💡Alright, so I've got this geometry problem here about a quadrilateral EFGH. Let me try to visualize it first. The problem says that EF is perpendicular to FH, and GH is equal to FH. Also, EF is 12 cm, FH is 7 cm, and there's a height from G perpendicular to FH that extends 5 cm beyond FH. Hmm, okay, let's break this down step by step.First, I need to figure out what shape EFGH is. Since EF is perpendicular to FH, and there's a height from G that extends beyond FH, it might be some kind of trapezoid or maybe a kite. But given that GH equals FH, which is 7 cm, and EF is 12 cm, it's probably not a regular shape. Maybe it's a right trapezoid because of the perpendicular sides.Let me sketch this out in my mind. Let's say point F is at the origin of a coordinate system for simplicity. Then, since EF is perpendicular to FH, and EF is 12 cm, I can place point E at (0, 12). Point H would be somewhere along the x-axis since FH is 7 cm, so H would be at (7, 0). Now, point G is connected to H, and GH is equal to FH, which is 7 cm. Also, there's a height from G perpendicular to FH that extends 5 cm beyond FH. That means if I draw a perpendicular line from G to FH, it goes 5 cm beyond point H.Wait, so if FH is 7 cm and the height extends 5 cm beyond, does that mean the total length of this perpendicular from G is 7 + 5 = 12 cm? Or is it that the extension beyond FH is 5 cm, making the total height from G to FH equal to 5 cm? Hmm, the problem says "extends 5 cm beyond FH," so I think it's the latter. That is, the height from G to FH is 5 cm beyond FH, meaning the height itself is 5 cm.But then, if GH is 7 cm and the height from G to FH is 5 cm, we can use the Pythagorean theorem to find the horizontal distance from G to H. Let me denote the foot of the perpendicular from G to FH as point K. So, triangle GHK is a right triangle with GH = 7 cm, GK = 5 cm, and HK as the horizontal distance. Using Pythagoras:HK = sqrt(GH^2 - GK^2) = sqrt(7^2 - 5^2) = sqrt(49 - 25) = sqrt(24) = 2*sqrt(6) cm.So, point G is 2*sqrt(6) cm to the right of H along the x-axis, and 5 cm above H. Since H is at (7, 0), G would be at (7 + 2*sqrt(6), 5).Now, I need to find the lengths of all sides to calculate the perimeter. We already know EF = 12 cm, FH = 7 cm, and GH = 7 cm. We need to find EG and maybe another side if I'm missing something.Wait, quadrilateral EFGH has sides EF, FG, GH, and HE. I have EF, FH, and GH. But FH is a diagonal, not a side. So, I need to find the lengths of FG and HE.Wait, no, EFGH is a quadrilateral, so the sides are EF, FG, GH, and HE. I have EF = 12 cm, GH = 7 cm. I need FG and HE.But I have coordinates for E, F, H, and G now. Let me write them down:- E: (0, 12)- F: (0, 0)- H: (7, 0)- G: (7 + 2*sqrt(6), 5)Now, let's find the lengths of FG and HE.First, FG is from F(0,0) to G(7 + 2*sqrt(6), 5). Using the distance formula:FG = sqrt[(7 + 2*sqrt(6) - 0)^2 + (5 - 0)^2] = sqrt[(7 + 2*sqrt(6))^2 + 25]Let me expand (7 + 2*sqrt(6))^2:= 7^2 + 2*7*2*sqrt(6) + (2*sqrt(6))^2= 49 + 28*sqrt(6) + 24= 73 + 28*sqrt(6)So, FG = sqrt(73 + 28*sqrt(6) + 25) = sqrt(98 + 28*sqrt(6))Hmm, that seems complicated. Maybe I made a mistake. Wait, no, the distance formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2], so I think I did it right.Now, for HE, which is from H(7,0) to E(0,12). Using the distance formula:HE = sqrt[(0 - 7)^2 + (12 - 0)^2] = sqrt[49 + 144] = sqrt[193] cm.Okay, so HE is sqrt(193) cm.Now, let's summarize the sides:- EF = 12 cm- FG = sqrt(98 + 28*sqrt(6)) cm- GH = 7 cm- HE = sqrt(193) cmWait, but the problem mentions that GH = FH, which is 7 cm, so that's correct. But I'm not sure if I need to find another side or if I'm missing something.Wait, maybe I misinterpreted the shape. Let me think again. If EF is perpendicular to FH, and GH = FH, and there's a height from G perpendicular to FH extending 5 cm beyond FH, perhaps EFGH is a kite with two sides equal.In a kite, two pairs of adjacent sides are equal. Given that GH = FH = 7 cm, and EF = 12 cm, maybe EF and EH are the other pair? But I'm not sure.Alternatively, maybe it's a trapezoid with EF and GH being the two parallel sides. But EF is 12 cm and GH is 7 cm, so they can't be parallel unless it's an isosceles trapezoid, but I don't think that's the case here.Wait, maybe I should consider the coordinates again. E is at (0,12), F at (0,0), H at (7,0), and G at (7 + 2*sqrt(6),5). So, the sides are:- EF: from (0,12) to (0,0) = 12 cm- FG: from (0,0) to (7 + 2*sqrt(6),5) = sqrt(98 + 28*sqrt(6)) cm- GH: from (7 + 2*sqrt(6),5) to (7,0) = 7 cm- HE: from (7,0) to (0,12) = sqrt(193) cmSo, the perimeter would be the sum of these four sides:Perimeter = 12 + sqrt(98 + 28*sqrt(6)) + 7 + sqrt(193)But that seems too complicated. Maybe there's a simpler way to find FG.Wait, let's think about triangle FGH. We have FH = 7 cm, GH = 7 cm, and the height from G to FH is 5 cm. So, triangle FGH is an isosceles triangle with sides FH = GH = 7 cm and height 5 cm. Therefore, the base FG can be found using the Pythagorean theorem.In an isosceles triangle, the height splits the base into two equal parts. So, if the height is 5 cm, then each half of the base is sqrt(7^2 - 5^2) = sqrt(49 - 25) = sqrt(24) = 2*sqrt(6) cm. Therefore, the full base FG is 2*2*sqrt(6) = 4*sqrt(6) cm.Wait, that makes more sense. So, FG = 4*sqrt(6) cm.Now, let's recalculate the perimeter:- EF = 12 cm- FG = 4*sqrt(6) cm- GH = 7 cm- HE = sqrt(193) cmSo, Perimeter = 12 + 4*sqrt(6) + 7 + sqrt(193) = 19 + 4*sqrt(6) + sqrt(193) cm.But wait, earlier I thought HE was sqrt(193) cm, which is correct because from (7,0) to (0,12) is sqrt(7^2 + 12^2) = sqrt(49 + 144) = sqrt(193).So, the perimeter is 19 + 4*sqrt(6) + sqrt(193) cm.But the problem didn't specify whether to simplify further or if there's a numerical value, but since it's asking for the perimeter in cm, and the answer is likely to be in terms of square roots, I think this is the final form.Wait, but in the initial thought process, I thought it was a kite, but now it seems more like a trapezoid with two sides equal. Maybe I should double-check.Alternatively, perhaps I can find EG instead of FG, but I think FG is the correct side to find.Wait, no, in the problem, it's quadrilateral EFGH, so the sides are EF, FG, GH, and HE. So, I think I have all the sides now.Therefore, the perimeter is 12 + 4*sqrt(6) + 7 + sqrt(193) = 19 + 4*sqrt(6) + sqrt(193) cm.But let me check if I can simplify this further or if I made a mistake in calculating FG.Earlier, I thought FG was sqrt(98 + 28*sqrt(6)), but then I realized that in the isosceles triangle FGH, FG should be 4*sqrt(6). Let me verify that.In triangle FGH, with FH = GH = 7 cm and height 5 cm, the base FG is split into two equal parts by the height. So, each half is sqrt(7^2 - 5^2) = sqrt(24) = 2*sqrt(6). Therefore, FG = 2*2*sqrt(6) = 4*sqrt(6). Yes, that's correct.So, FG = 4*sqrt(6) cm.Therefore, the perimeter is 12 + 4*sqrt(6) + 7 + sqrt(193) = 19 + 4*sqrt(6) + sqrt(193) cm.I think that's the correct perimeter.

