💡Okay, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. The problem says:"In the arithmetic sequence {a_n}, it is known that a_8 ≥ 15 and a_9 ≤ 13. The range of values for a_{12} is ______."Alright, so first, let me recall what an arithmetic sequence is. An arithmetic sequence is a sequence of numbers where each term after the first is obtained by adding a constant difference. So, if the first term is a_1 and the common difference is d, then the nth term is given by:a_n = a_1 + (n - 1)dGot that. So, for example, a_8 would be a_1 + 7d, and a_9 would be a_1 + 8d. The problem gives me two inequalities involving a_8 and a_9, and I need to find the range of possible values for a_{12}.Let me write down the given inequalities:1. a_8 ≥ 152. a_9 ≤ 13Since a_8 and a_9 are consecutive terms in the arithmetic sequence, I can express them in terms of a_1 and d:a_8 = a_1 + 7d ≥ 15a_9 = a_1 + 8d ≤ 13Hmm, so I have two inequalities here. Maybe I can subtract the first inequality from the second to find something about d?Let me try that. If I subtract the first inequality from the second, I get:(a_1 + 8d) - (a_1 + 7d) ≤ 13 - 15Simplifying the left side:a_1 + 8d - a_1 - 7d = dAnd the right side:13 - 15 = -2So, d ≤ -2Okay, so the common difference d is less than or equal to -2. That means the sequence is decreasing, right? Because the difference is negative.Now, I need to find the range of a_{12}. Let me write the expression for a_{12}:a_{12} = a_1 + 11dHmm, so I need to find the possible values of a_1 + 11d given the constraints on a_8 and a_9.I have two inequalities:1. a_1 + 7d ≥ 152. a_1 + 8d ≤ 13I can try to express a_1 from the first inequality and substitute it into the second inequality to find more information about d.From the first inequality:a_1 + 7d ≥ 15=> a_1 ≥ 15 - 7dFrom the second inequality:a_1 + 8d ≤ 13=> a_1 ≤ 13 - 8dSo, combining these two results:15 - 7d ≤ a_1 ≤ 13 - 8dHmm, so 15 - 7d ≤ 13 - 8dLet me solve this inequality for d.15 - 7d ≤ 13 - 8dSubtract 15 from both sides:-7d ≤ -2 - 8dAdd 8d to both sides:d ≤ -2Wait, that's the same result as before. So, d ≤ -2.So, the common difference is at most -2.Now, going back to a_{12} = a_1 + 11d.I need to express a_{12} in terms of the known inequalities.Let me see if I can write a_{12} using a_8 or a_9.Since a_8 = a_1 + 7d, and a_9 = a_1 + 8d, then a_{12} is a_1 + 11d.I can think of a_{12} as a_8 + 3d or a_9 + 2d.Let me try that.a_{12} = a_8 + 3dBut I know that a_8 ≥ 15, so:a_{12} = a_8 + 3d ≥ 15 + 3dSimilarly, a_{12} = a_9 + 2dAnd since a_9 ≤ 13,a_{12} = a_9 + 2d ≤ 13 + 2dSo, combining these two:15 + 3d ≤ a_{12} ≤ 13 + 2dHmm, but I also know that d ≤ -2.So, let me substitute d ≤ -2 into these inequalities.First, let's look at the lower bound: 15 + 3d.Since d is ≤ -2, the smallest value 3d can take is when d is as small as possible. But wait, d is just bounded above by -2, so d can be any number less than or equal to -2, but it can be more negative. So, 3d can be as small as negative infinity, but we have a lower bound on a_{12}.Wait, no, actually, since a_{12} is bounded below by 15 + 3d, and d is ≤ -2, 15 + 3d can be as small as possible. But we need to find the range of a_{12}, so perhaps we can find the maximum and minimum possible values.Wait, maybe I should express a_{12} in terms of a_8 and a_9.Alternatively, perhaps I can express a_{12} as a linear combination of a_8 and a_9.Let me think about that.Let me denote:a_8 = a_1 + 7da_9 = a_1 + 8dI need to find a_{12} = a_1 + 11d.Let me see if I can express a_{12} as a combination of a_8 and a_9.Let me set up equations:Let me suppose that a_{12} = x * a_8 + y * a_9So,a_1 + 11d = x(a_1 + 7d) + y(a_1 + 8d)Simplify the right side:= x a_1 + 7x d + y a_1 + 8y d= (x + y) a_1 + (7x + 8y) dSet this equal to a_1 + 11d:(x + y) a_1 + (7x + 8y) d = a_1 + 11dTherefore, we have the system of