Question

If 3+4 i3+4i is a root of x^{2}+p x+q=0x
2
+px+q=0 then (p, q)=(p,q)=

Answer 1

Given:

3 + 4i is a root of x^{2} + p x + q = 0

To find:

(p, q) = ?

Solution:

From given, we have,

a quadratic equation x^2 + p x + q = 0

Given that 3 + 4i  is a root of the above equation.

Therefore, this value should satisfy the given equation.

So, we have,

(3 + 4i)^2 + p (3 + 4i) + q = 0

9 - 16 + 2 × 3 × 4i + p (3 + 4i) + q = 0

-7 + 24i + p (3 + 4i) + q = 0

-7 + 24i + 3p + 4i p + q = 0

rearranging the terms, we get,

(-7 + 3p + q) + i(24 + 4p) = 0 + 0i

equating the respective, we get,

-7 + 3p + q = 0 ........(1)

24 + 4p = 0 ........(2)

Now, consider the equation (2),

24 + 4p = 0

4p = - 24

p = - 6

using the value of p in equation (1), we get,

-7 + 3p + q = 0

-7 + 3 (-6) + q = 0

-7 - 18 + q = 0

-25 + q = 0

q = 25

Therefore, (p, q) = (-6, 25)

Answer 2

Given :    3 + 4i    is a  root  of x² + px + q = 0

To find : (p , q)

Solution:

 3 + 4i    is a  root  of x² + px + q = 0

Complex roots are always in conjugate pairs

Hence another root would be

3 - 4 i

Sum of Roots  =  3 +  4i  + 3 -  4i    = 6

Product of Roots =   (3 +  4i) (3 -  4i)  = 9  - 16i²  = 9 + 16  = 25  ( ∵ i² = - 1)

x² + px + q = 0

Sum of Roots  = - p

Profduct of roots = q

Equating both

-p = 6  => p = - 6

& q = 25

(p,q) = (-6 , 25)

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