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📅 May 8, 2026 👤 scraface

May 08, 2026

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📌 5Question: Find the quadratic polynomial $ h(x) $ such that $ h(1) = 5 $, $ h(-1) = 3 $, and the sum of its roots is $ 4 $.

📌 Solution: Let $ h(x) = ax^2 + bx + c $. From $ h(1) = a + b + c = 5 $ and $ h(-1) = a - b + c = 3 $, adding gives $ 2a + 2c = 8 $, so $ a + c = 4 $. The sum of roots is $ -rac{b}{a} = 4 $, so $ b = -4a $. Substituting $ b = -4a $ into $ a + b + c = 5 $: $ a - 4a + c = 5 $ â $ -3a + c = 5 $. Since $ a + c = 4 $, subtracting gives $ -4a = 1 $, so $ a = -rac{1}{4} $. Then $ c = 4 - a = 4 + rac{1}{4} = rac{17}{4} $, and $ b = -4a = 1 $. Thus, $ h(x) = -rac{1}{4}x^2 + x + rac{17}{4} $. Multiplying by 4 to eliminate fractions: $ h(x) = -x^2 + 4x + 17 $. Verifying $ h(1) = -1 + 4 + 17 = 20 $? Wait, inconsistency. Rechecking: $ a = -1/4 $, $ c = 17/4 $, $ b = 1 $. Then $ h(1) = -1/4 + 1 + 17/4 = (-1 + 4 + 17)/4 = 20/4 = 5 $, correct. $ h(-1) = -1/4 -1 + 17/4 = ( -1 -4 + 17 )/4 = 12/4 = 3 $, correct. Sum of roots $ -b/a = -1 / (-1/4) = 4 $, correct. Final answer: $ oxed{-x^2 + 4x + \dfrac{17}{4}} $ or $ oxed{-\dfrac{1}{4}x^2 + x + \dfrac{17}{4}} $.

📌 Question: A science communicator observes that the number of views $ V(t) $ on a video grows quadratically over time $ t $ (in days). If $ V(1) = 120 $, $ V(2) = 200 $, and $ V(3) = 300 $, find $ V(4) $.

📌 Solution: Assume $ V(t) = at^2 + bt + c $. From $ V(1) = a + b + c = 120 $, $ V(2) = 4a + 2b + c = 200 $, $ V(3) = 9a + 3b + c = 300 $. Subtract first equation from the second: $ 3a + b = 80 $. Subtract second from the third: $ 5a + b = 100 $. Subtract these: $ 2a = 20 $ â $ a = 10 $. Then $ 3(10) + b = 80 $ â $ b = 50 $. From $ a + b + c = 120 $: $ 10 + 50 + c = 120 $ â $ c = 60 $. Thus, $ V(t) = 10t^2 + 50t + 60 $. For $ t = 4 $: $ V(4) = 10(16) + 50(4) + 60 = 160 + 200 + 60 = 420 $. Final answer: $ oxed{420} $.

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