y + p = k√(x + a)2. y + q = k√(x + b)I need to find y in terms of a, b, p, q. So, I have two equations with two unknowns: y and k. Maybe I can solve for k first and then find y.Let me subtract the original pension equation from both new equations to eliminate y. So, subtracting y = k√x from both sides:From the first equation: p = k√(x + a) - k√xFrom the second equation: q = k√(x + b) - k√xHmm, so p and q are expressed in terms of k and x. Maybe I can set up a system of equations here.Let me denote √x as t to simplify things. So, t = √x, which means x = t².Substituting back into the equations:p = k√(t² + a) - ktq = k√(t² + b) - ktSo, p = k(√(t² + a) - t)q = k(√(t² + b) - t)Now, I can write these as:p = k(√(t² + a) - t) ...(1)q = k(√(t² + b) - t) ...(2)I need to solve for k and t. Maybe I can divide equation (1) by equation (2) to eliminate k.So, p/q = [√(t² + a) - t] / [√(t² + b) - t]Hmm, that's a bit complicated, but let's see. Let me denote u = √(t² + a) and v = √(t² + b). Then, p/q = (u - t)/(v - t). But I'm not sure if that helps directly.Alternatively, maybe I can rationalize the denominators. Let's try that.Starting with equation (1):p = k(√(t² + a) - t)Multiply numerator and denominator by (√(t² + a) + t):p = k[(√(t² + a) - t)(√(t² + a) + t)] / (√(t² + a) + t)The numerator becomes (t² + a) - t² = aSo, p = k * a / (√(t² + a) + t)Similarly, for equation (2):q = k(√(t² + b) - t)Multiply numerator and denominator by (√(t² + b) + t):q = k * b / (√(t² + b) + t)So now, we have:p = (k a) / (√(t² + a) + t) ...(3)q = (k b) / (√(t² + b) + t) ...(4)Let me denote equation (3) as p = (k a)/(u + t) where u = √(t² + a)Similarly, equation (4) is q = (k b)/(v + t) where v = √(t² + b)But maybe instead, I can express k from both equations and set them equal.From equation (3):k = p (√(t² + a) + t)/aFrom equation (4):k = q (√(t² + b) + t)/bSince both equal k, set them equal:p (√(t² + a) + t)/a = q (√(t² + b) + t)/bSo,(p/a)(√(t² + a) + t) = (q/b)(√(t² + b) + t)Let me write this as:(p/a)√(t² + a) + (p/a)t = (q/b)√(t² + b) + (q/b)tBring all terms to one side:(p/a)√(t² + a) - (q/b)√(t² + b) + (p/a - q/b)t = 0This looks complicated. Maybe I can isolate the square roots.Let me denote:Let’s rearrange:(p/a)√(t² + a) - (q/b)√(t² + b) = (q/b - p/a)tHmm, still complicated. Maybe square both sides to eliminate the square roots. But that might get messy.Alternatively, let me denote s = t, so s = √x.Wait, maybe I can express √(t² + a) in terms of s. Wait, s is t, so √(t² + a) is √(s² + a). Maybe not helpful.Alternatively, let me think about expressing t in terms of k.From equation (1):p = k(√(t² + a) - t)Let me solve for √(t² + a):√(t² + a) = t + p/kSimilarly, from equation (2):√(t² + b) = t + q/kNow, square both sides:From equation (1):t² + a = (t + p/k)² = t² + 2tp/k + p²/k²Subtract t²:a = 2tp/k + p²/k²Similarly, from equation (2):t² + b = (t + q/k)² = t² + 2tq/k + q²/k²Subtract t²:b = 2tq/k + q²/k²So now, I have two equations:1. a = (2tp)/k + (p²)/k²2. b = (2tq)/k + (q²)/k²Let me write these as:a = (2tp)/k + (p²)/k² ...