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Home » Alternatively, suppose \( a = 2 \), but it’s given as 3. - go-checkin.com

Alternatively, suppose \( a = 2 \), but it’s given as 3. - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 x^2 + x^2 + 4x + 4 = 16

📌 2x^2 + 4x + 4 - 16 = 0

📌 2x^2 + 4x - 12 = 0

📌 x^2 + 2x - 6 = 0

📌 x = rac{-2 \pm \sqrt{4 + 24}}{2} = rac{-2 \pm \sqrt{28}}{2} = rac{-2 \pm 2\sqrt{7}}{2} = -1 \pm \sqrt{7}

📌 The corresponding \(y\)-coordinates are \(y = x + 2\), so:

📌 y = (-1 + \sqrt{7}) + 2 = 1 + \sqrt{7}, \quad y = (-1 - \sqrt{7}) + 2 = 1 - \sqrt{7}

📌 The points of intersection are \(A = (-1 + \sqrt{7}, 1 + \sqrt{7})\) and \(B = (-1 - \sqrt{7}, 1 - \sqrt{7})\).

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Thus, the polynomial is:
f(x) = -\frac{7}{6}x^3 + \frac{21}{2}x^2 - \frac{82}{3}x + 21
\boxed{-\frac{7}{6}x^3 + \frac{21}{2}x^2 - \frac{82}{3}x + 21}
Let \( a \) and \( b \) be complex numbers such that \( a + b = 4 + 2i...
We start with the identity for the square of a sum:
Rearranging gives:
a^2 + b^2 = (a + b)^2 - 2ab
Substitute the given values \( a + b = 4 + 2i \) and \( ab = 5 + 3i \)...
a^2 + b^2 = (4 + 2i)^2 - 2(5 + 3i)
First, compute \( (4 + 2i)^2 \):

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