💡Alright, so I have this quadratic function: f(x) = ax² + (2a - 1)x - 3, where a is not zero. The problem says that the maximum value of this function in the interval [-3/2, 2] is 1, and I need to find the value of a. Hmm, okay, let's break this down step by step.First, since it's a quadratic function, its graph is a parabola. The coefficient 'a' determines whether it opens upwards or downwards. If a is positive, it opens upwards, and if a is negative, it opens downwards. Since we're talking about a maximum value, that means the parabola must open downwards, right? So, a should be negative. Wait, no, hold on. The maximum value could also occur at one of the endpoints of the interval, even if the parabola opens upwards. Hmm, so maybe a could be positive or negative. I need to consider both possibilities.Let me recall that for a quadratic function f(x) = ax² + bx + c, the vertex occurs at x = -b/(2a). This is the point where the function reaches its maximum or minimum. Since we're dealing with a maximum, if the parabola opens downward, the vertex is the maximum point. If it opens upward, the maximum would be at one of the endpoints of the interval.So, in this case, the vertex is at x = -(2a - 1)/(2a). Let me write that down: x_vertex = -(2a - 1)/(2a). Simplifying that, it's x_vertex = (1 - 2a)/(2a). Okay, so that's the x-coordinate of the vertex.Now, depending on the value of 'a', this vertex could lie inside or outside the interval [-3/2, 2]. If the vertex is inside the interval, then the maximum value (if the parabola opens downward) would be at the vertex. If the vertex is outside the interval, then the maximum would be at one of the endpoints.So, I think I need to consider two cases:1. The vertex is inside the interval [-3/2, 2], and the maximum is at the vertex.2. The vertex is outside the interval, so the maximum is at one of the endpoints, either x = -3/2 or x = 2.Additionally, I should also consider whether the parabola opens upwards or downwards because that affects whether the vertex is a maximum or minimum.Let me first check if the vertex is inside the interval. So, I need to find when (1 - 2a)/(2a) is between -3/2 and 2.So, let's solve the inequality:-3/2 ≤ (1 - 2a)/(2a) ≤ 2But before that, I should note that a ≠ 0, as given.Let me solve the left inequality first:(1 - 2a)/(2a) ≥ -3/2Multiply both sides by 2a, but I have to be careful because if a is positive, the inequality sign remains the same, but if a is negative, it flips. Hmm, this could get complicated. Maybe I can approach this differently.Alternatively, let's consider the position of the vertex relative to the interval.Case 1: Vertex is inside the interval.Case 2: Vertex is to the left of the interval.Case 3: Vertex is to the right of the interval.So, let's define:Case 1: -3/2 ≤ (1 - 2a)/(2a) ≤ 2Case 2: (1 - 2a)/(2a) < -3/2Case 3: (1 - 2a)/(2a) > 2For each case, I can analyze where the maximum occurs.But perhaps it's better to first compute f(-3/2) and f(2) and set them equal to 1, since the maximum is 1, and see what values of 'a' satisfy that.So, let's compute f(-3/2):f(-3/2) = a*(-3/2)² + (2a - 1)*(-3/2) - 3Compute each term:(-3/2)² = 9/4, so first term is (9/4)aSecond term: (2a - 1)*(-3/2) = (-3/2)(2a) + (-3/2)(-1) = -3a + 3/2Third term: -3So, f(-3/2) = (9/4)a - 3a + 3/2 - 3Simplify:(9/4)a - 3a = (9/4 - 12/4)a = (-3/4)a3/2 - 3 = 3/2 - 6/2 = (-3/2)So, f(-3/2) = (-3/4)a - 3/2Set this equal to 1:(-3/4)a - 3/2 = 1Multiply both sides by 4 to eliminate denominators:-3a - 6 = 4-3a = 10a = -10/3Okay, so if a = -10/3, then f(-3/2) = 1.Now, let's compute f(2):f(2) = a*(2)² + (2a - 1)*(2) - 3Compute each term:2² = 4, so first term is 4aSecond term: (2a - 1)*2 = 4a - 2Third term: -3So, f(2) = 4a + 4a - 2 - 3 = 8a - 5Set this equal to 1:8a - 5 = 18a = 6a = 6/8 = 3/4So, if a = 3/4, then f(2) = 1.Now, so we have two possible values for 'a': -10/3 and 3/4.But we need to check whether these values actually result in the maximum being 1 in the interval.First, let's check a = -10/3.If a = -10/3, then the function is f(x) = (-10/3)x² + (2*(-10/3) - 1)x - 3Simplify:2*(-10/3) = -20/3, so 2a - 1 = -20/3 - 1 = -23/3So, f(x) = (-10/3)x² - (23/3)x - 3Since a is negative, the parabola opens downward, so the vertex is a maximum.We need to check if the vertex is within the interval [-3/2, 2].Compute x_vertex = (1 - 2a)/(2a)Plug in a = -10/3:x_vertex = (1 - 2*(-10/3))/(2*(-10/3)) = (1 + 20/3)/(-20/3) = (23/3)/(-20/3) = -23/20 = -1.15Wait, -1.15 is within the interval [-1.5, 2], right? Because -1.5 is -3/2, which is -1.5, and -1.15 is greater than -1.5, so it's inside the interval.So, the maximum occurs at x = -23/20, and f(-23/20) should be 1.But wait, we set f(-3/2) = 1, but the maximum is actually at x = -23/20, which is inside the interval. So, does f(-23/20) equal 1?Wait, let's compute f(-23/20) when a = -10/3.f(-23/20) = (-10/3)*(-23/20)² + (-23/3)*(-23/20) - 3First, compute (-23/20)² = (529/400)So, first term: (-10/3)*(529/400) = (-10*529)/(3*400) = (-5290)/1200 = -529/120 ≈ -4.408Second term: (-23/3)*(-23/20) = (529/60) ≈ 8.8167Third term: -3So, total f(-23/20) ≈ -4.408 + 8.8167 - 3 ≈ 1.4087Wait, that's approximately 1.4087, which is greater than 1. But the problem states that the maximum is 1. So, this is a contradiction. Therefore, a = -10/3 cannot be a solution because even though f(-3/2) = 1, the maximum at the vertex is higher than 1, which violates the condition.Therefore, a = -10/3 is not a valid solution.Now, let's check a = 3/4.So, a = 3/4, which is positive, so the parabola opens upward. Therefore, the vertex is a minimum, not a maximum. So, the maximum must occur at one of the endpoints.We already set f(2) = 1, so let's check f(-3/2) as well to ensure that it's not greater than 1.Compute f(-3/2) when a = 3/4:f(-3/2) = (3/4)*(-3/2)² + (2*(3/4) - 1)*(-3/2) - 3Compute each term:(-3/2)² = 9/4, so first term: (3/4)*(9/4) = 27/16 ≈ 1.6875Second term: (2*(3/4) - 1) = (3/2 - 1) = 1/2. So, (1/2)*(-3/2) = -3/4 = -0.75Third term: -3So, f(-3/2) = 27/16 - 3/4 - 3Convert to sixteenths:27/16 - 12/16 - 48/16 = (27 - 12 - 48)/16 = (-33)/16 ≈ -2.0625So, f(-3/2) ≈ -2.0625, which is less than 1. Therefore, the maximum is indeed at x = 2, which is 1.So, a = 3/4 is a valid solution.Wait, but earlier, when I considered the vertex, I thought that if the vertex is inside the interval, and the parabola opens downward, then the maximum is at the vertex. But in this case, a = 3/4 is positive, so the parabola opens upward, so the vertex is a minimum, and the maximum is at the endpoints.But let me also check if there's another case where the maximum is at the vertex when the parabola opens downward, i.e., a < 0, and the vertex is inside the interval, but f(vertex) = 1.Wait, earlier, when I set a = -10/3, the vertex was inside the interval, but f(vertex) was approximately 1.4087, which was higher than 1, so that was invalid. But maybe there's another value of 'a' where the vertex is inside the interval and f(vertex) = 1.So, let's consider that case.Case 3: The vertex is inside the interval, and f(vertex) = 1.So, we need to find 'a' such that:1. The vertex x_vertex = (1 - 2a)/(2a) is within [-3/2, 2].2. f(x_vertex) = 1.So, let's compute f(x_vertex).Since x_vertex = (1 - 2a)/(2a), let's plug that into f(x):f(x_vertex) = a*(x_vertex)^2 + (2a - 1)*x_vertex - 3 = 1Let me compute this expression.First, x_vertex = (1 - 2a)/(2a)Let me denote x_vertex as h for simplicity.h = (1 - 2a)/(2a)Compute f(h):f(h) = a*h² + (2a - 1)*h - 3 = 1So,a*h² + (2a - 1)*h - 4 = 0Now, substitute