(5)b = (2tq)/k + (q²)/k² ...(6)Let me denote u = t/k. Then, t = u k.Substituting into equation (5):a = 2(u k) p / k + (p²)/k² = 2 u p + p² / k²Similarly, equation (6):b = 2(u k) q / k + (q²)/k² = 2 u q + q² / k²So now, we have:a = 2 u p + (p²)/k² ...(7)b = 2 u q + (q²)/k² ...(8)Let me subtract equation (8) from equation (7):a - b = 2 u p - 2 u q + (p² - q²)/k²Factor:a - b = 2 u (p - q) + (p - q)(p + q)/k²Factor out (p - q):a - b = (p - q)(2 u + (p + q)/k²)Hmm, interesting. Let me write this as:(a - b) = (p - q)(2 u + (p + q)/k²)Let me solve for (2 u + (p + q)/k²):2 u + (p + q)/k² = (a - b)/(p - q)But from equation (7):a = 2 u p + (p²)/k²Let me express 2 u from equation (7):2 u = (a - (p²)/k²)/pSimilarly, from equation (8):2 u = (b - (q²)/k²)/qSo, (a - (p²)/k²)/p = (b - (q²)/k²)/qCross-multiplying:q(a - (p²)/k²) = p(b - (q²)/k²)Expand:qa - q p² /k² = pb - p q² /k²Bring all terms to one side:qa - pb = (q p² - p q²)/k²Factor the right side:qa - pb = pq (p - q)/k²So,k² = pq (p - q)/(qa - pb)Wait, let me write that:k² = [pq (p - q)] / (qa - pb)Note that qa - pb is the same as -(pb - qa), so:k² = [pq (p - q)] / ( - (pb - qa)) = [pq (p - q)] / (-1)(pb - qa) = - [pq (p - q)] / (pb - qa)But k² must be positive, so the negative sign must be accounted for. Let me see:Wait, let's double-check the algebra.From:qa - pb = (q p² - p q²)/k²Factor numerator on the right:= pq (p - q)/k²So,qa - pb = pq (p - q)/k²Then,k² = pq (p - q)/(qa - pb)But if I factor out a negative from the denominator:k² = pq (p - q)/( - (pb - qa)) = - pq (p - q)/(pb - qa)But since k² is positive, the numerator and denominator must have the same sign. So, perhaps I made a miscalculation.Wait, let's go back.We had:qa - pb = (q p² - p q²)/k²Factor numerator:= pq (p - q)/k²So,qa - pb = pq (p - q)/k²Therefore,k² = pq (p - q)/(qa - pb)But qa - pb can be written as -(pb - qa), so:k² = pq (p - q)/(- (pb - qa)) = - pq (p - q)/(pb - qa)But k² is positive, so the numerator and denominator must have opposite signs.Alternatively, maybe I should have kept the negative sign:k² = pq (p - q)/(qa - pb) = pq (p - q)/(- (pb - qa)) = - pq (p - q)/(pb - qa)But since k² is positive, the right side must be positive. Therefore, - pq (p - q)/(pb - qa) > 0Which implies that pq (p - q)/(pb - qa) < 0But I'm not sure if this is helpful. Maybe I can proceed without worrying about the sign for now.So, k² = [pq (p - q)] / (qa - pb)Let me write this as:k² = pq (p - q) / (qa - pb)So, k = sqrt[ pq (p - q) / (qa - pb) ]But I need to find y, which is k√x.But x is t², and t is related to u, which is t/k.Wait, this is getting too convoluted. Maybe I should find y directly.Wait, from equation (1):p = k(√(x + a) - √x)Similarly, from equation (2):q = k(√(x + b) - √x)Let me denote √x as t, so x = t².Then,p = k(√(t² + a) - t)q = k(√(t² + b) - t)Let me solve for k from both equations.From equation (1):k = p / (√(t² + a) - t)From equation (2):k = q / (√(t² + b) - t)Set them equal:p / (√(t² + a) - t) = q / (√(t² + b) - t)Cross-multiplying:p (√(t² + b) - t) = q (√(t² + a) - t)Let me expand both sides:p√(t² + b) - p t = q√(t² + a) - q tBring all terms to one side:p√(t² + b) - q√(t² + a) = (p - q) tHmm, this is similar to what I had before. Maybe I can square both sides to eliminate the square roots.So,[p√(t² + b) - q√(t² + a)]² = [(p - q) t]²Expanding the left side:p²(t² + b) + q²(t² + a) - 2 p q √{(t² + b)(t² + a)} = (p - q)² t²Simplify:p² t² + p² b + q² t² + q² a - 2 p q √{(t² + b)(t² + a)} = (p² - 2 p q + q²) t²Bring all terms to one side:p² t² + p² b + q² t² + q² a - 2 p q √{(t² + b)(t² + a)} - p² t² + 2 p q t² - q² t² = 0Simplify:(p² t² - p² t²) + (q² t² - q² t²) + p² b + q² a + 2 p q t² - 2 p q √{(t² + b)(t² + a)} = 0So,p² b + q² a + 2 p q t² - 2 p q √{(t² + b)(t² + a)} = 0Factor out 2 p q:2 p q [ t² - √{(t² + b)(t² + a)} ] + p² b + q² a = 0Let me isolate the square root term:2 p q [ t² - √{(t² + b)(t² + a)} ] = - (p² b + q² a)Divide both sides by 2 p q:t² - √{(t² + b)(t² + a)} = - (p² b + q² a)/(2 p q)Multiply both sides by -1:√{(t² + b)(t² + a)} - t² = (p² b + q² a)/(2 p q)Let me denote the left side as:√{(t² + b)(t² + a)} - t² = (p² b + q² a)/(2 p q)Let me compute the left side:√{(t² + b)(t² + a)} - t²Let me expand (t² + b)(t² + a):= t⁴ + (a + b) t² + a bSo,√(t⁴ + (a + b) t² + a b) - t²Let me denote s = t², so:√(s² + (a + b) s + a b) - sHmm, not sure if that helps. Alternatively, maybe I can write it as:√{(t² + b)(t² + a)} = t² + (p² b + q² a)/(2 p q)Let me square both sides:(t² + b)(t² + a) = [t² + (p² b + q² a)/(2 p q)]²Expand both sides:Left side: t⁴ + (a + b) t² + a bRight side: t⁴ + 2 t² (p² b + q² a)/(2 p q) + [(p² b + q² a)/(2 p q)]²Simplify right side:= t⁴ + (p² b + q² a)/p q * t² + (p² b + q² a)²/(4 p² q²)So, equate left and right:t⁴ + (a + b) t² + a b = t⁴ + [(p² b + q² a)/p q] t² + (p² b + q² a)²/(4 p² q²)Subtract t⁴ from both sides:(a + b) t² + a b = [(p² b + q² a)/p q] t² + (p² b + q² a)²/(4 p² q²)Bring all terms to one side:(a + b) t² + a b - [(p² b + q² a)/p q] t² - (p² b + q² a)²/(4 p² q²) = 0Factor t²:[ (a + b) - (p² b + q² a)/p q ] t² + a b - (p² b + q² a)²/(4 p² q²) = 0Let me compute the coefficient of t²:= (a + b) - (p² b + q² a)/p q= (a + b) - (p² b)/(p q) - (q² a)/(p q)= (a + b) - (p b)/q - (q a)/pSimilarly, the constant term:= a b - (p² b + q² a)²/(4 p² q²)This is getting really complicated. Maybe there's a better approach.Wait, going back to the original equations:y = k√xp = k(√(x + a) - √x)q = k(√(x + b) - √x)Let me denote √x as t, so x = t².Then,p = k(√(t² + a) - t)q = k(√(t² + b) - t)Let me solve for k from both equations:From p: k = p / (√(t² + a) - t)From q: k = q / (√(t² + b) - t)Set equal:p / (√(t² + a) - t) = q / (√(t² + b) - t)Cross-multiplying:p (√(t² + b) - t) = q (√(t² + a) - t)Expand:p√(t² + b) - p t = q√(t² + a) - q tBring like terms together:p√(t² + b) - q√(t² + a) = (p - q) tLet me square both sides:[p√(t² + b) - q√(t² + a)]² = (p - q)² t²Expand left side:p²(t² + b) + q²(t² + a) - 2 p q √{(t² + b)(t² + a)} = (p² - 2 p q + q²) t²Simplify:p² t² + p² b + q² t² + q² a - 2 p q √{(t² + b)(t² + a)} = p² t² - 2 p q t² + q² t²Subtract p² t² + q² t² from both sides:p² b + q² a - 2 p q √{(t² + b)(t² + a)} = -2 p q t²Bring all terms to one side:p² b + q² a + 2 p q t² - 2 p q √{(t² + b)(t² + a)} = 0Factor out 2 p q:2 p q [ t² - √{(t² + b)(t² + a)} ] + p² b + q² a = 0Let me isolate the square root term:2 p q [ t² - √{(t² + b)(t² + a)} ] = - (p² b + q² a)Divide both sides by 2 p q:t² - √{(t² + b)(t² + a)} = - (p² b + q² a)/(2 p q)Multiply both sides by -1:√{(t² + b)(t² + a)} - t² = (p² b + q² a)/(2 p q)Let me denote the left side as:√{(t² + b)(t² + a)} - t² = (p² b + q² a)/(2 p q)Let me compute the left side:√{(t² + b)(t² + a)} - t²Let me expand (t² + b)(t² + a):= t⁴ + (a + b) t² + a bSo,√(t⁴ + (a + b) t² + a b) - t²Let me denote s = t², so:√(s² + (a + b) s + a b) - sHmm, not sure if that helps. Alternatively, maybe I can write it as:√{(t² + b)(t² + a)} = t² + (p² b + q² a)/(2 p q)Let me square both sides:(t² + b)(t² + a) = [t² + (p² b + q² a)/(2 p q)]²Expand both sides:Left side: t⁴ + (a + b) t² + a bRight side: t⁴ + 2 t² (p² b + q² a)/(2 p q) + [(p² b + q² a)/(2 p q)]²Simplify right side:= t⁴ + (p² b + q² a)/p q * t² + (p² b + q² a)²/(4 p² q²)So, equate left and right:t⁴ + (a + b) t² + a b = t⁴ + [(p² b + q² a)/p q] t² + (p² b + q² a)²/(4 p² q²)Subtract t⁴ from both sides:(a + b) t² + a b = [(p² b + q² a)/p q] t² + (p² b + q² a)²/(4 p² q²)Bring all terms to one side:(a + b) t² + a b - [(p² b + q² a)/p q] t² - (p² b + q² a)²/(4 p² q²) = 0Factor t²:[ (a + b) - (p² b + q² a)/p q ] t² + a b - (p² b + q² a)²/(4 p² q²) = 0Let me compute the coefficient of t²:= (a + b) - (p² b + q² a)/p q= (a + b) - (p² b)/(p q) - (q² a)/(p q)= (a + b) - (p b)/q - (q a)/pSimilarly, the constant term:= a b - (p² b + q² a)²/(4 p² q²)This is getting really complicated. Maybe I can find a relationship between t and k.Wait, from equation (1):p = k(√(t² + a) - t)Let me solve for √(t² + a):√(t² + a) = t + p/kSimilarly, from equation (2):√(t² + b) = t + q/kNow, square both sides:From equation (1):t² + a = t² + 2 t p/k + p²/k²Subtract t²:a = 2 t p/k + p²/k²Similarly, from equation (2):t² + b = t² + 2 t q/k + q²/k²Subtract t²:b = 2 t q/k + q²/k²So now, I have:a = (2 t p)/k + (p²)/k² ...(5)b = (2 t q)/k + (q²)/k² ...(6)Let me denote u = t/k. Then, t = u k.Substituting into equation (5):a = 2 (u k) p / k + (p²)/k² = 2 u p + p² / k²Similarly, equation (6):b = 2 (u k) q / k + (q²)/k² = 2 u q + q² / k²So now, we have:a = 2 u p + (p²)/k² ...(7)b = 2 u q + (q²)/k² ...