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Algebra Level 3

Let $f( x ) ={ x }^{ 4 }+{ ax }^{ 3 }+{ bx }^{ 2 }+{ cx }+d$ be a polynomial with all real roots.

If $| f(i) | = 1$ , where $i =\sqrt{-1}$ , find $a+b-c+d$ .

1 solution

Prajwal Krishna
Dec 16, 2016

Let $\alpha ,\beta ,\gamma \quad and\quad \delta$ be roots of equation f(x)

Now $f\left( x \right) =\quad (x-\alpha )(x-\beta )(x-\gamma )(x-\delta )$

Further $f\left( i \right) =\quad (i-\alpha )(i-\beta )(i-\gamma )(i-\delta )$

Taking modulus both side

$\Rightarrow \quad 1=\sqrt { 1+{ \alpha }^{ 2 } } \sqrt { 1+{ \beta }^{ 2 } } \sqrt { 1+{ \gamma }^{ 2 } } \sqrt { 1+{ \delta }^{ 2 } }$

$\alpha ,\beta ,\gamma \quad and\quad \delta$ are real so min value $\sqrt { (1+{ \alpha }^{ 2 }) }$ is 1 at $\alpha$ =0

Hence $\alpha ,\beta ,\gamma \quad and\quad \delta$ = a

Hence a=b=c=d=0

Why you considered minimum value ???

Kushal Bose - 4 years, 6 months ago

Because if all numbers in a product are greater than 1 then there product would be greater than 1 ,

Prajwal Krishna - 4 years, 5 months ago

Minimum (1+a^2 ) is 1 so all must be one such that there product is 1

Prajwal Krishna - 4 years, 5 months ago