Question

If 3+4i is a root of the equation
x + pr+q=0 (p, q are real numbers), then
(a)p=6,q=25
(b)p=6,q=1
(c)p=-6,q=-7
(d)p=-6,q=25​

Answer 1

Answer:

If 3+4i is a root of x^2+px +q= 0 , then other root will be 3–4i .

Sum of roots = 3+4i +3–4i = - coeff. Of x/ coeff.of x^2= - p/1.

or 6 = - p/1 , or p= -6 .Answer.

Product of roors =(3+4i) .(3–4i) = constant term/ oeff. of x^2.

or 9 - 16i^2= q/1 =>9+16= q or q = 25 . Answer.

Answer 2

Step-by-step explanation:

given

root of an equation=3+4i

sum of roots of and equation=-b/a

given equation=x^2+px+q=0

$\sf{sum\:of \: the \: root \: \: = \frac{ - p}{q} }\sf{if\:one\:the \:root\:is \:3 + 4i}\\ \sf{then\:another\:root\:is\: 3 - 4i} \\sum of roots=-p=6$

$\sf{\bold{\rightarrow}\:product\:of\:roots\: = (3 + 4i)(3 - 4i) = 9 - 16i^2 = 25 = q}$