equations:x + y = 17x + 8y = 11Let me solve this system.From the first equation: x = 1 - ySubstitute into the second equation:7(1 - y) + 8y = 117 - 7y + 8y = 117 + y = 11y = 11 - 7 = 4Then, x = 1 - y = 1 - 4 = -3So, a_{12} = -3 a_8 + 4 a_9Interesting. So, a_{12} can be expressed as -3 times a_8 plus 4 times a_9.Given that, and knowing that a_8 ≥ 15 and a_9 ≤ 13, let's plug these into the expression for a_{12}.So,a_{12} = -3 a_8 + 4 a_9Since a_8 ≥ 15, -3 a_8 ≤ -3 * 15 = -45Similarly, since a_9 ≤ 13, 4 a_9 ≤ 4 * 13 = 52Therefore, adding these two inequalities:-3 a_8 + 4 a_9 ≤ -45 + 52 = 7So, a_{12} ≤ 7But is there a lower bound?Wait, let me think. Since a_8 can be as large as possible (since it's only bounded below by 15), and a_9 can be as small as possible (since it's only bounded above by 13), but in reality, a_8 and a_9 are related through the common difference d.Wait, perhaps I need to consider the relationship between a_8 and a_9.We know that a_9 = a_8 + dSo, d = a_9 - a_8From the inequalities:a_8 ≥ 15a_9 ≤ 13Therefore, d = a_9 - a_8 ≤ 13 - 15 = -2Which is consistent with what we found earlier, that d ≤ -2.So, d is at most -2.Now, going back to a_{12} = a_1 + 11dBut since a_1 = a_8 - 7d, from a_8 = a_1 + 7dSo, a_{12} = (a_8 - 7d) + 11d = a_8 + 4dSimilarly, since a_9 = a_1 + 8d, so a_1 = a_9 - 8dThus, a_{12} = a_9 - 8d + 11d = a_9 + 3dSo, a_{12} can be expressed as a_8 + 4d or a_9 + 3d.Given that, and knowing that d ≤ -2, let's see.From a_{12} = a_8 + 4dSince a_8 ≥ 15 and d ≤ -2,a_{12} = a_8 + 4d ≥ 15 + 4*(-2) = 15 - 8 = 7Wait, that's interesting. So, a_{12} is greater than or equal to 7?But earlier, I found that a_{12} ≤ 7.Wait, that seems contradictory. Let me check my calculations.Wait, if a_{12} = a_8 + 4d, and a_8 ≥ 15, d ≤ -2,Then, a_{12} = a_8 + 4d ≥ 15 + 4*(-2) = 15 - 8 = 7But also, from the other expression, a_{12} = -3a_8 + 4a_9 ≤ 7So, combining these two results:a_{12} ≥ 7 and a_{12} ≤ 7Therefore, a_{12} must equal 7.Wait, that can't be right because the problem says "the range of values for a_{12}", implying it's not a single value.Hmm, perhaps I made a mistake in my reasoning.Let me go back.I have:a_{12} = a_8 + 4dSince a_8 ≥ 15 and d ≤ -2,So, a_{12} ≥ 15 + 4*(-2) = 15 - 8 = 7But also, a_{12} = a_9 + 3dSince a_9 ≤ 13 and d ≤ -2,Then, a_{12} = a_9 + 3d ≤ 13 + 3*(-2) = 13 - 6 = 7So, from both expressions, a_{12} is both ≥7 and ≤7, so a_{12} must equal 7.But that seems too restrictive. Let me check with specific values.Suppose d = -2, which is the maximum possible value for d.Then, from a_8 = a_1 + 7d ≥15,a_1 + 7*(-2) ≥15a_1 -14 ≥15a_1 ≥29Similarly, a_9 = a_1 +8d ≤13a_1 +8*(-2) ≤13a_1 -16 ≤13a_1 ≤29So, a_1 must be exactly 29.Therefore, a_{12} = a_1 +11d =29 +11*(-2)=29 -22=7So, in this case, a_{12}=7.Now, what if d is less than -2, say d=-3.Then, a_8 = a_1 +7*(-3)=a_1 -21 ≥15 => a_1 ≥36a_9 =a_1 +8*(-3)=a_1 -24 ≤13 => a_1 ≤37So, a_1 is between 36 and 37.Therefore, a_{12}=a_1 +11*(-3)=a_1 -33Since a_1 is between 36 and 37, a_{12} is between 36-33=3 and 37-33=4.So, a_{12} can be between 3 and 4.Wait, that's different from the previous result.So, when d=-2, a_{12}=7When d=-3, a_{12} is between 3 and 4So, the value of a_{12} can vary depending on d.Wait, so my earlier conclusion that a_{12}=7 was incorrect because I didn't consider that a_8 and a_9 have their own constraints which affect a_1 and d.So, perhaps I need to find the maximum and minimum possible values of a_{12} given the constraints.Let me try a different approach.We have:a_8 = a_1 +7d ≥15a_9 = a_1 +8d ≤13We can write these as:1. a_1 +7d ≥152. a_1 +8d ≤13Let me subtract the first inequality from the second:(a_1 +8d) - (a_1 +7d) ≤13 -15Which simplifies to:d ≤-2So, d is at most -2.Now, let's express a_{12} in terms of a_8 and d.a_{12} = a_8 +4dWe know that a_8 ≥15 and d ≤-2.So, to find the minimum value of a_{12}, we