Question

if alpha beta gamma delta at the root of the equation X ki power 4 minus K x cube + L X + m=0 where KLM are real number then the minimum value of Alpha squire Plus beta squire plus Gama squire plus delta square is

Answer 1

Answer:

minimum value is 0

Step-by-step explanation:

hope this answer helped you

Answer 2

We need to recall the concept of the roots of a quartic equation.

• If $\alpha ,\beta ,\gamma, \delta$ are the roots of a quartic equation $ax^4+bx^3+cx^2+dx+e=0$ then,

  $\alpha +\beta +\gamma+\delta=\frac{-b}{a}$ ,  $\sum \alpha \beta =\frac{c}{a}$ ,  $\sum \alpha \beta \gamma=\frac{-d}{a}$ , $\alpha \beta \gamma \delta =\frac{e}{a}$

Given:

$\alpha ,\beta ,\gamma, \delta$ are the roots of a quartic equation $x^4-Kx^3+Lx+M=0$.

From the relation between the roots and coefficients, we get

$\sum\alpha =\alpha +\beta +\gamma+\delta=K$ , $\sum \alpha \beta =0$ , $\sum \alpha \beta \gamma=-L$ , $\alpha \beta \gamma \delta =M$

Now,

$\alpha^2 +\beta^2 +\gamma^2+\delta^2$

$=(\sum \alpha)^2 -2(\sum\alpha \beta )$

$=(K)^2 -2(0 )$

$=K^2$

Since $K^2\geq 0$.

So, the minimum value of $K^2$ is $0$.

Hence, the minimum value of $\alpha^2 +\beta^2 +\gamma^2+\delta^2$ is $0$.