Question

If 2+3i is a root of equation x2 + ax + b =0, then (a, b) is
1) (4,-13
2) (-4,13)
3) (-4,-13)
4) (4,13)​​

Answer 1

$\Large\mathtt\green{ }\huge\underline\mathtt\red{Answer : }$

(2+root3i)^2+p×2+roit3 i)+q=0

4-3+4root3i+2ptroot3pitq=0

(1+2ptg)4root3+root3p=0

2ptq-1

p-4. sorry i use a,b instead p,q

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Answer 2

(2+root3i)^2+p×2+roit3 i)+q=0

(2+root3i)^2+p×2+roit3 i)+q=04-3+4root3i+2ptroot3pitq=0

(2+root3i)^2+p×2+roit3 i)+q=04-3+4root3i+2ptroot3pitq=0(1+2ptg)4root3+root3p=0

(2+root3i)^2+p×2+roit3 i)+q=04-3+4root3i+2ptroot3pitq=0(1+2ptg)4root3+root3p=02ptq-1

(2+root3i)^2+p×2+roit3 i)+q=04-3+4root3i+2ptroot3pitq=0(1+2ptg)4root3+root3p=02ptq-1p-4. sorry i use a,b instead p,q