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

Step-by-step explanation:

(2+root3i)^2+p*2+roit3 i)+q=0

4-3+4root3i+2p+root3pi+q=0

(1+2p+q)(4root3+root3p=0

2p+q=-1

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

Answer 2

Answer:

( a, b) = ( -4, 13)

Step-by-step explanation:

given equation is :

x² + ax + b = 0

we need to find the values of 'a' and 'b' where one root of the above equation is ( 2 + 3i )

so, substitute value of x= 2+ 3i  in  x² + ax + b = 0

(2+3i)² + a(2+3i) + b =0

2² + (3i)² + 2a + 3ai + b = 0

i² = (i)(i) = -1

4 + 9i² + 2a + 3ai + b = 0

4 - 9 + 2a + 3ai + b=0

a(2+3i) + b - 5 = 0

if we could substitute value of a,b as (-4 , 13)

we get;

(-4)(2+3i) + 13 - 5 =0

-8 - 12i + 8 = 0

-12i ≈ 0

where i = √-1

nearly 0

so ( a, b) = ( -4, 13)