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Exercise 3.2 · Q2

Q.Find a polynomial equation of minimum degree with rational coefficients, having 2+3 i2+\sqrt3\,i as a root.

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Step 1. Identify the automatic second root. Since the required coefficients are rational, and 2+3i2+\sqrt3i is a root, its conjugate 2−3i2-\sqrt3i (Complex Conjugate Root Theorem) is also a root.

Step 2. Compute sum and product. Sum =(2+3i)+(2−3i)=4=(2+\sqrt3i)+(2-\sqrt3i)=4. Product =(2)2+(3)2=4+3=7=(2)^2+(\sqrt3)^2=4+3=7 (using (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2 with a=2,b=3a=2,b=\sqrt3).

Step 3. Form the quadratic. x2−(sum)x+(product)=0  ⟹  x2−4x+7=0x^2-(\text{sum})x+(\text{product})=0 \implies x^2-4x+7=0.

✓Final answer

x2−4x+7=0\boxed{x^2-4x+7=0}

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