Question

Theory of equations

Answer 1

x^2 +x +1 is a factor of ax^3 +bx^2 +cx +d =0 , then the real root of the above equation is (a , b, c ,d belongs to R)

Answer 2

Options are:-

(a) -d/a
(b)d/a
(c)(b-a)/a
(d)(a-b)/a

[Has Multiple options correct]

Answer 3

\[\large ax^3 + bx^2 + cx + d = (x^2 + x + 1)(px+q)\]

Answer 4

distribute right hand side, and compare coefficients. you get :
\(\large p = a\)
\(\large q = d\)

So, \(\large ax + d\) is the linear factor \(\iff \) \(\large -\frac{d}{a}\) is a real root

Answer 5

Yes (A) is given the other root given is (a-b)/a

Answer 6

wym ? im not gettin u..

Answer 7

This question has multiple options correct

Answer 8

hmm a cubic has exactly 3 roots, so...
whats the correct answer ?

Answer 9

(a-b)/a

Answer 10

Okay its not a second root, they're equivalent

Answer 11

compare the coefficients, you would get set of equations - from that you can figure out the right options

Answer 12

Oks... thanks.

Answer 13

I got
p=a
q=d
q+p=b
q+p=c

------> How do i figure out that (a-b)/a is correct

Answer 14

take the 3rd equation

Answer 15

q+p = b
substitute p and q values :
a+d = b

Answer 16

q+p =a +d Like this

Answer 17

q+p = b
substitute p and q values :
a+d = b
\(\implies d = b-a\)

Answer 18

so \(\large px+q = ax+d = ax + (b-a)\)

Answer 19

Ah! that's makes sense

Answer 20

they have just eliminated the \(d\) to cookup that second correct option ^

Answer 21

Yup , its just equivalent

Answer 22

Alternate way given that x^2+x+1=0 is the factor of the given polynomial
then the roots of x^2+x+1=0 are also the roots of the given cubic
now the roots of x^2+x+1=0 are w,w^2 (such that w+w^2=0 and w^3=1)

thus the root of the cubic are w,w^2,l (where l is the reqd root which we need to find)
now the product of the roots is given by -d/a
here thus
w*w^2*l= -d/a
so l =-d/a

Answer 23

wow , nice thinking !