Question

Help with two questions? (Questions Inside)

Answer 1

@rishavraj quick help?

Answer 2

find out atleast one real root by hit and trial method ...according to options real roots can be 2 or -2....
check which one is correct first of all

Answer 3

You should probably first use your Rational Root Theorem and find the Rational Roots.

Answer 4

see u will get tht only 2 satisfies the equation......and bingo u got only one option with 2 as only real root :))

Answer 5

Yes I got 2 as well, what do you think it'd be?

Answer 6

@rishavraj

Answer 7

it means theres only one real root ....check out ur options theres only one option havng tht

Answer 8

and obviously tht real root is 2

Answer 9

B?

Answer 10

yup.... )))

Answer 11

Cool, lets move onto the next one?

Answer 12

okkk

Answer 13

on manipulation the equation becomes
\[x^3 + 3x^2 + 6x + 4 = 0\]

Answer 14

go for hit and trial to get real root.... and as option suggest it might be 1 or -1 so try them

Answer 15

@LetsLearn2000 check this out...would be really helpful
http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-cubicequations-2009-1.pdf

Answer 16

Would it be -1? and ill check the link out rn

Answer 17

because in these questions u got such options which make then really easy..

Answer 18

yup its -1

Answer 19

because u see if the options were given in such a way tht all the real roots were same in all options thn u were supposed to solve it further

Answer 20

and one more this see there are two options having only -1 as a real root...so keep in mind tht complex roots come in conjugate pair

Answer 21

i just mentioned complex roots appear with their conjugate

Answer 22

A?

Answer 23

Or wait would it be B?

Answer 24

@LetsLearn2000 oops my bad no wi noticed the options....lol u need to check between option A and B.....

Answer 25

so now u need to solve ..lol

Answer 26

so u see one root was -1 ....so divide the equation with (x + 1) ...study tht link for sure.....would be helpful.....