 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        The sum of non-real roots of the polynomial equation x3 + 3x2 + 3x + 3 = 0.	1.    lies between 0 and 1 	2.    equals 0 	3.    lies between -1 and 0 	4.    has absolute value bigger than 1 `
3 years ago

```							Let the the roots be a,b,c....in which b and c be the 2 non real roots...we need to find b+c...On the other hand we have:::a+b+c = -3....ab+bc+ca = 3...abc = -3...Now we use these 3 equations...bc = -3/a...putting this in 2nd eq...we get....-> a(b+c) - 3/a = 3-> a² (b+c) - 3a - 3 = 0...which is a quadratic in `a`..Discriminant of this should be 0 because only one possible value of a can be root of the given cubic equation...hence on doing so...-> 9 + 12(b+c) = 0-> (b+c) = -3/4....ans....option 3...
```
3 years ago
```							Write the equation as (x+1)3 = -2. Then the roots are (-2)1/3-1,  (-2)1/3w-1, and  (-2)1/3w2 – 1, where w is the primitive cube root of unity. Then sum of non-real roots = [ (-2)1/3w-1]+ [ (-2)1/3w2 – 1] = -[ (2)1/3w +(2)1/3w2] -2 = 21/3 - 2 [Since 1+w+w2= 0] = -[2-21/3]. 2>21/3 >1. Hence 1>2-21/3>0 and hence -11/3 - 2 Hence the option C is correct
```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions