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Algebra Level 5

Let $f(x)=x^4+ax^3+bx^2+cx+d$ be a polynomial with real coefficients and real zeroes. If $|f(i)|=1$ , where $i=\sqrt{-1}$ , then find the value of $a+b+c+d.$

More than one value exist -1 1 Can not be determined 0

1 solution

Sandeep Bhardwaj
Feb 19, 2015

Assume that $4$ real zeroes of $f(x)$ be $x_1,x_2,x_3,x_4$ , then $f(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4).$

Now evaluating the value of $f(x)$ at $x=i$ , we get $f(i) =(i-x_1)(i-x_2)(i-x_3)(i-x_4)$ $\therefore |f(i)| =\sqrt{1+{x_1}^2}\sqrt{1+{x_2}^2}\sqrt{1+{x_3}^2}\sqrt{1+{x_4}^2}=1$ $\Rightarrow x_1=x_2=x_3=x_4=0$ $\Rightarrow f(x)=x^4$ $\Rightarrow a=b=c=d=0 .$

Adding $a,b,c$ and $d$ we get $a+b+c+d=0 . \square$ enjoy!

How you reached the 3rd step ? Does modulus means positive value or something else here?..

Umesh garg - 6 years, 3 months ago

modulus means magnitude that is the distance of any quantity from origin.

U Z - 6 years, 3 months ago

THANK U SIR

aman khandal - 5 years, 2 months ago

Overrated problem

Ronak Agarwal - 6 years, 3 months ago

yeah .. why it's level 5 solved, without pen and paper it's level 1

A Former Brilliant Member - 4 years, 9 months ago

How did you arrive at step 3 ? abso value f (i) = ....

Radhika Nair - 6 years, 3 months ago

absolute value of a quantity means it's magnitude from the origin.

U Z - 6 years, 3 months ago

Lol guessed a,b,c,d are 0.

Rushikesh Joshi - 6 years, 2 months ago

How from f(i) become |f(i)| ???

Suparjo Tamin - 6 years, 2 months ago

Square f(i) then square root it. Absolute value is more or less just 'making it positive'. Square some number, number must become positive, but now we dont want the 'squared number', we just want the number, then square root.

Jeffrey Li - 6 years ago

Just in case consider the polynomial $f(x) = x^4 + x^3 + x^2 + x +1$ or $a=b=c=d=1$

Now putting $x = i$ we get

$f(i) = 1 \rightarrow |f(i)| = 1$

Umm.. ain't this a contradiction?

neelesh vij - 5 years, 2 months ago

Its zeroes are not all real .

Aakash Khandelwal - 5 years, 2 months ago

Oh got it! Thanks!

neelesh vij - 5 years, 2 months ago

But all the zeroes of the polynomial $f(x)$ are not real in this case. @neelesh vij

Sandeep Bhardwaj - 5 years, 2 months ago

Same way...!!!!

A Former Brilliant Member - 4 years ago

If abs(f(i)) is 1 then won't (x^2 + 1) be a factor of f(x) ??? Then by that we can approach the answer.

Soutrik Bandyopadhyay - 6 years, 3 months ago

Thank you. My question was about the step next to that .

Radhika Nair - 6 years, 3 months ago

$\Longrightarrow { x }_{ 1 }={ x }_{ 2 }={ x }_{ 3 }={ x }_{ 4 }=0$

Well, to make the RHS and LHS equal, There's only one value of roots possible, i.e. 0, you can check for yourself. As soon as we put the value of all 4 roots equal to 0, LHS = RHS becomes true.

Adarsh Pathak - 6 years, 3 months ago

Should mention that mod denotes the mod in complex numbers.

Sahil Silare - 4 years, 4 months ago

Let's check out your elegance

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