💡Okay, so I have this problem to solve: I need to calculate the absolute value of (2 + 2i) raised to the sixth power plus 3. That is, |(2 + 2i)^6 + 3|. Hmm, that seems a bit complicated, but I think I can break it down step by step.First, I remember that when dealing with complex numbers, especially when raising them to powers, it's often helpful to convert them into polar form. Polar form represents a complex number in terms of its modulus (or magnitude) and its argument (or angle). This is because raising a complex number to a power is much simpler in polar form thanks to De Moivre's theorem.So, let me start by converting the complex number 2 + 2i into polar form. To do that, I need to find its modulus and its argument.The modulus of a complex number a + bi is given by sqrt(a^2 + b^2). For 2 + 2i, that would be sqrt(2^2 + 2^2) = sqrt(4 + 4) = sqrt(8). Simplifying sqrt(8), I get 2*sqrt(2). So, the modulus is 2*sqrt(2).Next, the argument is the angle that the complex number makes with the positive real axis. Since both the real and imaginary parts are positive, the complex number lies in the first quadrant. The argument θ can be found using tanθ = b/a. So, tanθ = 2/2 = 1. Therefore, θ = arctan(1) = π/4 radians or 45 degrees.So, in polar form, 2 + 2i can be written as 2*sqrt(2) * (cos(π/4) + i*sin(π/4)).Now, I need to raise this complex number to the sixth power. Using De Moivre's theorem, which states that (r*(cosθ + i*sinθ))^n = r^n * (cos(nθ) + i*sin(nθ)), I can compute (2 + 2i)^6.First, let's compute r^6. Since r = 2*sqrt(2), r^6 = (2*sqrt(2))^6. Let me calculate that:(2)^6 = 64(sqrt(2))^6 = (2^(1/2))^6 = 2^(3) = 8So, (2*sqrt(2))^6 = 64 * 8 = 512.Next, let's compute the angle nθ. Here, n = 6 and θ = π/4, so 6*(π/4) = (6π)/4 = (3π)/2.Therefore, (2 + 2i)^6 in polar form is 512*(cos(3π/2) + i*sin(3π/2)).Now, let's convert this back to rectangular form to make it easier to add 3. The cosine of 3π/2 is 0, and the sine of 3π/2 is -1. So, 512*(0 + i*(-1)) = -512i.So, (2 + 2i)^6 = -512i.Now, the problem asks for |(2 + 2i)^6 + 3|. So, I need to add 3 to -512i. Adding a real number to a purely imaginary number results in a complex number with a real part and an imaginary part. Specifically, 3 + (-512i) = 3 - 512i.Now, I need to find the modulus of this complex number, 3 - 512i. The modulus of a complex number a + bi is sqrt(a^2 + b^2). So, let's compute that:a = 3, b = -512a^2 = 9b^2 = (-512)^2 = 262,144So, a^2 + b^2 = 9 + 262,144 = 262,153Therefore, the modulus is sqrt(262,153). Let me compute that. Hmm, sqrt(262,153). I can try to see if this is a perfect square or approximate it.Let me check 512^2 = 262,144, which is very close to 262,153. So, 512^2 = 262,144, and 513^2 = (512 + 1)^2 = 512^2 + 2*512 + 1 = 262,144 + 1,024 + 1 = 263,169. Wait, that's too high. Wait, 512^2 is 262,144, and 512.005^2 would be approximately 262,144 + 2*512*0.005 + (0.005)^2 ≈ 262,144 + 5.12 + 0.000025 ≈ 262,149.120025. Hmm, but our value is 262,153, which is higher than that.Wait, maybe I made a mistake. Let me recalculate 512^2. 512*512: 500*500=250,000, 500*12=6,000, 12*500=6,000, 12*12=144. So, 250,000 + 6,000 + 6,000 + 144 = 262,144. Correct.So, 512^2 = 262,144, and 513^2 = 263,169. So, 262,153 is between 512^2 and 513^2. Let's see how much more it is: 262,153 - 262,144 = 9. So, it's 512^2 + 9. Therefore, sqrt(262,153) = sqrt(512^2 + 3^2). Hmm, that's interesting. Wait, 512^2 + 3^2 = (512)^2 + (3)^2, which is the sum of squares. So, sqrt(512^2 + 3^2) is the hypotenuse of a right triangle with sides 512 and 3.But I don't think that helps me simplify it further. So, I can either leave it as sqrt(262,153) or approximate it numerically.Let me try to approximate sqrt(262,153). Since 512^2 = 262,144, and 512.005^2 ≈ 262,144 + 2*512*0.005 + (0.005)^2 ≈ 262,144 + 5.12 + 0.000025 ≈ 262,149.120025. Wait, but our value is 262,153, which is 262,144 + 9. So, 512^2 + 9 = (512)^2 + (3)^2. So, sqrt(512^2 + 3^2) = sqrt(262,144 + 9) = sqrt(262,153).Alternatively, I can think of it as sqrt(512^2 + 3^2) = sqrt((512)^2 + (3)^2). Hmm, but that doesn't simplify it further.Wait, maybe I can factor out 512^2: sqrt(512^2 + 3^2) = 512*sqrt(1 + (3/512)^2). Let's compute (3/512)^2: 9/262,144 ≈ 0.00003435. So, sqrt(1 + 0.00003435) ≈ 1 + 0.000017175 (using the approximation sqrt(1 + x) ≈ 1 + x/2 for small x). Therefore, sqrt(262,153) ≈ 512*(1 + 0.000017175) ≈ 512 + 512*0.000017175 ≈ 512 + 0.0088 ≈ 512.0088.So, approximately, sqrt(262,153) ≈ 512.0088.But since the problem asks for the exact value, I should probably leave it as sqrt(262,153). However, let me check if 262,153 is a perfect square. Let me see:Let me try 512^2 = 262,144513^2 = 263,169So, 262,153 is between these two. Let me check 512.005^2 as before, but that's not helpful.Alternatively, maybe I can factor 262,153 to see if it's a perfect square. Let's try dividing by small primes:262,153 ÷ 7: 7*37,450 = 262,150, so 262,153 - 262,150 = 3, so remainder 3. Not divisible by 7.262,153 ÷ 13: 13*20,165 = 262,145, remainder 8. Not divisible by 13.262,153 ÷ 3: 2+6+2+1+5+3=19, which is not divisible by 3, so 262,153 not divisible by 3.262,153 ÷ 5: ends with 3, so no.262,153 ÷ 11: Let's use the divisibility rule for 11: (2 + 2 + 5) - (6 + 1 + 3) = (9) - (10) = -1, which is not divisible by 11, so no.262,153 ÷ 17: Let's see, 17*15,420 = 262,140, remainder 13. Not divisible by 17.262,153 ÷ 19: 19*13,797 = 262,143, remainder 10. Not divisible by 19.262,153 ÷ 23: 23*11,397 = 262,131, remainder 22. Not divisible by 23.262,153 ÷ 29: 29*9,039 = 262,131, remainder 22. Not divisible by 29.262,153 ÷ 31: 31*8,456 = 262,136, remainder 17. Not divisible by 31.262,153 ÷ 37: 37*7,085 = 262,145, remainder 8. Not divisible by 37.262,153 ÷ 43: 43*6,100 = 262,300, which is higher, so 43*6,096 = 262,128, remainder 25. Not divisible by 43.262,153 ÷ 47: 47*5,577 = 262,119, remainder 34. Not divisible by 47.262,153 ÷ 53: 53*4,946 = 262,138, remainder 15. Not divisible by 53.262,153 ÷ 59: 59*4,443 = 262,137, remainder 16. Not divisible by 59.262,153 ÷ 61: 61*4,297 = 262,177, which is higher, so 61*4,296 = 262,116, remainder 37. Not divisible by 61.262,153 ÷ 67: 67*3,912 = 262,104, remainder 49. Not divisible by 67.262,153 ÷ 71: 71*3,692 = 262,132, remainder 21. Not divisible by 71.262,153 ÷ 73: 73*3,590 = 262,070, remainder 83. Not divisible by 73.262,153 ÷ 79: 79*3,318 = 262,122, remainder 31. Not divisible by 79.262,153 ÷ 83: 83*3,158 = 262,114, remainder 39. Not divisible by 83.262,153 ÷ 89: 89*2,945 = 262,105, remainder 48. Not divisible by 89.262,153 ÷ 97: 97*2,702 = 262,094, remainder 59. Not divisible by 97.Hmm, this is getting tedious. Maybe 262,153 is a prime number? Or perhaps it's a square of a larger number. Wait, but 512^2 is 262,144, and 513^2 is 263,169, so 262,153 is between them, and it's not a perfect square. Therefore, sqrt(262,153) is irrational and cannot be simplified further. So, the exact value is sqrt(262,153), which is approximately 512.0088.But wait, maybe I made a mistake earlier. Let me double-check my steps.First, converting 2 + 2i to polar form: modulus is sqrt(8) = 2*sqrt(2), argument is π/4. That seems correct.Raising to the sixth power: (2*sqrt(2))^6 = 2^6*(sqrt(2))^6 = 64*8 = 512. Correct.Angle: 6*(π/4) = 3π/2. Correct.So, (2 + 2i)^6 = 512*(cos(3π/2) + i*sin(3π/2)) = 512*(0 - i) = -512i. Correct.Adding 3: 3 + (-512i) = 3 - 512i. Correct.Modulus: sqrt(3^2 + (-512)^2) = sqrt(9 + 262,144) = sqrt(262,153). Correct.So, I think my steps are correct. Therefore, the exact value is sqrt(262,153), which is approximately 512.0088.But wait, the problem asks for the absolute value, which is a real number. So, I can either leave it as sqrt(262,153) or provide the approximate decimal value. However, since the problem didn't specify, I think providing the exact value is better, so sqrt(262,153).Wait, but let me check if 262,153 can be factored further or if it's a perfect square. Maybe I missed something.Let me try to see if 262,153 is divisible by 13: 13*20,165 = 262,145, remainder 8. Not divisible.Divisible by 7: 7*37,450 = 262,150, remainder 3. Not divisible.Divisible by 3: Sum of digits is 2+6+2+1+5+3=19, not divisible by 3.Divisible by 5: Ends with 3, no.Divisible by 11: (2+2+5) - (6+1+3) = 9 - 10 = -1, not divisible.Divisible by 17: 17*15,420 = 262,140, remainder 13. Not divisible.Divisible by 19: 19*13,797 = 262,143, remainder 10. Not divisible.Divisible by 23: 23*11,397 = 262,131, remainder 22. Not divisible.Divisible by 29: 29*9,039 = 262,131, remainder 22. Not divisible.Divisible by 31: 31*8,456 = 262,136, remainder 17. Not divisible.Divisible by 37: 37*7,085 = 262,145, remainder 8. Not divisible.Divisible by 43: 43*6,100 = 262,300, which is higher, so 43*6,096 = 262,128, remainder 25. Not divisible.Divisible by 47: 47*5,577 = 262,119, remainder 34. Not divisible.Divisible by 53: 53*4,946 = 262,138, remainder 15. Not divisible.Divisible by 59: 59*4,443 = 262,137, remainder 16. Not divisible.Divisible by 61: 61*4,297 = 262,177, which is higher, so 61*4,296 = 262,116, remainder 37. Not divisible.Divisible by 67: 67*3,912 = 262,104, remainder 49. Not divisible.Divisible by 71: 71*3,692 = 262,132, remainder 21. Not divisible.Divisible by 73: 73*3,590 = 262,070, remainder 83. Not divisible.Divisible by 79: 79*3,318 = 262,122, remainder 31. Not divisible.Divisible by 83: 83*3,158 = 262,114, remainder 39. Not divisible.Divisible by 89: 89*2,945 = 262,105, remainder 48. Not divisible.Divisible by 97: 97*2,702 = 262,094, remainder 59. Not divisible.So, it seems that 262,153 is a prime number or at least doesn't have any small prime factors. Therefore, sqrt(262,153) cannot be simplified further, and the exact value is sqrt(262,153).Alternatively, if I want to write it in terms of 512 and 3, I can note that sqrt(512^2 + 3^2) = sqrt(262,144 + 9) = sqrt(262,153), which is the same as before.So, in conclusion, the absolute value |(2 + 2i)^6 + 3| is equal to sqrt(262,153). If I need to provide a numerical approximation, it's approximately 512.0088, but since the problem likely expects an exact value, I'll stick with sqrt(262,153).Wait, but let me double-check my calculations one more time to make sure I didn't make any mistakes.Starting from (2 + 2i)^6:1. Convert 2 + 2i to polar form: modulus = 2*sqrt(2), argument = π/4. Correct.2. Raise to the sixth power: (2*sqrt(2))^6 = 512, angle = 6*(π/4) = 3π/2. Correct.3. So, (2 + 2i)^6 = 512*(cos(3π/2) + i*sin(3π/2)) = 512*(0 - i) = -512i. Correct.4. Adding 3: 3 + (-512i) = 3 - 512i. Correct.5. Modulus: sqrt(3^2 + (-512)^2) = sqrt(9 + 262,144) = sqrt(262,153). Correct.Yes, everything checks out. So, the final answer is sqrt(262,153).

💡To determine for how many integers ( x ) the expression ( x^4 - 49x^2 + 50 ) is negative, we can follow these steps:1. **Substitute**: Let ( y = x^2 ). The expression becomes ( y^2 - 49y + 50 ).2. **Factor the Quadratic**: Factor ( y^2 - 49y + 50 ) as ( (y - 1)(y - 50) ).3. **Determine When the Expression is Negative**: The product ( (y - 1)(y - 50) ) is negative when ( y ) is between 1 and 50 (i.e., ( 1 < y < 50 )).4. **Identify Perfect Squares in the Interval**: The perfect squares between 1 and 50 are 4, 9, 16, 25, 36, and 49.5. **Count Corresponding ( x ) Values**: Each perfect square ( y ) corresponds to two integers ( x ) (positive and negative). Therefore, the integers ( x ) are ( pm 2, pm 3, pm 4, pm 5, pm 6, pm 7 ).6. **Total Count**: There are 12 such integers.Thus, the number of integers ( x ) for which ( x^4 - 49x^2 + 50 ) is negative is 12.

💡Alright, let's try to tackle this problem step by step. So, we have two sets, A and B, defined by quadratic inequalities. We need to find the range of values for 'a' such that the intersection of A and B contains exactly one integer. Hmm, okay, let's break this down.First, let's understand what set A is. Set A is defined by the inequality (x^2 + 2x - 8 > 0). To find the values of x that satisfy this, I think I should solve the quadratic equation (x^2 + 2x - 8 = 0). Let me try factoring this. Hmm, looking for two numbers that multiply to -8 and add to 2. That would be 4 and -2. So, the equation factors to ((x + 4)(x - 2) = 0). Therefore, the roots are x = -4 and x = 2.Since the quadratic opens upwards (the coefficient of (x^2) is positive), the inequality (x^2 + 2x - 8 > 0) will hold true when x is less than -4 or greater than 2. So, set A is all real numbers x such that x < -4 or x > 2.Okay, moving on to set B, which is defined by (x^2 - 2ax + 4 leq 0). This is another quadratic inequality. To find the values of x that satisfy this, I need to analyze the quadratic equation (x^2 - 2ax + 4 = 0). The solutions to this equation will give me the critical points that define the intervals where the inequality holds.First, let's find the discriminant of this quadratic to see if it has real roots. The discriminant D is given by (D = ( -2a )^2 - 4 * 1 * 4 = 4a^2 - 16). For the quadratic to have real roots, the discriminant must be non-negative. So, (4a^2 - 16 geq 0), which simplifies to (a^2 geq 4). Since a is given to be positive, this means (a geq 2). So, for a >= 2, the quadratic has real roots, and the inequality (x^2 - 2ax + 4 leq 0) will hold between these two roots.Let me find the roots. Using the quadratic formula, the roots are:(x = [2a pm sqrt{(2a)^2 - 16}]/2 = [2a pm sqrt{4a^2 - 16}]/2 = a pm sqrt{a^2 - 4}).So, the roots are (x = a + sqrt{a^2 - 4}) and (x = a - sqrt{a^2 - 4}). Since a >= 2, both roots are real and distinct. The quadratic opens upwards, so the inequality (x^2 - 2ax + 4 leq 0) holds between these two roots. Therefore, set B is the interval ([a - sqrt{a^2 - 4}, a + sqrt{a^2 - 4}]).Now, we need to find the intersection of sets A and B, which is (A cap B). Set A is (x < -4) or (x > 2), and set B is the interval between (a - sqrt{a^2 - 4}) and (a + sqrt{a^2 - 4}). So, the intersection will be the overlap between these intervals.Since a is positive and greater than or equal to 2, the interval B is centered around a and extends to the left and right by (sqrt{a^2 - 4}). Let's see where this interval B lies in relation to set A.Given that set A is x < -4 or x > 2, and set B is an interval around a positive a (since a >= 2), the intersection will likely be in the region x > 2. The left part of set A (x < -4) is probably not overlapping with set B because B is around positive a.So, the intersection (A cap B) will be the overlap between (x > 2) and the interval B. Therefore, the intersection is ([2, a + sqrt{a^2 - 4}]) intersected with B, but actually, since B is a closed interval, it's the overlap between B and (2, ∞).Wait, actually, set A is x < -4 or x > 2, so the intersection with B would be the part of B that is greater than 2. So, if the lower bound of B is less than 2, then the intersection would start at 2, otherwise, it would start at the lower bound of B. But since a >= 2, let's see:The lower bound of B is (a - sqrt{a^2 - 4}). Let's compute this:(a - sqrt{a^2 - 4} = frac{(a - sqrt{a^2 - 4})(a + sqrt{a^2 - 4})}{a + sqrt{a^2 - 4}}} = frac{a^2 - (a^2 - 4)}{a + sqrt{a^2 - 4}}} = frac{4}{a + sqrt{a^2 - 4}}).Since a >= 2, the denominator is positive, so the lower bound is positive. Therefore, the lower bound of B is positive, and since a >= 2, the lower bound is at least:When a = 2, (2 - sqrt{4 - 4} = 2 - 0 = 2). So, when a = 2, the lower bound is 2. For a > 2, the lower bound is greater than 2.Wait, let me check that. If a = 3, then the lower bound is 3 - sqrt(9 - 4) = 3 - sqrt(5) ≈ 3 - 2.236 ≈ 0.764, which is less than 2. Hmm, that contradicts my earlier thought. So, actually, for a > 2, the lower bound can be less than 2.Wait, let me compute it again. The lower bound is (a - sqrt{a^2 - 4}). Let's see:For a = 2: 2 - sqrt(4 - 4) = 2 - 0 = 2.For a = 3: 3 - sqrt(9 - 4) = 3 - sqrt(5) ≈ 3 - 2.236 ≈ 0.764.For a = 4: 4 - sqrt(16 - 4) = 4 - sqrt(12) ≈ 4 - 3.464 ≈ 0.536.So, as a increases, the lower bound of B decreases towards 0. Therefore, for a > 2, the lower bound is less than 2, and the upper bound is greater than a.Therefore, the intersection (A cap B) will be the part of B that is greater than 2. So, the intersection is [2, a + sqrt(a^2 - 4)].Wait, but set A is x > 2, so the intersection is (2, a + sqrt(a^2 - 4)]. But since B is a closed interval, it's [2, a + sqrt(a^2 - 4)].But actually, when a = 2, the lower bound is 2, and the upper bound is 2 + sqrt(4 - 4) = 2. So, B becomes just the single point {2} when a = 2. But since set A is x > 2, the intersection would be empty when a = 2 because B is just {2}, which is not in A (since A is x > 2, not including 2). Wait, but the original definition of A is x^2 + 2x -8 > 0, which at x=2 is 4 + 4 -8 = 0, so x=2 is not included in A. So, when a=2, B is {2}, which doesn't intersect with A, so the intersection is empty.But the problem states that the intersection contains exactly one integer. So, when a=2, the intersection is empty, which doesn't satisfy the condition. Therefore, a must be greater than 2.So, for a > 2, the intersection (A cap B) is [2, a + sqrt(a^2 - 4)]. But wait, actually, since B is [a - sqrt(a^2 - 4), a + sqrt(a^2 - 4)], and A is x > 2, the intersection is [max(2, a - sqrt(a^2 - 4)), a + sqrt(a^2 - 4)].But as we saw earlier, for a > 2, a - sqrt(a^2 - 4) is less than 2. Therefore, the intersection is [2, a + sqrt(a^2 - 4)].Now, we need this intersection to contain exactly one integer. So, the interval [2, a + sqrt(a^2 - 4)] must contain exactly one integer. Let's think about what integers are in this interval.The interval starts at 2 and goes up to a + sqrt(a^2 - 4). So, the integers in this interval would be 2, 3, 4, etc., depending on how large a is.But we need exactly one integer in this interval. So, the interval must contain exactly one integer, which is 3. Because if it contains 2, then it would have to include 2 and 3, which would be two integers. But wait, 2 is the lower bound, and the interval is [2, ...). So, if the upper bound is just above 3, then the interval would include 2 and 3, which are two integers. But we need exactly one integer. Therefore, the interval must include only 3, not 2 or 4.Wait, but 2 is included in the interval because the interval is [2, ...). So, if the interval is [2, something), and we need exactly one integer, that integer must be 3, and the interval must not include 4. So, the interval must be [2, 4), but not including 4. But wait, if it's [2, 4), then it includes 2, 3, which are two integers. Hmm, this is confusing.Wait, maybe I'm approaching this wrong. Let's think about it differently. The intersection (A cap B) is [2, a + sqrt(a^2 - 4)]. We need this interval to contain exactly one integer. So, the length of the interval is a + sqrt(a^2 - 4) - 2. We need this length to be such that only one integer is inside.But actually, the interval starts at 2, which is an integer. So, if the interval is [2, c), where c is between 3 and 4, then the integers in the interval would be 2 and 3, which is two integers. But we need exactly one integer. Therefore, perhaps the interval should not include 2? But wait, the intersection is [2, c], so 2 is included.Wait, maybe I'm misunderstanding. The problem says the intersection contains exactly one integer. So, perhaps the interval [2, c] contains exactly one integer, which would mean that c is between 3 and 4, but not including 4. Because if c is between 3 and 4, then the integers in [2, c] would be 2 and 3, which is two integers. Hmm, this is conflicting.Wait, perhaps the integer is 3, and the interval must include 3 but not include 2 or 4. But since the interval starts at 2, it's impossible to exclude 2. So, maybe the only way is that the interval includes 3 but not 4, and since it starts at 2, it includes 2 and 3, which is two integers. Therefore, perhaps the problem is that the intersection must contain exactly one integer, but since the interval starts at 2, which is an integer, the only way is that the interval is [2, 3), which would include only 2. But 2 is not in A, because A is x > 2. Wait, no, A is x^2 + 2x -8 > 0, which at x=2 is 0, so x=2 is not in A. Therefore, the intersection (A cap B) is (2, a + sqrt(a^2 - 4)]. So, it's an open interval at 2.Therefore, the integers in this interval would be the integers greater than 2 and less than or equal to a + sqrt(a^2 - 4). So, we need exactly one integer in (2, a + sqrt(a^2 - 4)]. That integer must be 3, and a + sqrt(a^2 - 4) must be less than 4. Because if it's 4 or more, then 4 would also be included, making two integers (3 and 4). So, to have exactly one integer, 3, the upper bound must be between 3 and 4.So, we need:3 <= a + sqrt(a^2 - 4) < 4.But also, since the interval starts just above 2, we need to ensure that the lower bound is just above 2, but since a > 2, the lower bound is a - sqrt(a^2 - 4), which is less than 2, so the intersection is (2, a + sqrt(a^2 - 4)].Therefore, to have exactly one integer, 3, in this interval, we need:3 <= a + sqrt(a^2 - 4) < 4.So, let's solve these inequalities.First, let's solve 3 <= a + sqrt(a^2 - 4).Let me denote sqrt(a^2 - 4) as s. Then, s = sqrt(a^2 - 4), so s >= 0.So, 3 <= a + s.But s = sqrt(a^2 - 4), so:3 <= a + sqrt(a^2 - 4).Let me isolate sqrt(a^2 - 4):sqrt(a^2 - 4) >= 3 - a.But since sqrt(a^2 - 4) is always non-negative, and 3 - a could be positive or negative depending on a.If a >= 3, then 3 - a <= 0, so sqrt(a^2 - 4) >= negative number, which is always true.If a < 3, then 3 - a > 0, so sqrt(a^2 - 4) >= 3 - a.But let's consider a >= 2.Case 1: a >= 3.In this case, 3 - a <= 0, so sqrt(a^2 - 4) >= 3 - a is always true because sqrt(a^2 - 4) >= 0 >= 3 - a.So, the inequality 3 <= a + sqrt(a^2 - 4) is automatically satisfied for a >= 3.Case 2: 2 < a < 3.Here, 3 - a > 0, so we have sqrt(a^2 - 4) >= 3 - a.Let's square both sides to eliminate the square root:a^2 - 4 >= (3 - a)^2.Expanding the right side:(3 - a)^2 = 9 - 6a + a^2.So, a^2 - 4 >= 9 - 6a + a^2.Subtract a^2 from both sides:-4 >= 9 - 6a.Subtract 9 from both sides:-13 >= -6a.Divide both sides by -6, remembering to reverse the inequality:13/6 <= a.So, in this case, a >= 13/6 ≈ 2.1667.But since we are in the case where 2 < a < 3, the solution here is 13/6 <= a < 3.Combining both cases, the solution to 3 <= a + sqrt(a^2 - 4) is a >= 13/6.Now, let's solve the upper inequality: a + sqrt(a^2 - 4) < 4.Again, let s = sqrt(a^2 - 4).So, a + s < 4.Isolate s:s < 4 - a.Since s = sqrt(a^2 - 4) >= 0, and 4 - a must be positive, so 4 - a > 0 => a < 4.So, a < 4.Now, let's square both sides of s < 4 - a:sqrt(a^2 - 4) < 4 - a.Square both sides:a^2 - 4 < (4 - a)^2.Expand the right side:(4 - a)^2 = 16 - 8a + a^2.So, a^2 - 4 < 16 - 8a + a^2.Subtract a^2 from both sides:-4 < 16 - 8a.Subtract 16 from both sides:-20 < -8a.Divide both sides by -8, reversing the inequality:20/8 > a => 5/2 > a => a < 5/2 = 2.5.So, combining this with the earlier condition that a >= 13/6 ≈ 2.1667, we have:13/6 <= a < 5/2.Therefore, the range of values for a is [13/6, 5/2).Let me double-check this.When a = 13/6 ≈ 2.1667, let's compute a + sqrt(a^2 - 4):a = 13/6 ≈ 2.1667a^2 = (169/36) ≈ 4.6944a^2 - 4 ≈ 0.6944sqrt(a^2 - 4) ≈ 0.8333So, a + sqrt(a^2 - 4) ≈ 2.1667 + 0.8333 ≈ 3.So, the upper bound is exactly 3, meaning the interval is (2, 3]. So, the integers in this interval are 3, which is exactly one integer.When a approaches 5/2 = 2.5 from below, let's compute a + sqrt(a^2 - 4):a = 2.5a^2 = 6.25a^2 - 4 = 2.25sqrt(a^2 - 4) = 1.5So, a + sqrt(a^2 - 4) = 2.5 + 1.5 = 4.But since a < 5/2, let's take a = 2.499:a ≈ 2.499a^2 ≈ 6.245a^2 - 4 ≈ 2.245sqrt(a^2 - 4) ≈ 1.498So, a + sqrt(a^2 - 4) ≈ 2.499 + 1.498 ≈ 3.997, which is just below 4.Therefore, the interval is (2, ~4), which includes integers 3 and 4. But wait, we need exactly one integer. So, when a approaches 5/2 from below, the upper bound approaches 4, so the interval approaches (2, 4), which includes 3 and 4, which are two integers. Therefore, to ensure that only 3 is included, the upper bound must be less than 4. Hence, a must be less than 5/2.Therefore, the range of a is [13/6, 5/2).I think this makes sense. So, the final answer is that a must be between 13/6 and 5/2, including 13/6 but not including 5/2.