h = (1 - 2a)/(2a):a*[(1 - 2a)/(2a)]² + (2a - 1)*[(1 - 2a)/(2a)] - 4 = 0Let me compute each term step by step.First term: a*[(1 - 2a)/(2a)]²= a*(1 - 4a + 4a²)/(4a²)= (1 - 4a + 4a²)/(4a)Second term: (2a - 1)*[(1 - 2a)/(2a)]= (2a - 1)(1 - 2a)/(2a)Note that (2a - 1)(1 - 2a) = -(2a - 1)^2So,= - (2a - 1)^2 / (2a)Third term: -4So, putting it all together:(1 - 4a + 4a²)/(4a) - (2a - 1)^2/(2a) - 4 = 0Multiply through by 4a to eliminate denominators:(1 - 4a + 4a²) - 2*(2a - 1)^2 - 16a = 0Now, expand each term:First term: 1 - 4a + 4a²Second term: -2*(4a² - 4a + 1) = -8a² + 8a - 2Third term: -16aCombine all terms:1 - 4a + 4a² -8a² + 8a - 2 -16a = 0Combine like terms:(4a² -8a²) + (-4a +8a -16a) + (1 -2) = 0(-4a²) + (-12a) + (-1) = 0So,-4a² -12a -1 = 0Multiply both sides by -1:4a² + 12a + 1 = 0Now, solve for 'a' using quadratic formula:a = [-12 ± sqrt(144 - 16)] / 8= [-12 ± sqrt(128)] / 8= [-12 ± 8*sqrt(2)] / 8= [-3 ± 2*sqrt(2)] / 2So, two possible solutions:a = [-3 + 2√2]/2 ≈ (-3 + 2.828)/2 ≈ (-0.172)/2 ≈ -0.086a = [-3 - 2√2]/2 ≈ (-3 - 2.828)/2 ≈ (-5.828)/2 ≈ -2.914Now, let's check if these values of 'a' satisfy the condition that the vertex is inside the interval [-3/2, 2].First, compute x_vertex for a = [-3 + 2√2]/2 ≈ -0.086x_vertex = (1 - 2a)/(2a)Plug in a ≈ -0.086:1 - 2*(-0.086) = 1 + 0.172 = 1.1722a ≈ 2*(-0.086) ≈ -0.172So, x_vertex ≈ 1.172 / (-0.172) ≈ -6.81Which is way outside the interval [-1.5, 2]. So, this solution is invalid.Now, check a = [-3 - 2√2]/2 ≈ -2.914Compute x_vertex:1 - 2a = 1 - 2*(-2.914) = 1 + 5.828 ≈ 6.8282a ≈ 2*(-2.914) ≈ -5.828So, x_vertex ≈ 6.828 / (-5.828) ≈ -1.17Which is within the interval [-1.5, 2]. So, this is a valid solution.Wait, but earlier, when I set a = -10/3, the vertex was inside the interval, but f(vertex) was higher than 1, which was invalid. Now, with a ≈ -2.914, which is approximately -3, let's compute f(vertex) to check if it's 1.Wait, but I already set f(vertex) = 1 when solving for 'a', so theoretically, it should be 1. But let me verify numerically.Compute a = [-3 - 2√2]/2 ≈ (-3 - 2.828)/2 ≈ -2.914Compute x_vertex ≈ -1.17Compute f(x_vertex):f(x) = ax² + (2a -1)x -3So, f(-1.17) ≈ (-2.914)*(-1.17)^2 + (2*(-2.914) -1)*(-1.17) -3First, (-1.17)^2 ≈ 1.3689So, first term: (-2.914)*(1.3689) ≈ -3.984Second term: (2*(-2.914) -1) = (-5.828 -1) = -6.828Multiply by (-1.17): (-6.828)*(-1.17) ≈ 7.979Third term: -3So, total f(-1.17) ≈ -3.984 + 7.979 -3 ≈ 0.995 ≈ 1Which is approximately 1, so that checks out.But wait, earlier, when I set a = -10/3, the vertex was inside the interval, but f(vertex) was higher than 1, which was invalid. However, in this case, with a ≈ -2.914, the vertex is inside the interval, and f(vertex) = 1, which is exactly the maximum value. So, this is a valid solution.But wait, I thought earlier that a = 3/4 was the only solution, but now I have another solution with a ≈ -2.914.But let me express a exactly.We had a = [-3 ± 2√2]/2So, a = (-3 + 2√2)/2 and a = (-3 - 2√2)/2We saw that a = (-3 + 2√2)/2 ≈ -0.086, which resulted in x_vertex ≈ -6.81, outside the interval, so invalid.a = (-3 - 2√2)/2 ≈ -2.914, which resulted in x_vertex ≈ -1.17, inside the interval, so valid.Therefore, a = (-3 - 2√2)/2 is another solution.But wait, earlier, when I set f(-3/2) = 1, I got a = -10/3, which was invalid because the vertex was inside the interval and f(vertex) > 1.But now, with a = (-3 - 2√2)/2, the vertex is inside the interval, and f(vertex) = 1, which is the maximum.So, does this mean that there are two solutions: a = 3/4 and a = (-3 - 2√2)/2?Wait, let me check if a = (-3 - 2√2)/2 is equal to -10/3.Compute (-3 - 2√2)/2 ≈ (-3 - 2.828)/2 ≈ (-5.828)/2 ≈ -2.914-10/3 ≈ -3.333, so they are different.So, perhaps there are two solutions: a = 3/4 and a = (-3 - 2√2)/2.But wait, when