need to consider the smallest possible a_8 and the largest possible d (since d is negative, the largest d is -2).Wait, no. To minimize a_{12}, since d is negative, increasing d (making it less negative) would decrease a_{12}.Wait, no, let's think carefully.a_{12} = a_8 +4dGiven that a_8 is at least 15, and d is at most -2.To find the minimum value of a_{12}, we need to minimize a_8 +4d.Since a_8 is as small as possible (15) and d is as small as possible (which is negative infinity, but in reality, d is bounded by the other inequality).Wait, but d is bounded by both a_8 and a_9.Wait, perhaps I need to express a_{12} in terms of a_8 and a_9.Earlier, I found that a_{12} = -3a_8 +4a_9.Given that, and knowing that a_8 ≥15 and a_9 ≤13,So, substituting the maximum of a_8 and the minimum of a_9 would give the minimum of a_{12}.Wait, but a_8 and a_9 are related.Because a_9 = a_8 +d, and d ≤-2.So, a_9 = a_8 +d ≤ a_8 -2Given that a_8 ≥15,So, a_9 ≤15 -2=13, which is consistent with the given a_9 ≤13.So, a_9 is at most 13, and a_8 is at least 15.But a_9 can be as low as possible, depending on d.Wait, but a_9 is also related to a_1.Wait, perhaps I need to find the relationship between a_8 and a_9.Since a_9 = a_8 +d, and d ≤-2,So, a_9 ≤ a_8 -2But a_8 ≥15,So, a_9 ≤15 -2=13, which is given.So, a_9 can be as low as possible, but in reality, a_9 is also related to a_1.Wait, maybe I need to express a_1 in terms of a_8 and d.From a_8 = a_1 +7d,So, a_1 =a_8 -7dSimilarly, from a_9 =a_1 +8d,So, a_9 = a_8 -7d +8d =a_8 +dWhich is consistent.So, a_9 =a_8 +dGiven that, and a_9 ≤13,So, a_8 +d ≤13But a_8 ≥15,So, 15 +d ≤13Which implies d ≤-2, which we already know.So, to find the range of a_{12}, let's express it in terms of a_8 and d.a_{12}=a_8 +4dWe know that a_8 ≥15 and d ≤-2.But also, from a_9 =a_8 +d ≤13,So, d ≤13 -a_8But since a_8 ≥15,d ≤13 -15= -2Which is consistent.So, to find the range of a_{12}=a_8 +4d,We can express it as:a_{12}=a_8 +4dBut since a_9 =a_8 +d ≤13,So, d ≤13 -a_8Therefore, substituting into a_{12}:a_{12}=a_8 +4d ≤a_8 +4*(13 -a_8)=a_8 +52 -4a_8= -3a_8 +52But a_8 ≥15,So, -3a_8 +52 ≤-3*15 +52= -45 +52=7So, a_{12} ≤7Similarly, to find the lower bound, we need to find the minimum value of a_{12}=a_8 +4d.But since d can be as small as possible (more negative), a_{12} can be as small as possible.Wait, but we have constraints on a_8 and a_9.Wait, let's see.From a_8 =a_1 +7d ≥15,And a_9 =a_1 +8d ≤13,We can express a_1 from the first inequality:a_1 ≥15 -7dFrom the second inequality:a_1 ≤13 -8dSo, combining these:15 -7d ≤a_1 ≤13 -8dWhich implies:15 -7d ≤13 -8dSolving for d:15 -7d ≤13 -8dAdd 8d to both sides:15 +d ≤13Subtract 15:d ≤-2Which we already know.So, now, a_{12}=a_1 +11dBut a_1 is between 15 -7d and13 -8d.So, substituting the lower bound of a_1:a_{12} ≥(15 -7d) +11d=15 +4dAnd substituting the upper bound of a_1:a_{12} ≤(13 -8d) +11d=13 +3dSo, 15 +4d ≤a_{12} ≤13 +3dNow, we need to find the range of a_{12} given that d ≤-2.So, let's analyze the lower bound:15 +4dSince d ≤-2,15 +4d ≤15 +4*(-2)=15 -8=7Wait, but that's the upper bound.Wait, no, 15 +4d is the lower bound for a_{12}.But since d is negative, 4d is negative, so 15 +4d is less than 15.But we also have an upper bound for a_{12} as 13 +3d.Since d ≤-2,13 +3d ≤13 +3*(-2)=13 -6=7So, both the lower and upper bounds for a_{12} are ≤7.But that can't be right because a_{12} is between 15 +4d and13 +3d, and both of these are ≤7.Wait, but when d is more negative, 15 +4d becomes smaller, and 13 +3d also becomes smaller.So, the range of a_{12} is from negative infinity up to 7.But that can't be, because a_{12} is a term in the arithmetic sequence, which is linear.Wait, perhaps I made a mistake in interpreting the bounds.Wait, let's think about it differently.We have:15 +4d ≤a_{12} ≤13 +3dBut since d ≤-2,Let me consider d as a variable ≤-2.Let me define d = -k, where k ≥2.So, d = -k, k ≥2.Then, substituting into the inequalities:15 +4*(-k) ≤a_{12} ≤13 +3*(-k)Which simplifies to:15 -4k ≤a_{12} ≤13 -3kNow, since k ≥2,Let me see how these bounds behave as k increases.When k=2,15 -8=7 ≤a_{12} ≤13 -6=7So, a_{12}=7When k=3,15 -12=3 ≤a_{12} ≤13 -9=4So, a_{12} is between 3 and4When k=4,15 -16=-1 ≤a_{12} ≤13 -12=1So, a_{12} is between -1 and1When k=5,15 -20=-5 ≤a_{12} ≤13 -15=-2Wait, that can't be, because the lower bound is -5 and the upper bound is -2, which is a valid range.Wait, but when k increases, the lower bound becomes more negative, and the upper bound also becomes more negative, but the upper bound is always greater than the lower bound.Wait, but in the case when k=5,a_{12} is between -5 and -2But earlier, when k=3, a_{12} is between 3 and4When k=2, a_{12}=7So, as k increases from 2 onwards, the range of a_{12} shifts from 7 down to negative infinity.But that can't be, because the problem is asking for the range of a_{12}, which should be bounded.Wait, perhaps I need to consider that a_1 must satisfy both inequalities:15 -7d ≤a_1 ≤13 -8dSo, for a_1 to exist, the lower bound must be ≤ upper bound:15 -7d ≤13 -8dWhich simplifies to:15 -7d ≤13 -8dAdd 8d to both sides:15 +d ≤13So, d ≤-2Which is consistent.But also, a_1 must be a real number, so the lower bound must be ≤ upper bound.So, as long as d ≤-2, a_1 exists.Therefore, a_{12} can take any value from negative infinity up to 7.But that seems counterintuitive because in the arithmetic sequence, the terms are linear, so they can decrease indefinitely if d is negative.But the problem is asking for the range of a_{12}, so it's possible that a_{12} can be any real number less than or equal to7.But wait, in the earlier example when d=-2, a_{12}=7When d=-3, a_{12} is between3 and4When d=-4, a_{12} is between-1 and1When d=-5, a_{12} is between-5 and-2So, as d becomes more negative, a_{12} decreases without bound.Therefore, the range of a_{12} is all real numbers less than or equal to7.But the problem is asking for the range of values for a_{12}, so it's (-∞,7]But in the initial problem, it's given as a fill-in-the-blank, so maybe it's expressed as a_{12} ≤7But let me check the initial problem again.It says:"In the arithmetic sequence {a_n}, it is known that a_8 ≥15 and a_9 ≤13. The range of values for a_{12} is ______."So, the answer is a_{12} ≤7But earlier, when I considered d=-2, a_{12}=7And for d < -2, a_{12} <7So, the maximum value of a_{12} is7, and it can be any number less than7.Therefore, the range is (-∞,7]But in the initial problem, the user wrote:"the range of values for a_{12} is ______."So, perhaps the answer is a_{12} ≤7But in the initial assistant answer, it was written as boxed{leq 7}But in the thought process, I realized that a_{12} can be as small as possible, so the range is all real numbers less than or equal to7.But perhaps the problem expects a specific interval, but since it's an arithmetic sequence, and d can be any real number ≤-2, a_{12} can be any real number ≤7.So, the range is (-∞,7]But in the initial answer, it was written as boxed{leq 7}Which is correct, but perhaps more precise to write the interval.But since the problem didn't specify whether to write it as an inequality or an interval, and in the initial answer, it was written as boxed{leq 7}, which is acceptable.But in the thought process, I realized that a_{12} can be any real number less than or equal to7, so the range is all real numbers less than or equal to7.Therefore, the range is (-∞,7]But since the problem is in Chinese, and the initial answer was boxed{leq 7}, perhaps that's acceptable.Alternatively, if the problem expects a specific interval, but since it's not bounded below, it's better to express it as a_{12} ≤7.So, the range of values for a_{12} is all real numbers less than or equal to7.

💡Okay, so I have this problem where I need to find the sum of the squares of the roots of the polynomial ( x^4 + 6x^3 + 12x^2 + 9x + 2 ). Hmm, that sounds a bit complicated, but I think I can break it down step by step.First, I remember that for any polynomial, the sum and product of its roots can be found using Vieta's formulas. Since this is a quartic polynomial (degree 4), there are four roots, let's call them ( r_1, r_2, r_3, ) and ( r_4 ). Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots.The general form of a quartic polynomial is ( x^4 + a x^3 + b x^2 + c x + d ). Comparing this to the given polynomial, I can identify the coefficients:- ( a = 6 )- ( b = 12 )- ( c = 9 )- ( d = 2 )From Vieta's formulas, I know that:- The sum of the roots ( r_1 + r_2 + r_3 + r_4 = -a = -6 )- The sum of the products of the roots taken two at a time ( r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4 = b = 12 )- The sum of the products of the roots taken three at a time ( r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4 = -c = -9 )- The product of the roots ( r_1r_2r_3r_4 = d = 2 )But I need the sum of the squares of the roots, which is ( r_1^2 + r_2^2 + r_3^2 + r_4^2 ). I recall that there's a formula that relates this to the sum of the roots and the sum of the products of the roots taken two at a time. Specifically, the formula is:[(r_1 + r_2 + r_3 + r_4)^2 = r_1^2 + r_2^2 + r_3^2 + r_4^2 + 2(r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4)]So, if I rearrange this equation to solve for the sum of the squares, I get:[r_1^2 + r_2^2 + r_3^2 + r_4^2 = (r_1 + r_2 + r_3 + r_4)^2 - 2(r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4)]Plugging in the values from Vieta's formulas:[r_1^2 + r_2^2 + r_3^2 + r_4^2 = (-6)^2 - 2(12) = 36 - 24 = 12]Wait, that seems straightforward. But just to make sure I didn't make a mistake, let me double-check the steps.1. I identified the coefficients correctly: ( a = 6 ), ( b = 12 ), etc.2. Applied Vieta's formulas correctly to find the sum of roots and sum of products two at a time.3. Used the identity for the square of the sum to express the sum of squares in terms of the known sums.Everything seems to add up. So, the sum of the squares of the roots should be 12.But, just to be thorough, maybe I can factor the polynomial and find the roots explicitly, then square them and add. Let's see if that's feasible.Looking at the polynomial ( x^4 + 6x^3 + 12x^2 + 9x + 2 ), I wonder if it can be factored into simpler polynomials. Maybe it's a product of quadratics or even linear factors.Let me try to factor it. I'll look for rational roots using the Rational Root Theorem, which states that any possible rational root, expressed as a fraction ( frac{p}{q} ), has ( p ) as a factor of the constant term and ( q ) as a factor of the leading coefficient. Here, the constant term is 2 and the leading coefficient is 1, so possible rational roots are ( pm1, pm2 ).Let me test ( x = -1 ):[(-1)^4 + 6(-1)^3 + 12(-1)^2 + 9(-1) + 2 = 1 - 6 + 12 - 9 + 2 = 0]Hey, ( x = -1 ) is a root! So, ( (x + 1) ) is a factor.Now, let's perform polynomial division or use synthetic division to factor out ( (x + 1) ) from the polynomial.Using synthetic division with ( x = -1 ):- Coefficients: 1 (x^4), 6 (x^3), 12 (x^2), 9 (x), 2 (constant)- Bring down the 1.- Multiply by -1: 1 * (-1) = -1. Add to next coefficient: 6 + (-1) = 5.- Multiply by -1: 5 * (-1) = -5. Add to next coefficient: 12 + (-5) = 7.- Multiply by -1: 7 * (-1) = -7. Add to next coefficient: 9 + (-7) = 2.- Multiply by -1: 2 * (-1) = -2. Add to last coefficient: 2 + (-2) = 0.So, the result of the division is ( x^3 + 5x^2 + 7x + 2 ).Now, let's factor this cubic polynomial ( x^3 + 5x^2 + 7x + 2 ). Again, using the Rational Root Theorem, possible roots are ( pm1, pm2 ).Testing ( x = -1 ):[(-1)^3 + 5(-1)^2 + 7(-1) + 2 = -1 + 5 - 7 + 2 = -1]Not zero. Testing ( x = -2 ):[(-2)^3 + 5(-2)^2 + 7(-2) + 2 = -8 + 20 - 14 + 2 = 0]Great, ( x = -2 ) is a root. So, ( (x + 2) ) is a factor.Performing synthetic division on ( x^3 + 5x^2 + 7x + 2 ) with ( x = -2 ):- Coefficients: 1 (x^3), 5 (x^2), 7 (x), 2 (constant)- Bring down the 1.- Multiply by -2: 1 * (-2) = -2. Add to next coefficient: 5 + (-2) = 3.- Multiply by -2: 3 * (-2) = -6. Add to next coefficient: 7 + (-6) = 1.- Multiply by -2: 1 * (-2) = -2. Add to last coefficient: 2 + (-2) = 0.So, the result is ( x^2 + 3x + 1 ).Now, the original polynomial factors as:[(x + 1)(x + 2)(x^2 + 3x + 1)]The quadratic ( x^2 + 3x + 1 ) doesn't factor nicely with rational roots, so I'll need to find its roots using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, ( a = 1 ), ( b = 3 ), ( c = 1 ):[x = frac{-3 pm sqrt{9 - 4}}{2} = frac{-3 pm sqrt{5}}{2}]So, the four roots of the polynomial are:- ( x = -1 )- ( x = -2 )- ( x = frac{-3 + sqrt{5}}{2} )- ( x = frac{-3 - sqrt{5}}{2} )Now, let's compute the squares of each root and add them up.1. ( (-1)^2 = 1 )2. ( (-2)^2 = 4 )3. ( left( frac{-3 + sqrt{5}}{2} right)^2 = frac{9 - 6sqrt{5} + 5}{4} = frac{14 - 6sqrt{5}}{4} = frac{7 - 3sqrt{5}}{2} )4. ( left( frac{-3 - sqrt{5}}{2} right)^2 = frac{9 + 6sqrt{5} + 5}{4} = frac{14 + 6sqrt{5}}{4} = frac{7 + 3sqrt{5}}{2} )Adding these up:[1 + 4 + frac{7 - 3sqrt{5}}{2} + frac{7 + 3sqrt{5}}{2}]Combine the fractions:[1 + 4 + frac{7 - 3sqrt{5} + 7 + 3sqrt{5}}{2} = 1 + 4 + frac{14}{2} = 1 + 4 + 7 = 12]So, the sum of the squares of the roots is indeed 12. That matches the result I got earlier using Vieta's formulas. It's reassuring to see that both methods give the same answer.I think I've covered all the steps thoroughly. I first used Vieta's formulas to find the sum of the squares directly, then I factored the polynomial to find the roots explicitly and squared them to confirm the result. Both approaches led me to the same conclusion, so I'm confident that the value of ( A ) is 12.**Final Answer**The value of ( A ) is boxed{12}.

💡Okay, so I have this equation to solve:[frac{1}{x - 1} + frac{2}{x - 3} + frac{3}{x - 5} + dots + frac{50}{x - 99} = x.]Wow, that's a lot of terms! It goes all the way up to 50 in the numerator and 99 in the denominator. Hmm, I need to find the number of real solutions to this equation. First, let me try to understand the structure of the equation. On the left side, there are 50 fractions, each with a numerator that increases by 1 and a denominator that increases by 2 each time. So, the denominators are all odd numbers starting from 1 up to 99. That makes sense because 1 + 2*(50-1) = 99. So, each term is of the form (frac{k}{x - (2k - 1)}) where (k) ranges from 1 to 50.So, the function on the left side is:[f(x) = sum_{k=1}^{50} frac{k}{x - (2k - 1)}.]And we need to find when this equals (x), so we're looking for the solutions to (f(x) = x).Let me think about the behavior of (f(x)). Each term (frac{k}{x - (2k - 1)}) is a hyperbola with a vertical asymptote at (x = 2k - 1). So, the function (f(x)) has vertical asymptotes at all the odd integers from 1 to 99. That's 50 vertical asymptotes in total.Between each pair of consecutive asymptotes, the function (f(x)) will go from negative infinity to positive infinity or vice versa. Let me check that. For each term (frac{k}{x - (2k - 1)}), as (x) approaches (2k - 1) from the left, the term tends to negative infinity, and as (x) approaches from the right, it tends to positive infinity. So, each term contributes to the behavior near its asymptote.But since all these terms are added together, the overall function (f(x)) will have similar behavior near each asymptote. So, near each (x = 2k - 1), (f(x)) will go from negative infinity to positive infinity as (x) crosses the asymptote from left to right.Now, let's think about the behavior of (f(x)) as (x) approaches infinity or negative infinity. Each term (frac{k}{x - (2k - 1)}) behaves like (frac{k}{x}) for large (x), so the entire sum behaves like (sum_{k=1}^{50} frac{k}{x}), which is (frac{1 + 2 + 3 + dots + 50}{x}). The sum of the first 50 natural numbers is (frac{50*51}{2} = 1275), so as (x) approaches infinity or negative infinity, (f(x)) approaches 0.So, as (x) goes to positive infinity, (f(x)) approaches 0 from above because all the terms are positive when (x) is large. Similarly, as (x) goes to negative infinity, (f(x)) approaches 0 from below because each term (frac{k}{x - (2k - 1)}) becomes negative and their sum is negative.Now, let's consider the function (g(x) = f(x) - x). We need to find the number of real solutions to (g(x) = 0). To analyze (g(x)), let's look at its behavior in different intervals. The vertical asymptotes divide the real line into 51 intervals:1. ((-infty, 1))2. ((1, 3))3. ((3, 5))...50. ((97, 99))51. ((99, infty))In each of these intervals, (f(x)) is continuous because there are no asymptotes within the interval. Also, (f(x)) is strictly decreasing or increasing? Hmm, let me think.Wait, actually, each term (frac{k}{x - (2k - 1)}) is a hyperbola, and its derivative is (-frac{k}{(x - (2k - 1))^2}), which is always negative. So, each term is decreasing wherever it's defined. Therefore, the sum (f(x)) is also a decreasing function in each interval where it's defined because it's the sum of decreasing functions.So, in each interval between two consecutive asymptotes, (f(x)) is strictly decreasing from negative infinity to positive infinity. Therefore, (f(x)) will cross the line (y = x) exactly once in each interval because it goes from negative infinity to positive infinity, and (y = x) is a straight line.Wait, but (f(x)) is decreasing, so it goes from positive infinity to negative infinity as (x) increases through the interval? Or is it the other way around?Hold on, let's clarify. For each term (frac{k}{x - a}), as (x) approaches (a) from the left, the term goes to negative infinity, and as (x) approaches (a) from the right, it goes to positive infinity. So, for each interval between two asymptotes, say between (a) and (b), as (x) approaches (a) from the right, (f(x)) approaches positive infinity, and as (x) approaches (b) from the left, (f(x)) approaches negative infinity. Therefore, in each interval, (f(x)) is decreasing from positive infinity to negative infinity.So, in each interval, (f(x)) starts at positive infinity, decreases, and ends at negative infinity. The line (y = x) is a straight line with a slope of 1. So, in each interval, since (f(x)) goes from positive infinity to negative infinity, and (y = x) is a straight line, they must intersect exactly once in each interval.Therefore, in