💡Alright, so I've got this quadratic inequality to solve: 2x² + 7x + 3 > 0. Hmm, okay, quadratic inequalities can be tricky, but I think I remember some steps from my algebra class. Let me try to work through this step by step.First, I know that to solve a quadratic inequality like this, it's helpful to find the roots of the corresponding quadratic equation. So, I should start by solving 2x² + 7x + 3 = 0. That should give me the critical points where the quadratic expression equals zero, and those points will help me determine the intervals to test for the inequality.Alright, let's solve 2x² + 7x + 3 = 0. I can use the quadratic formula here, which is x = [-b ± √(b² - 4ac)] / (2a). In this equation, a = 2, b = 7, and c = 3. Plugging those values into the formula:x = [-7 ± √(7² - 4*2*3)] / (2*2)x = [-7 ± √(49 - 24)] / 4x = [-7 ± √25] / 4x = [-7 ± 5] / 4So, that gives me two solutions:x = (-7 + 5)/4 = (-2)/4 = -0.5x = (-7 - 5)/4 = (-12)/4 = -3Okay, so the roots are x = -3 and x = -0.5. These are the points where the quadratic expression equals zero. Now, since the quadratic coefficient (a = 2) is positive, the parabola opens upwards. That means the graph of the quadratic will be below the x-axis between the two roots and above the x-axis outside of them.But wait, the inequality is 2x² + 7x + 3 > 0, which means we're looking for the values of x where the quadratic expression is positive. Since the parabola opens upwards, it will be positive outside the interval between the roots. So, the solution should be x < -3 or x > -0.5.Let me double-check that. If I pick a value less than -3, say x = -4, and plug it into the expression:2*(-4)² + 7*(-4) + 3 = 2*16 - 28 + 3 = 32 - 28 + 3 = 7, which is positive. Good.If I pick a value between -3 and -0.5, say x = -2:2*(-2)² + 7*(-2) + 3 = 2*4 - 14 + 3 = 8 - 14 + 3 = -3, which is negative. So, that interval doesn't satisfy the inequality.If I pick a value greater than -0.5, say x = 0:2*(0)² + 7*0 + 3 = 0 + 0 + 3 = 3, which is positive. Perfect.So, putting it all together, the solution set is all real numbers less than -3 or greater than -0.5. That matches option B.Wait, just to make sure I didn't make any mistakes in my calculations. Let me go through the quadratic formula again:x = [-7 ± √(49 - 24)] / 4x = [-7 ± √25] / 4x = [-7 ± 5] / 4Yes, that gives x = (-7 + 5)/4 = -2/4 = -0.5 and x = (-7 - 5)/4 = -12/4 = -3. Correct.And since the coefficient of x² is positive, the parabola opens upwards, so the expression is positive outside the roots. So, x < -3 or x > -0.5 is indeed the solution.I think I'm confident with that answer now.

💡Alright, so I've got this problem here about natural numbers (a_1, a_2, ldots, a_n) where each number doesn't exceed its index, meaning (a_k leq k), and the total sum of all these numbers is even. I need to prove that there's a way to assign plus or minus signs to each number such that the resulting sum is zero. Hmm, okay, let's try to unpack this.First, let me make sure I understand the problem correctly. We have a sequence of natural numbers where each term is at most its position in the sequence. For example, (a_1) can only be 1, (a_2) can be 1 or 2, (a_3) can be 1, 2, or 3, and so on. The sum of all these numbers is even, and I need to show that by flipping the signs of some of them, I can make the total sum zero.I remember something about the Pigeonhole Principle and maybe induction being useful in these kinds of problems. Let me think about induction first. Maybe I can prove this statement for (n = 1), then assume it's true for (n) and show it's true for (n + 1). But wait, for (n = 1), we just have (a_1 = 1), and the sum is 1, which is odd. But the problem states that the sum is even, so maybe the base case isn't (n = 1). Let me check (n = 2).For (n = 2), (a_1) must be 1, and (a_2) can be 1 or 2. The total sum is either (1 + 1 = 2) or (1 + 2 = 3). Since the sum needs to be even, we only consider (a_1 = 1) and (a_2 = 1). Then, the possible sums with signs are (1 + 1 = 2) and (1 - 1 = 0). So, yes, we can get zero here. That works.Okay, so the base case for (n = 2) holds. Now, let's assume that for some (n), the statement is true. That is, if we have numbers (a_1, a_2, ldots, a_n) with each (a_k leq k) and the sum is even, then there's a way to assign signs to make the sum zero. Now, I need to show it's true for (n + 1).So, consider (n + 1) numbers (a_1, a_2, ldots, a_{n+1}) with each (a_k leq k) and the total sum is even. I need to show that there's a combination of signs such that the sum is zero.Hmm, maybe I can consider two cases: one where (a_{n+1}) is equal to (a_n) and another where it's different. Wait, that might not be the right approach. Let me think differently.I recall that if the total sum is even, then it's possible to partition the set into two subsets with equal sums. Maybe that's related. But in this case, it's not exactly partitioning; it's assigning signs. But assigning a plus or minus is similar to partitioning into two subsets: one with positive signs and one with negative signs. The difference between the sums of these two subsets should be zero, meaning their sums are equal.So, if I can partition the set ({a_1, a_2, ldots, a_n}) into two subsets with equal sums, then I can assign positive signs to one subset and negative signs to the other, resulting in a total sum of zero. That makes sense.But how do I ensure that such a partition exists? The total sum is even, so it's possible that it can be divided into two equal parts. But I also have the condition that each (a_k leq k). Maybe this condition ensures that the numbers aren't too large, making such a partition possible.Wait, maybe I can use induction here. Suppose for (n) numbers, the statement holds. Now, for (n + 1), I can consider the number (a_{n+1}). Since (a_{n+1} leq n + 1), it's not too large compared to the previous numbers. Maybe I can adjust the previous partition to include (a_{n+1}) in one of the subsets.Alternatively, maybe I can use the Pigeonhole Principle. The number of possible subsets is (2^n), and the number of possible sums is limited. If the total sum is even, then there must be two subsets with the same sum, and their symmetric difference would give a subset that sums to zero. But I'm not sure if that directly applies here.Wait, another approach: consider the partial sums. Start with (a_1), then (a_1 pm a_2), then (a_1 pm a_2 pm a_3), and so on. If at any point a partial sum is zero, we're done. If not, since there are (2^{n}) possible partial sums and only a limited number of possible values, by