(8)Let me subtract equation (8) from equation (7):a - b = 2 u p - 2 u q + (p² - q²)/k²Factor:a - b = 2 u (p - q) + (p - q)(p + q)/k²Factor out (p - q):a - b = (p - q)(2 u + (p + q)/k²)Let me solve for (2 u + (p + q)/k²):2 u + (p + q)/k² = (a - b)/(p - q)But from equation (7):a = 2 u p + (p²)/k²Let me express 2 u from equation (7):2 u = (a - (p²)/k²)/pSimilarly, from equation (8):2 u = (b - (q²)/k²)/qSo, (a - (p²)/k²)/p = (b - (q²)/k²)/qCross-multiplying:q(a - (p²)/k²) = p(b - (q²)/k²)Expand:qa - q p² /k² = pb - p q² /k²Bring all terms to one side:qa - pb = (q p² - p q²)/k²Factor the right side:= pq (p - q)/k²So,qa - pb = pq (p - q)/k²Therefore,k² = pq (p - q)/(qa - pb)But k² must be positive, so the denominator and numerator must have the same sign.Now, let me find y = k√x = k tFrom equation (7):a = 2 u p + (p²)/k²But u = t/k, so:a = 2 (t/k) p + (p²)/k²Multiply both sides by k²:a k² = 2 p t k + p²Similarly, from equation (8):b k² = 2 q t k + q²But t = y/k, since y = k t.So, t = y/kSubstitute into equation (7):a k² = 2 p (y/k) k + p² = 2 p y + p²Similarly, equation (8):b k² = 2 q y + q²So now, we have:a k² = 2 p y + p² ...(9)b k² = 2 q y + q² ...(10)Let me solve equation (9) for y:2 p y = a k² - p²So,y = (a k² - p²)/(2 p)Similarly, from equation (10):2 q y = b k² - q²So,y = (b k² - q²)/(2 q)Since both equal y, set them equal:(a k² - p²)/(2 p) = (b k² - q²)/(2 q)Multiply both sides by 2:(a k² - p²)/p = (b k² - q²)/qCross-multiplying:q(a k² - p²) = p(b k² - q²)Expand:a q k² - q p² = b p k² - p q²Bring all terms to one side:a q k² - b p k² = q p² - p q²Factor:k²(a q - b p) = p q (p - q)So,k² = [p q (p - q)] / (a q - b p)But from earlier, we had:k² = pq (p - q)/(qa - pb)Which is the same as:k² = [p q (p - q)] / (a q - b p)So, consistent.Now, let me substitute k² back into the expression for y.From equation (9):y = (a k² - p²)/(2 p)Substitute k²:y = [a * (p q (p - q)/(a q - b p)) - p²]/(2 p)Simplify numerator:= [ (a p q (p - q))/(a q - b p) - p² ] / (2 p)Factor p in numerator:= p [ (a q (p - q))/(a q - b p) - p ] / (2 p)Cancel p:= [ (a q (p - q))/(a q - b p) - p ] / 2Combine terms:= [ (a q (p - q) - p (a q - b p)) / (a q - b p) ] / 2Expand numerator:= [ a q p - a q² - a q p + b p² ] / (a q - b p) / 2Simplify:= [ - a q² + b p² ] / (a q - b p) / 2Factor numerator:= [ b p² - a q² ] / (a q - b p) / 2Note that (a q - b p) = - (b p - a q), so:= [ b p² - a q² ] / (- (b p - a q)) / 2= - [ b p² - a q² ] / (b p - a q) / 2= [ a q² - b p² ] / (b p - a q) / 2So,y = (a q² - b p²)/(2 (b p - a q))Which can be written as:y = (a q² - b p²)/(2 (b p - a q))Alternatively, factoring out a negative from numerator and denominator:= ( - (b p² - a q²) ) / ( - (a q - b p) ) / 2= (b p² - a q²)/(a q - b p)/2But the original expression is fine.So, the annual pension y is:y = (a q² - b p²)/(2 (b p - a q))Which can also be written as:y = (a q² - b p²)/(2 (b p - a q))Alternatively, factoring out the negative sign:y = (a q² - b p²)/(2 (b p - a q)) = (b p² - a q²)/(2 (a q - b p))But both are equivalent.So, the final answer is:y = (a q² - b p²)/(2 (b p - a q))Which can be boxed as:boxed{dfrac{a q^2 - b p^2}{2(bp - aq)}}

