Question

If x^3+ax+1=0 and x^4+ax^2+1=0 have a common root, then no. of values of a is ??

Answer 1

assume that common root is \(t\) now subtract 2 equations and see what happens

Answer 2

the equation u obtain from subtracting that 2 equation must has only one real root

Answer 3

the eq obtained -

Answer 4

try to factor it completely

Answer 5

\[t^{4}-t^{3}+at^{2}-at=0\]??

Answer 6

\( t^3(t-1)+at(t-1)=0 \)

Answer 7

t^3+at=0

Answer 8

t(t^2+a)=0
t=0

Answer 9

correct?

Answer 10

actually am kinda stuck here

we have \(t(t-1)(t^2+a)=0 \) and we want to this equation has a one root

Answer 11

but x=0 is not a root for equations am i right?

Answer 12

yes

Answer 13

are you getting it?

Answer 14

and x=1 is a commen root am i right?

Answer 15

how?

Answer 16

see the equation for t

Answer 17

but no wait

Answer 18

yes

Answer 19

what is the answer?

Answer 20

yes putting 1 in both the eq. a=-1
ans. no. of values of a = 1.

Answer 21

putting 1 in both the eq. gives a=-2

Answer 22

1^3+a(1)+1=0
1^4+a(1)^2+1=0

??

Answer 23

yes

Answer 24

see only possible commen roots for 2 equations are t=0 , t=1 or t^2=-a am i right?

Answer 25

yes

Answer 26

but t=0 and t^2=-a will not work try to work it out by substituting

Answer 27

yes