each of the 50 intervals between the asymptotes, there is exactly one real solution. Now, what about the intervals outside the asymptotes, that is, ((-infty, 1)) and ((99, infty))?In the interval ((-infty, 1)), (f(x)) approaches 0 from below as (x) approaches negative infinity, and as (x) approaches 1 from the left, (f(x)) approaches negative infinity. So, (f(x)) is decreasing from 0 to negative infinity in this interval. The line (y = x) is a straight line that goes from negative infinity to 1 as (x) goes from negative infinity to 1. So, in this interval, (f(x)) starts at 0 and decreases to negative infinity, while (y = x) starts at negative infinity and increases to 1. Therefore, they must cross exactly once in this interval.Similarly, in the interval ((99, infty)), (f(x)) approaches 0 from above as (x) approaches infinity, and as (x) approaches 99 from the right, (f(x)) approaches positive infinity. So, (f(x)) is decreasing from positive infinity to 0 in this interval. The line (y = x) is a straight line that goes from 99 to infinity as (x) goes from 99 to infinity. So, (f(x)) starts at positive infinity and decreases to 0, while (y = x) starts at 99 and increases to infinity. Therefore, they must cross exactly once in this interval as well.So, in total, we have 50 intervals between the asymptotes, each contributing one real solution, plus two more intervals on the ends, each contributing one real solution. That gives us 50 + 2 = 52 real solutions.Wait, let me double-check. Each interval between asymptotes is 50, and then two more on the ends, so 52. That seems right.But just to be thorough, let me consider the behavior in each interval:1. ((-infty, 1)): (f(x)) goes from 0 to negative infinity, (y = x) goes from negative infinity to 1. They cross once.2. Each ((2k - 1, 2k + 1)) for (k = 1) to (49): (f(x)) goes from positive infinity to negative infinity, (y = x) is a straight line. They cross once.3. ((99, infty)): (f(x)) goes from positive infinity to 0, (y = x) goes from 99 to infinity. They cross once.So, that's 1 + 50 + 1 = 52 intervals, each contributing exactly one real solution. Therefore, the equation has 52 real solutions.I think that's solid. I can't think of any reason why this wouldn't hold. Maybe I should consider if (f(x)) could ever be equal to (x) more than once in an interval, but since (f(x)) is strictly decreasing in each interval and (y = x) is strictly increasing, they can only intersect once in each interval. Yeah, that makes sense.So, I'm confident that the number of real solutions is 52.

💡Okay, so I have this problem here: the domain of the function f(x) is [-3,1], and I need to find the domain of the function f(2x - 1). Hmm, let me think about how to approach this.First, I remember that the domain of a function is all the possible input values (x-values) that will output a real number. So, for f(x), the inputs can be any real number from -3 to 1, inclusive. That means if I plug in any x between -3 and 1 into f(x), it will work, but anything outside of that range won't.Now, the question is about f(2x - 1). This is a transformation of the original function f(x). Specifically, it's a horizontal scaling and shift. I think scaling affects the domain because it stretches or compresses the graph horizontally, which changes the input values needed to cover the same range as the original function.Let me write down what I know:- The domain of f(x) is [-3, 1].- We need to find the domain of f(2x - 1).So, for f(2x - 1) to be defined, the expression inside the function, which is 2x - 1, must lie within the domain of f(x). That means:-3 ≤ 2x - 1 ≤ 1Okay, so I need to solve this inequality for x to find the new domain. Let me do that step by step.First, I'll add 1 to all parts of the inequality to isolate the term with x:-3 + 1 ≤ 2x - 1 + 1 ≤ 1 + 1Simplifying that:-2 ≤ 2x ≤ 2Now, I need to solve for x by dividing all parts of the inequality by 2:-2 / 2 ≤ 2x / 2 ≤ 2 / 2Which simplifies to:-1 ≤ x ≤ 1So, the domain of f(2x - 1) is all real numbers x such that x is between -1 and 1, inclusive. That means the domain is [-1, 1].Wait, let me double-check that. If I plug in x = -1 into 2x - 1, I get 2*(-1) - 1 = -2 - 1 = -3, which is the lower bound of the original domain. Similarly, if I plug in x = 1, I get 2*1 - 1 = 2 - 1 = 1, which is the upper bound of the original domain. So, that makes sense. The transformation 2x - 1 scales the input by a factor of 2 and shifts it by -1, which effectively compresses the domain from [-3,1] to [-1,1].I think that's correct. So, the domain of f(2x - 1) is [-1,1].