the Pigeonhole Principle, some partial sums must repeat, and their difference would give a subset that sums to zero.But I need to make sure that the partial sums are bounded in a way that the Pigeonhole Principle applies. Given that each (a_k leq k), the maximum possible partial sum after (k) terms is (1 + 2 + ldots + k = frac{k(k + 1)}{2}). So, the number of possible partial sums is limited, and as we consider more terms, we must eventually get a repeat or hit zero.Wait, but the problem is about assigning signs to all terms, not just finding a subset. So, maybe I need to consider all possible sign combinations. There are (2^n) possible sign combinations, and each combination gives a sum. The possible sums range from (-S) to (S), where (S) is the total sum. Since (S) is even, the number of possible sums is (2S + 1). If (2^n > 2S + 1), then by the Pigeonhole Principle, some sums must repeat, but I'm not sure if that helps directly.Alternatively, maybe I can use the fact that the total sum is even and each (a_k leq k) to ensure that the partial sums cover all residues modulo something, forcing a zero sum.Wait, another idea: consider the problem modulo 2. Since the total sum is even, the sum modulo 2 is zero. If I can show that there's a subset with an even sum, then maybe I can adjust the signs accordingly. But I'm not sure.Wait, maybe I should look for a specific construction. Suppose I start assigning all positive signs. If the sum is zero, we're done. If not, maybe I can flip the sign of some (a_k) to reduce the sum towards zero. But I need to ensure that flipping signs can actually reach zero.Alternatively, think about the problem as a graph where each node represents a partial sum, and edges represent adding or subtracting the next term. Starting from zero, we want to reach zero again after all terms. But I'm not sure how to formalize that.Wait, maybe I can use induction more carefully. Suppose for (n) terms, the statement holds. Now, for (n + 1) terms, consider the total sum (S = a_1 + a_2 + ldots + a_{n+1}), which is even. If I can find a subset of the first (n) terms that sums to (a_{n+1}), then I can assign positive signs to that subset and negative signs to the rest, including (a_{n+1}), resulting in a total sum of zero.But how do I ensure that such a subset exists? Since (a_{n+1} leq n + 1), and the sum of the first (n) terms is (S - a_{n+1}), which is also even because (S) is even and (a_{n+1}) is either even or odd. Wait, no, (S - a_{n+1}) could be even or odd depending on (a_{n+1}). Hmm, maybe that's not the right approach.Wait, another idea: consider the partial sums of the first (n) terms. There are (2^n) possible sign combinations, leading to partial sums ranging from (-T) to (T), where (T = a_1 + a_2 + ldots + a_n). If (T geq a_{n+1}), then maybe I can adjust the partial sum to be equal to (a_{n+1}) by flipping signs, and then subtract (a_{n+1}) to get zero. But I'm not sure.Alternatively, since (a_{n+1} leq n + 1), and the sum of the first (n) terms is (S - a_{n+1}), which is even, maybe I can find a subset of the first (n) terms that sums to (a_{n+1}), allowing me to balance the equation.Wait, I'm getting a bit stuck here. Maybe I should look for a different approach. Let's think about the problem in terms of parity. Since the total sum is even, any subset sum will have the same parity as the total sum. So, if I can find a subset with sum equal to half of the total sum, then assigning positive signs to that subset and negative signs to the rest will give a total sum of zero.But how do I ensure that such a subset exists? This is similar to the partition problem, which is NP-hard, but in this case, the constraints might make it possible. Given that each (a_k leq k), maybe the numbers are small enough to guarantee such a partition.Wait, another thought: maybe I can use the fact that the numbers are bounded by their indices to ensure that the partial sums cover all residues modulo something, forcing a zero sum. For example, if I consider the partial sums modulo (a_{n+1}), there are (a_{n+1}) possible residues. If I have more partial sums than residues, then by the Pigeonhole Principle, two partial sums must be congruent modulo (a_{n+1}), and their difference would be a multiple of (a_{n+1}). But I'm not sure how that helps directly.Alternatively, maybe I can use the fact that the total sum is even and each (a_k leq k) to ensure that the numbers are small enough to allow for a zero sum through sign assignment.Wait, I think I need to try a different strategy. Let's consider the problem for small values of (n) to see if I can spot a pattern or come up with an inductive step.For (n = 2), as I saw earlier, it works. For (n = 3), let's see. Suppose (a_1 = 1), (a_2 = 1), (a_3 = 1). The total sum is 3, which is odd, so we don't consider it. If (a_3 = 2), the total sum is 4, which is even. Then, possible sign combinations:1 + 1 + 2 = 41 + 1 - 2 = 0So, we can get zero. If (a_3 = 3), the total sum is 5, which is odd, so we don't consider it. So, for (n = 3), it works.For (n = 4), let's take (a_1 = 1), (a_2 = 1), (a_3 = 1), (a_4 = 1). Total sum is 4, even. Possible sign combinations:1 + 1 + 1 + 1 = 41 + 1 + 1 - 1 = 21 + 1 - 1 + 1 = 21 + 1 - 1 - 1 = 0So, we can get zero. If (a_4 = 2), total sum is 5, which is odd, so we don't consider it. If (a_4 = 3), total sum is 6, even. Then, possible sign combinations:1 + 1 + 1 + 3 = 61 + 1 + 1 - 3 = 0So, we can get zero. If (a_4 = 4), total sum is 7, which is odd, so we don't consider it.Okay, so for (n = 4), it also works. Maybe induction is the way to go.Assume that for (n) numbers, the statement holds. Now, for (n + 1), consider the total sum (S = a_1 + a_2 + ldots + a_{n+1}), which is even. If I can find a subset of the first (n) numbers that sums to (a_{n+1}), then I can assign positive signs to that subset and negative signs to the rest, including (a_{n+1}), resulting in a total sum of zero.But how do I ensure that such a subset exists? Since (a_{n+1} leq n + 1), and the sum of the first (n) numbers is (S - a_{n+1}), which is even, maybe I can use the induction hypothesis on the first (n) numbers to find such a subset.Wait, the induction hypothesis says that for the first (n) numbers, there's a way to assign signs to get zero. But I need a subset that sums to (a_{n+1}), not necessarily zero. Hmm, maybe I need to adjust the induction hypothesis.Alternatively, maybe I can consider the problem differently. Let's think about the possible partial sums when assigning signs to the first (n) numbers. There are (2^n) possible sums, ranging from (-T) to (T), where (T = a_1 + a_2 + ldots + a_n). Since (T) is even (because (S) is even and (a_{n+1}) is either even or odd, but (S - a_{n+1}) is even), the number of possible partial sums is (2T + 1). If (2^n > 2T + 1), then by the Pigeonhole Principle, some sums must repeat, but I'm not sure how that helps.Wait, another idea: consider the partial sums modulo (a_{n+1}). There are (a_{n+1}) possible residues. If I have more partial sums than residues, then by the Pigeonhole Principle, two partial sums must be congruent modulo (a_{n+1}), and their difference would be a multiple of (a_{n+1}). If that difference is exactly (a_{n+1}), then I can adjust the signs accordingly.But I'm not sure if this guarantees that the difference is exactly (a_{n+1}). It could be a multiple, but not necessarily the exact value.Wait, maybe I can use the fact that (a_{n+1} leq n + 1) and the sum of the first (n) numbers is (S - a_{n+1}), which is even. If I can find a subset of the first (n) numbers that sums to (a_{n+1}), then I can assign positive signs to that subset and negative signs to the rest, including (a_{n+1}), resulting in a total sum of zero.But how do I ensure that such a subset exists? Maybe by considering the partial sums and using the Pigeonhole Principle again. If the sum of the first (n) numbers is (T = S - a_{n+1}), which is even, and (a_{n+1} leq n + 1), then the number of possible subset sums is large enough to cover (a_{n+1}).Wait, I'm going in circles here. Maybe I should try a different approach altogether. Let's think about the problem in terms of linear algebra. We're looking for a vector of signs (epsilon_i in {+1, -1}) such that (sum_{i=1}^{n+1} epsilon_i a_i = 0). This is equivalent to finding a solution to the equation (sum_{i=1}^{n+1} epsilon_i a_i = 0).Since the total sum (S) is even, we can write (S = 2k) for some integer (k). We need to find a subset of the (a_i)s that sums to (k), and assign positive signs to that subset and negative signs to the rest. This would give a total sum of (k - (S - k) = 2k - S = 0).So, the problem reduces to finding a subset of the (a_i)s that sums to (k). Given that each (a_i leq i), maybe this ensures that such a subset exists.Wait, I think I'm onto something. Since each (a_i leq i), the numbers are relatively small compared to their indices, which might make it easier to find such a subset. Maybe I can use a greedy algorithm approach, selecting the largest possible numbers first to reach (k).But I'm not sure if that always works. Maybe I need a more rigorous approach. Let's consider the partial sums again. If I consider all possible subset sums of the first (n) numbers, there are (2^n) possible sums. The maximum possible sum is (T = a_1 + a_2 + ldots + a_n), which is (S - a_{n+1}). Since (S) is even and (a_{n+1}) is either even or odd, (T) is even or odd accordingly.But I need to find a subset sum equal to (a_{n+1}). Since (a_{n+1} leq n + 1) and (T = S - a_{n+1}) is even, maybe the number of possible subset sums is large enough to cover (a_{n+1}).Wait, another idea: consider the set of all possible subset sums of the first (n) numbers. If this set includes (a_{n+1}), then we're done. If not, maybe we can adjust the signs in a way that allows us to reach zero.Alternatively, maybe I can use the fact that the numbers are bounded by their indices to ensure that the subset sums cover all residues modulo something, forcing the existence of a subset sum equal to (a_{n+1}).I'm still not entirely sure, but I think the key idea is to use induction and the fact that the numbers are bounded by their indices to ensure that a subset with the desired sum exists. Maybe I can formalize this by considering the partial sums and using the Pigeonhole Principle to show that a subset sum equal to (a_{n+1}) must exist.In summary, I think the approach is to use induction, assuming the statement holds for (n) and then showing it holds for (n + 1) by finding a subset of the first (n) numbers that sums to (a_{n+1}), allowing us to assign signs to achieve a total sum of zero. The boundedness of each (a_k) by (k) ensures that such a subset exists, possibly through the Pigeonhole Principle or another combinatorial argument.

💡Okay, so I have this problem about an isosceles triangle where two sides are 13 cm and 6 cm, and I need to find the length of the third side. The options are A: 7 cm, B: 13 cm, C: 6 cm, and D: 8 cm. Hmm, let me think about this step by step.First, I remember that in an isosceles triangle, two sides are equal in length. So, there are two possibilities here: either the two equal sides are both 13 cm, or both 6 cm. Let me explore both scenarios.**Case 1: The two equal sides are 13 cm each.**If that's the case, then the sides of the triangle would be 13 cm, 13 cm, and 6 cm. Now, I need to check if this satisfies the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side.Let's check:- 13 + 13 > 6? Yes, because 26 > 6.- 13 + 6 > 13? Yes, because 19 > 13.- 13 + 6 > 13? Again, yes, same as above.So, this configuration works. The third side is 6 cm.**Case 2: The two equal sides are 6 cm each.**In this case, the sides would be 6 cm, 6 cm, and 13 cm. Let me check the triangle inequality here too.- 6 + 6 > 13? Hmm, 12 > 13? No, that's not true. 12 is not greater than 13.- 6 + 13 > 6? Yes, 19 > 6.- 6 + 13 > 6? Yes, same as above.Since one of the inequalities doesn't hold (12 is not greater than 13), this configuration doesn't form a valid triangle. So, this case is not possible.Therefore, the only valid configuration is when the two equal sides are 13 cm each, making the third side 6 cm. Looking back at the options, that's option C: 6 cm.Wait a minute, but I just realized something. The problem says it's an isosceles triangle with two sides of 13 cm and 6 cm. Does that mean the two equal sides are 13 cm and 6 cm? That doesn't make sense because in an isosceles triangle, the two equal sides must be the same length. So, either both are 13 cm or both are 6 cm.Given that, and based on the triangle inequality, the only valid option is that the two equal sides are 13 cm, making the third side 6 cm. So, the answer should be C: 6 cm.But wait, let me double-check. If the two equal sides are 13 cm, then the third side is 6 cm, which is shorter. That seems okay because the triangle inequality holds. If I had tried to make the two equal sides 6 cm, the third side would have been too long, violating the triangle inequality.So, yeah, I'm pretty confident now that the third side is 6 cm, which is option C.