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Home » Sum of roots \( = -racba = 3 + (-2) = 1 \). - Abu Waleed Tea

Sum of roots \( = -racba = 3 + (-2) = 1 \). - Abu Waleed Tea

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Substitute \( d = 3 \) into the equations:

📌 \[ a \sin(c) + 3 = 2 \quad \Rightarrow \quad a \sin(c) = -1 \]

📌 \[ a \cos(c) + 3 = 4 \quad \Rightarrow \quad a \cos(c) = 1 \]

📌 Square and add the equations:

📌 \[ (a \sin(c))^2 + (a \cos(c))^2 = (-1)^2 + 1^2 \]

📌 \[ a^2 (\sin^2(c) + \cos^2(c)) = 1 + 1 = 2 \]

📌 \[ a^2 = 2 \quad \Rightarrow \quad a = \sqrt{2} \]

📌 With \( a = \sqrt{2} \), use:

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\[ (27p + 9q + 3r + s) - (8p + 4q + 2r + s) = 39 - 14 \]
\[ 19p + 5q + r = 25 \quad \text{(E2)} \]
Subtract equation (3) from (4):
\[ (64p + 16q + 4r + s) - (27p + 9q + 3r + s) = 84 - 39 \]
\[ 37p + 7q + r = 45 \quad \text{(E3)} \]
Now subtract E1 from E2:
\[ (19p + 5q + r) - (7p + 3q + r) = 25 - 11 \]
\[ 12p + 2q = 14 \]
\[ 6p + q = 7 \quad \text{(E4)} \]
Subtract E2 from E3:

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