Question

If $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation $x^4-K{x}^3+K{x}^2+Lx+M=0$, where $K,L$ & $M$ are real numbers, then the minimum value of ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is B $-1$

Given, $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation $x^4-K{x}^3+K{x}^2+Lx+M=0$.

Then from the relation between the roots and co-efficient we get,

$\sum \alpha =\alpha +\beta +\gamma +\delta =K$.....(1) and $\sum \alpha .\beta =K$......(2), $\sum \alpha .\beta .\gamma =-L$ and $\alpha .\beta .\gamma .\delta =M$.

Now, ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2$

$=(\sum \alpha {)}^2-2(\sum \alpha .\beta )$

$=K^2-2K$ [ Using (1) and (2)]

$=K^2-2K+1-1$

$=(K-1{)}^2-1$.

We have minimum value of $(K-1{)}^2$ is $0$ as $(K-1{)}^2\ge 0$.

So the minimum value of ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2$ is $-1$.