💡Alright, so I have this problem here: If the minimum of {2x - 5, x + 1} is equal to -1, then what is the value of x? The options are A: -3, B: 2, C: -2, D: 2 or -2.Okay, let me try to understand what this is asking. The min function here takes two numbers, 2x - 5 and x + 1, and returns the smaller one. So, we're told that the smaller one is -1. That means either 2x - 5 is -1, or x + 1 is -1, but not both necessarily, right? Because if both were equal to -1, then they would be the same, but the problem says m and n are real numbers and m ≠ n, so they can't be the same.Wait, actually, the problem says "m and n are real numbers, and m ≠ n." Hmm, does that mean that 2x - 5 and x + 1 must be different? So, they can't be equal? That's an important point. So, 2x - 5 ≠ x + 1. Let me write that down:2x - 5 ≠ x + 1If I solve that inequality, subtract x from both sides:x - 5 ≠ 1Then add 5 to both sides:x ≠ 6So, x cannot be 6. But in the options given, 6 isn't even an option, so maybe that's not directly relevant here. But it's good to note.Now, back to the original problem. The minimum of {2x - 5, x + 1} is -1. So, either 2x - 5 is the smaller one and equals -1, or x + 1 is the smaller one and equals -1.Let me consider both cases.**Case 1:** 2x - 5 is the minimum, so 2x - 5 = -1.Let's solve for x:2x - 5 = -1Add 5 to both sides:2x = 4Divide both sides by 2:x = 2Okay, so x could be 2. But I need to check if this makes sense. If x is 2, then let's compute both expressions:2x - 5 = 2*2 - 5 = 4 - 5 = -1x + 1 = 2 + 1 = 3So, the minimum of {-1, 3} is indeed -1. That works.**Case 2:** x + 1 is the minimum, so x + 1 = -1.Let's solve for x:x + 1 = -1Subtract 1 from both sides:x = -2So, x could also be -2. Let's check this as well.If x is -2, then:2x - 5 = 2*(-2) - 5 = -4 - 5 = -9x + 1 = -2 + 1 = -1So, the minimum of {-9, -1} is -9, which is not equal to -1. Wait, that's a problem. Because we were supposed to have the minimum equal to -1, but in this case, the minimum is -9, which is less than -1. That means our assumption that x + 1 is the minimum is incorrect in this case because 2x - 5 is actually smaller.So, even though x = -2 satisfies x + 1 = -1, it doesn't satisfy the condition that the minimum of {2x - 5, x + 1} is -1 because 2x - 5 is smaller.Therefore, x = -2 is not a valid solution.So, only x = 2 works.Wait, but let me think again. Maybe I made a mistake. If x = -2, then 2x - 5 is -9, which is less than -1, so the minimum would be -9, not -1. So, indeed, x = -2 doesn't satisfy the condition that the minimum is -1.Therefore, the only solution is x = 2.But wait, the options include D: 2 or -2. So, why is that an option? Maybe I need to consider if both could be solutions under different conditions.Let me think. If x = 2, then 2x - 5 = -1 and x + 1 = 3, so the minimum is -1. If x = -2, then 2x - 5 = -9 and x + 1 = -1, so the minimum is -9, which is not equal to -1. Therefore, x = -2 is not a valid solution.So, only x = 2 is the correct answer.But wait, maybe I should consider if there's a range of x where the minimum is -1. Let me think about the functions 2x - 5 and x + 1.Let me plot these two linear functions mentally. 2x - 5 has a steeper slope than x + 1. They intersect where 2x - 5 = x + 1, which is at x = 6, as we saw earlier. So, for x < 6, 2x - 5 < x + 1, and for x > 6, 2x - 5 > x + 1.So, for x < 6, the minimum is 2x - 5, and for x > 6, the minimum is x + 1.We are told that the minimum is -1. So, let's consider two cases:1. For x < 6, the minimum is 2x - 5 = -1. So, 2x - 5 = -1 => x = 2. Since 2 < 6, this is valid.2. For x > 6, the minimum is x + 1 = -1. So, x + 1 = -1 => x = -2. But wait, x = -2 is less than 6, which contradicts our assumption that x > 6. Therefore, this solution is invalid.Therefore, only x = 2 is valid.So, the answer is B: 2.But wait, the options include D: 2 or -2. Maybe I need to consider if both could be solutions in some way.Wait, if x = -2, then the minimum is -9, which is less than -1, so it doesn't satisfy the condition. Therefore, x = -2 is not a solution.Therefore, the only solution is x = 2.So, the answer is B: 2.

💡First, I need to determine the time each train travels until they meet. The first train starts at 7 a.m. and meets the second train at 12 p.m., so it travels for 5 hours.Next, I'll calculate the distance the first train covers in those 5 hours by multiplying its speed by the time: 20 km/h * 5 h = 100 km.Since the total distance between the two stations is 200 km, the second train must cover the remaining distance: 200 km - 100 km = 100 km.Now, I'll determine how long the second train travels by dividing the distance it covers by its speed: 100 km / 25 km/h = 4 hours.Finally, to find out when the second train started, I'll subtract the travel time from the meeting time: 12 p.m. - 4 hours = 8 a.m.