💡Okay, so I have this problem where I'm given two equations involving three variables x, y, and z. The equations are:1. ((x + y + z)(xy + xz + yz) = 36)2. (x^2(y + z) + y^2(x + z) + z^2(x + y) = 10)And I need to find the value of (xyz). Hmm, okay. Let me try to figure this out step by step.First, I remember that when dealing with symmetric equations like these, sometimes expanding them can help relate them to each other. Let me start by expanding the first equation.So, expanding ((x + y + z)(xy + xz + yz)). Let me distribute each term in the first parenthesis to each term in the second parenthesis.That would be:- (x cdot xy = x^2y)- (x cdot xz = x^2z)- (x cdot yz = xyz)- (y cdot xy = xy^2)- (y cdot xz = xyz)- (y cdot yz = y^2z)- (z cdot xy = xyz)- (z cdot xz = xz^2)- (z cdot yz = yz^2)So, putting it all together, the expansion is:(x^2y + x^2z + xyz + xy^2 + xyz + y^2z + xyz + xz^2 + yz^2)Now, let's combine like terms. I notice that there are three terms of (xyz), so that's (3xyz). The other terms are all the quadratic terms like (x^2y), (x^2z), etc. So, the expanded form is:(3xyz + x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2)Okay, so the first equation becomes:(3xyz + x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2 = 36)Now, looking at the second equation:(x^2(y + z) + y^2(x + z) + z^2(x + y) = 10)Let me expand this as well. Distribute each squared term:- (x^2 cdot y = x^2y)- (x^2 cdot z = x^2z)- (y^2 cdot x = xy^2)- (y^2 cdot z = y^2z)- (z^2 cdot x = xz^2)- (z^2 cdot y = yz^2)So, the expansion is:(x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2 = 10)Wait a minute, that's interesting. The expanded form of the second equation is exactly the same as the non-xyz terms in the first equation. So, in the first equation, we have (3xyz + [x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2] = 36). And from the second equation, we know that ([x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2] = 10).So, substituting the second equation into the first, we get:(3xyz + 10 = 36)Now, solving for (xyz), subtract 10 from both sides:(3xyz = 36 - 10)(3xyz = 26)Then, divide both sides by 3:(xyz = frac{26}{3})So, the value of (xyz) is (frac{26}{3}).Wait, let me double-check my steps to make sure I didn't make a mistake. I expanded both equations correctly, right? The first equation expanded to (3xyz) plus the sum of all those quadratic terms, and the second equation was exactly that sum of quadratic terms. So substituting 10 into the first equation gives me (3xyz + 10 = 36), which leads to (xyz = frac{26}{3}). Yeah, that seems right.I don't think I made any calculation errors. The logic flows correctly, and the substitution makes sense. So, I'm confident that (xyz = frac{26}{3}) is the correct answer.**Final Answer**The value of (xyz) is boxed{dfrac{26}{3}}.