a = (-3 - 2√2)/2, the parabola opens downward, so the vertex is a maximum, which is 1, and it's inside the interval, so that's valid.When a = 3/4, the parabola opens upward, so the maximum is at the endpoint x = 2, which is 1, and the other endpoint is lower, so that's also valid.Therefore, there are two solutions: a = 3/4 and a = (-3 - 2√2)/2.But wait, let me check if a = (-3 - 2√2)/2 is indeed a solution.Compute f(-3/2) when a = (-3 - 2√2)/2.f(-3/2) = a*(9/4) + (2a -1)*(-3/2) -3Plug in a = (-3 - 2√2)/2:First term: [(-3 - 2√2)/2]*(9/4) = [(-3 - 2√2)*9]/8 = (-27 - 18√2)/8Second term: [2*(-3 - 2√2)/2 -1]*(-3/2) = [(-3 - 2√2) -1]*(-3/2) = (-4 - 2√2)*(-3/2) = (12 + 6√2)/2 = 6 + 3√2Third term: -3So, f(-3/2) = (-27 - 18√2)/8 + 6 + 3√2 -3Simplify:Convert all terms to eighths:(-27 - 18√2)/8 + 48/8 + 24√2/8 -24/8Combine:(-27 -18√2 + 48 +24√2 -24)/8 = ( (-27 +48 -24) + (-18√2 +24√2) )/8 = ( -3 +6√2 )/8Which is approximately (-3 +8.485)/8 ≈ 5.485/8 ≈ 0.685, which is less than 1.Similarly, compute f(2) when a = (-3 - 2√2)/2:f(2) = a*4 + (2a -1)*2 -3= 4a + 4a -2 -3 = 8a -5Plug in a = (-3 - 2√2)/2:8a = 8*(-3 - 2√2)/2 = 4*(-3 - 2√2) = -12 -8√2So, f(2) = -12 -8√2 -5 = -17 -8√2 ≈ -17 -11.314 ≈ -28.314, which is much less than 1.Therefore, the maximum is indeed at the vertex, which is 1, so this is a valid solution.Therefore, we have two solutions: a = 3/4 and a = (-3 - 2√2)/2.But wait, earlier, when I set f(-3/2) =1, I got a = -10/3, which was invalid because the vertex was inside the interval and f(vertex) >1. So, that's not a solution.But when I set f(vertex)=1, I got a = (-3 - 2√2)/2, which is valid.So, in total, there are two solutions: a = 3/4 and a = (-3 - 2√2)/2.But let me check if a = (-3 - 2√2)/2 is the same as a = -10/3.Compute (-3 - 2√2)/2 ≈ (-3 -2.828)/2 ≈ -5.828/2 ≈ -2.914-10/3 ≈ -3.333, so they are different.Therefore, the two solutions are a = 3/4 and a = (-3 - 2√2)/2.But let me write them in exact form:a = 3/4 and a = (-3 - 2√2)/2But let me simplify (-3 - 2√2)/2:= (-3/2) - √2So, a = -3/2 - √2Alternatively, a = (-3 - 2√2)/2So, both forms are acceptable.Therefore, the values of 'a' are 3/4 and (-3 - 2√2)/2.But wait, let me check if a = (-3 - 2√2)/2 is indeed a solution.We already did that numerically, but let me verify algebraically.We had:4a² +12a +1 =0Solutions: a = [-12 ± sqrt(144 -16)]/8 = [-12 ± sqrt(128)]/8 = [-12 ± 8√2]/8 = [-3 ± 2√2]/2So, yes, that's correct.Therefore, the two solutions are a = 3/4 and a = (-3 - 2√2)/2.But wait, in the initial problem, it's stated that a ≠0, which is satisfied by both solutions.Therefore, the answer is a = 3/4 and a = (-3 - 2√2)/2.But let me check if the problem expects multiple answers or just one.The problem says "Find the value of a", so perhaps both are acceptable.But in the initial thought process, I thought only a = 3/4 was the solution, but upon further analysis, there's another solution when the vertex is inside the interval and the maximum is at the vertex.Therefore, the final answer is a = 3/4 and a = (-3 - 2√2)/2.But to express them neatly:a = boxed{dfrac{3}{4}} and a = boxed{dfrac{-3 - 2sqrt{2}}{2}}But wait, the problem might expect only one answer, but from the analysis, there are two solutions.Alternatively, perhaps I made a mistake in considering both cases.Wait, when a = (-3 - 2√2)/2, the vertex is inside the interval, and f(vertex)=1, which is the maximum.When a = 3/4, the vertex is outside the interval, and the maximum is at x=2, which is 1.Therefore, both are valid solutions.So, the answer is a = 3/4 and a = (-3 - 2√2)/2.But let me check if a = (-3 - 2√2)/2 is indeed different from a = -10/3.Yes, as we saw numerically, they are different.Therefore, the two solutions are correct.So, to conclude, the values of 'a' are 3/4 and (-3 - 2√2)/2.

