Question

The equaqtion \(ax^3+bx^2+cx+d\) is known to’ve three distinct real roots .How many roots are there of the equation :
\(4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\)

Answer 1

Hi! Back again with a good question ?

Answer 2

:)

Answer 3

How do you think we should solve this question?

Answer 4

Any idea ?

Answer 5

Okay one thing that I have in my mind is this:

If the roots of the polynomial are say \(x_1, ~x_2, ~x_3\). Then,

\[\large{ax^3 + bx^2 + cx + d = a(x-x_1)(x-x_2)(x-x_3)}\]

Answer 6

You can also use calculus to solve this problem but I would discuss it later on

Answer 7

Now plugging in this in the problem we can come up with something

Answer 8

Also, we know that:

\[\large{\sum x_1 = -\cfrac{b}{a}}\]

Answer 9

\[\large{\sum x_1 x_2 = \cfrac{c}{a}}\]

\[\large{x_1x_2x_3 = -\cfrac{d}{a}}\]

Answer 10

Now lets plug in the polynomial in the equation given in the problem

Answer 11

Or better would be this:

\[\large{b = -a\sum x_1} \tag{1}\]

\[\large{c = a\sum x_1 x_2}\tag{2}\]

\[\large{d = -ax_1x_2x_3}\tag{3}\]

Answer 12

\[\Large 4(ax^3+bx^2+cx+d)(3ax+b)=(3ax^2+2bx+c)^2\]

\[\large{4(3a^2x^4+4abx^3+x^2(b^2+3ac)+x(bc+3ad)+bd) = (3ax^2+2bx+c)^2}\]

Answer 13

\[{4(3a^2x^4+4abx^3+x^2(b^2+3ac)+x(bc+3ad)+bd) = (3ax^2+2bx+c)^2}\]

Answer 14

Simplifying this using wolfram

Answer 15

\[\large{6a^2x^4+8abx^3+x^2(3ac+3b^2)+x(3bc-3ad)+c^2-bd= 0}\]

Answer 16

*sigh* I am tired :D

Answer 17

Lets plug in the values of b, c and d from (1), (2) and (3)

Answer 18

Use the De' Carte's condition of real (positive and negative roots)
We know that if there are 2n sign changes in f(x) starting from the first term, then it may have 2n or 2n-2 or 2n-4 so on upto 0 number of POSITIVE roots.And if there are 2n+1 sign changes then there could be 2n+1 or 2n-1 or 2n-3 so on up to 1 positive roots. And applying the same thing to f(-x) gives the number of negative roots.

Answer 19

I am just trying to come up with something :)

Answer 20

I am thinking of using the De' Carte's condition for ax^3 + bx^2 + cx + d first and come up with the signs of a,b,c and d

Answer 21

Any ideas anyone ?

Answer 22

@hartnn any idea ?

Answer 23

seems like a B^2 -4AC =0 format

(ax^3+bx^2+cx+d) X^2 + (3ax^2+2bx+c)X + (3ax+b) =0
has EQUAL roots....

Answer 24

Sure? I have not checked actually

Answer 25

Yeah correct..

Answer 26

Not sure but I think the answer may be 2 solutions. @JungHyunRan can you check it?

Answer 27

Did it work @JungHyunRan ?

Answer 28

I think (and am pretty sure) that the number of roots = 2 (just an intution)

Also, I think the problem is not correctly stated or something is missing.

Answer 29

yeah! It has 2 solutions
Good job! @vishweshshrimali5 :)

Answer 30

Thanks :)

It worked !! :D

Answer 31

Another of your tricky questions @JungHyunRan . I really appreciate your trying to solve questions which increase the skills of thinking out of the box which is the most required skill for betterment in maths.

Best of luck!

Answer 32

^_^ thank you