 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

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
```
X*4+ax*3++bx*2+cx+d be a polynomial equation with real cofficients and real zeroes.if |f(i)|=1.find a+b+c+d.i=root of -1

```
6 years ago IIT Delhi
174 Points
```							f(i)=1-ai-b-ci+d=11-b-d=real part=1-i(a+c)=complex part =0b+d=0a+c=0a+b+c+d=0sher mohammadb.tech, iit delhi
```
6 years ago
```							If p,q,r,s are the real roots of the polynomial, then f(i) = (i-p)(i-q)(i-r)(i-s)
and hence |f(i)| = (1+p^2)(1+q^2)(1+r^2)(1+s^2) >=1.

But |f(i)| = 1 and hence we must have p=q=r=s=0
i.e. a=b=c=d=0=a+b+c+d
```
6 years ago
```							@sher md. its mod of f(i).
```
6 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