💡Alright, so I've got this problem here: I need to find the coefficient of ( x^4 ) in the expansion of ( left(x^2 + frac{2}{x}right)^5 ). The options are A: 10, B: 20, C: 40, D: 80. Hmm, okay. I remember something about the binomial theorem from algebra class, so maybe I can use that here.Let me recall the binomial theorem. It says that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k ). So, in this case, ( a ) is ( x^2 ), ( b ) is ( frac{2}{x} ), and ( n ) is 5. That means each term in the expansion will look like ( binom{5}{k} (x^2)^{5 - k} left(frac{2}{x}right)^k ).Okay, so I need to find the term where the power of ( x ) is 4. Let me figure out what exponent of ( x ) each part contributes. The term ( (x^2)^{5 - k} ) will give me ( x^{2(5 - k)} ), which simplifies to ( x^{10 - 2k} ). Then, ( left(frac{2}{x}right)^k ) is the same as ( 2^k cdot x^{-k} ). So, combining these, the exponent of ( x ) in each term is ( 10 - 2k - k = 10 - 3k ).I need this exponent to be 4 because I'm looking for the coefficient of ( x^4 ). So, I can set up the equation ( 10 - 3k = 4 ) and solve for ( k ). Let's do that:( 10 - 3k = 4 )Subtract 10 from both sides:( -3k = -6 )Divide both sides by -3:( k = 2 )Okay, so the term where ( k = 2 ) will have ( x^4 ). Now, let's find the coefficient of that term. The general term is ( binom{5}{k} (x^2)^{5 - k} left(frac{2}{x}right)^k ). Plugging in ( k = 2 ):First, calculate the binomial coefficient ( binom{5}{2} ). I think that's 10 because ( binom{5}{2} = frac{5!}{2!(5-2)!} = frac{120}{2 times 6} = 10 ).Next, ( (x^2)^{5 - 2} = (x^2)^3 = x^6 ).Then, ( left(frac{2}{x}right)^2 = frac{4}{x^2} = 4x^{-2} ).Multiplying all these together: ( 10 times x^6 times 4x^{-2} = 10 times 4 times x^{6 - 2} = 40x^4 ).So, the coefficient of ( x^4 ) is 40. Looking back at the options, that's option C.Wait, let me double-check to make sure I didn't make a mistake. So, ( k = 2 ), binomial coefficient is 10, ( 2^2 = 4 ), so 10 times 4 is 40. Yeah, that seems right. The exponent calculation: ( 10 - 3k = 4 ) gives ( k = 2 ), which is correct. So, I think I did it right.I guess another way to think about it is to actually expand the expression step by step, but that might take longer. Let's see, ( left(x^2 + frac{2}{x}right)^5 ) can be expanded as:First, ( (x^2 + 2/x)^5 ). Each term is a combination of multiplying ( x^2 ) and ( 2/x ) five times. So, the general term is ( binom{5}{k} (x^2)^{5 - k} (2/x)^k ). Which is what I had before.So, yeah, I think 40 is the right coefficient. So, the answer should be C: 40.**Final Answer**boxed{C}

💡Okay, so I have this problem here: The sum of the areas of two squares is 85, and the difference in their areas is 41. I need to find the sum of their perimeters. Hmm, let me think about how to approach this.First, I know that the area of a square is calculated by squaring its side length. So, if I let the side length of the larger square be ( x ) and the side length of the smaller square be ( y ), then their areas would be ( x^2 ) and ( y^2 ) respectively.According to the problem, the sum of these areas is 85. So, I can write that as:[ x^2 + y^2 = 85 ]Also, the difference in their areas is 41. Since ( x ) is the larger square, its area minus the smaller one should be 41:[ x^2 - y^2 = 41 ]Now, I have two equations:1. ( x^2 + y^2 = 85 )2. ( x^2 - y^2 = 41 )I think I can solve these two equations simultaneously to find the values of ( x ) and ( y ). Maybe by adding or subtracting the equations?Let me try adding the two equations together:[ (x^2 + y^2) + (x^2 - y^2) = 85 + 41 ]Simplifying the left side:[ x^2 + y^2 + x^2 - y^2 = 2x^2 ]And the right side:[ 85 + 41 = 126 ]So, I get:[ 2x^2 = 126 ]Dividing both sides by 2:[ x^2 = 63 ]Taking the square root of both sides:[ x = sqrt{63} ]Hmm, ( sqrt{63} ) can be simplified since 63 is 9 times 7:[ x = sqrt{9 times 7} = 3sqrt{7} ]Okay, so the side length of the larger square is ( 3sqrt{7} ). Now, I need to find the side length of the smaller square, ( y ). I can use one of the original equations to find ( y ). Let's use the first equation:[ x^2 + y^2 = 85 ]We already know ( x^2 = 63 ), so:[ 63 + y^2 = 85 ]Subtracting 63 from both sides:[ y^2 = 22 ]Taking the square root:[ y = sqrt{22} ]Alright, so the side lengths are ( 3sqrt{7} ) and ( sqrt{22} ). Now, I need to find the perimeters of both squares and then add them together.The perimeter of a square is 4 times the side length. So, for the larger square:[ text{Perimeter}_1 = 4x = 4 times 3sqrt{7} = 12sqrt{7} ]And for the smaller square:[ text{Perimeter}_2 = 4y = 4 times sqrt{22} = 4sqrt{22} ]Adding these two perimeters together:[ text{Total Perimeter} = 12sqrt{7} + 4sqrt{22} ]I think that's the final answer, but let me double-check my steps to make sure I didn't make any mistakes.1. I set up the equations correctly based on the given areas.2. Adding the equations to eliminate ( y^2 ) gave me ( x^2 = 63 ), which seems right.3. Substituting back to find ( y^2 = 22 ) also makes sense.4. Calculating the perimeters by multiplying each side length by 4 is correct.So, I think my answer is correct. The sum of their perimeters is ( 12sqrt{7} + 4sqrt{22} ).

💡Okay, so I have this problem about two lines, l₁ and l₂, and I need to figure out whether the condition "a = -1" is sufficient, necessary, both, or neither for the lines to be perpendicular. Let me try to break this down step by step.First, I remember that two lines are perpendicular if the product of their slopes is -1. So, I need to find the slopes of l₁ and l₂. The equations are given as:l₁: x + a y - 2 = 0 l₂: x - a y - 1 = 0Hmm, these are in general form. I think it might be easier to find the slopes if I convert them into slope-intercept form, which is y = m x + b, where m is the slope.Starting with l₁: x + a y - 2 = 0. Let me solve for y.x + a y - 2 = 0 a y = -x + 2 y = (-1/a) x + 2/aSo, the slope of l₁ is -1/a.Now, for l₂: x - a y - 1 = 0. Let me solve for y here as well.x - a y - 1 = 0 - a y = -x + 1 y = (1/a) x - 1/aSo, the slope of l₂ is 1/a.Okay, now that I have both slopes, I can set up the condition for perpendicularity. The product of the slopes should be -1.Slope of l₁ * Slope of l₂ = (-1/a) * (1/a) = (-1)/(a²)This product should equal -1 for the lines to be perpendicular.So, (-1)/(a²) = -1Let me solve this equation for a.Multiply both sides by a²:-1 = -1 * a² -1 = -a²Multiply both sides by -1:1 = a²So, a² = 1Taking square roots of both sides:a = ±1Therefore, a can be either 1 or -1 for the lines to be perpendicular.Wait, so the problem states that "a = -1" is a condition for perpendicularity. But from my calculation, a can be either 1 or -1. So, does that mean that a = -1 is just one of the possible values? It seems like it's not the only condition.So, if a = -1 is one of the possible values that make the lines perpendicular, then it's a sufficient condition because if a is -1, the lines will definitely be perpendicular. But it's not necessary because a could also be 1, and the lines would still be perpendicular.Let me think again to make sure I didn't make a mistake. I converted both equations to slope-intercept form correctly, right?For l₁: x + a y - 2 = 0 a y = -x + 2 y = (-1/a) x + 2/a Yes, that seems correct.For l₂: x - a y - 1 = 0 - a y = -x + 1 y = (1/a) x - 1/a That also looks correct.Then, the product of the slopes is (-1/a)*(1/a) = -1/a². Setting that equal to -1 gives -1/a² = -1, so 1/a² = 1, which leads to a² = 1, so a = ±1. That seems solid.Therefore, a = -1 is sufficient because it ensures perpendicularity, but it's not necessary because a could also be 1. So, the condition is sufficient but not necessary.I think that makes sense. So, the answer should be option A: Sufficient but not necessary.**Final Answer**boxed{A}

💡First, I need to determine how much money Pete and Raymond each received from their grandmother. Pete spends 4 nickels. Since each nickel is worth 5 cents, Pete spends 4 * 5 = 20 cents.Raymond has 7 dimes left after spending. Each dime is worth 10 cents, so Raymond has 7 * 10 = 70 cents remaining.Together, they spent a total of 200 cents. This means that the sum of the money they received minus the money they have left equals 200 cents.Let P be the amount Pete received and R be the amount Raymond received. The equation can be written as:(P - 20) + (R - 70) = 200Simplifying this equation:P + R - 90 = 200P + R = 290This means that the total amount of money they received together is 290 cents. However, without additional information, I cannot determine the exact amount each of them received individually.