Question

If α,β,γ,δ are root of equation $x^{4} +7x^{3}-x+6=0$ then find α^2 + β^2 + γ^2 + δ^2.


Please give proper explanation
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Answer 1

Topic :-

Theory of Equations

Given :-

$\mathtt { \alpha,\beta,\gamma\:and\:\delta\:are\:roots\:of\:the\:equation :-}$

$\mathtt{x^4+7x^3-x+6=0}$

To Find :-

$\mathtt{Value\:of\:\:\alpha^2+\beta^2+\gamma^2+\delta^2.}$

Concept Used :-

Theory of Equations

$\mathtt{If\:\alpha,\beta,\gamma\:and\:\delta\:are\:roots\:of\:the\:equation\:\:ax^4+bx^3+cx^2+dx+e=0,then}$

$\mathtt{\alpha+\beta+\gamma+\delta=\dfrac{-b}{a}}$

$\mathtt{\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\dfrac{c}{a}}$

Solution :-

$\mathtt{On\:comparing\:given\:equation\:with\:ax^4+bx^3+cx^2+dx+e=0,we\:get,}$

$\mathtt{a=1}$

$\mathtt{b=7}$

$\mathtt{c=0}$

So, we can write,

$\mathtt{\alpha+\beta+\gamma+\delta=\dfrac{-7}{1}=-7}$

$\mathtt{\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\dfrac{0}{1}=0}$

Now,

$\mathtt {(\alpha+\beta+\gamma+\delta)^2=\alpha^2+\beta^2+\gamma^2+\delta^2+2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)}$

Put values accordingly,

$\mathtt {(-7)^2=\alpha^2+\beta^2+\gamma^2+\delta^2+2(0)}$

$\mathtt {49=\alpha^2+\beta^2+\gamma^2+\delta^2}$

Answer :-

$\mathtt {So,\:\alpha^2+\beta^2+\gamma^2+\delta^2=\bold{49}.}$

Additional Information :-

Every equation of nth degree ( n ≥ 1 ) has exactly 'n' roots.

Imaginary roots occur in conjugate pairs.

Answer 2

Concept Used

Let us consider a bi- quadratic equation

$\sf \: {ax}^{4} + {bx}^{3} + {cx}^{2} + dx + e = 0 \: having \: zeroes$

$\sf \: \alpha, \beta, \gamma \: and \: \delta \: then$

$\bull \tt \: S_1 = \alpha + \beta + \gamma + \delta = - \dfrac{b}{a}$

$\bull \tt \: S_2 = \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \dfrac{c}{a}$

$\bull \tt \: S_3 = \alpha \beta \gamma + \alpha \beta \delta + \beta \gamma \delta + \alpha \gamma \delta = - \dfrac{d}{a}$

$\bull \tt \: S_4 = \alpha \beta \gamma \delta = - \: \dfrac{e}{a}$

$\large\underline{\bold{Solution :- }}$

• Given bi quadratic equation is

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: {x}^{4} + {7x}^{3} - x + 6 = 0$

So,

• On comparing with

$\: \: \: \: \: \: \: \: \: \: \: \: \sf \: {ax}^{4} + {bx}^{3} + {cx}^{2} + dx + e = 0$

we have

• a = 1

• b = 7

• c = 0

• d = - 1

• e = 6

Now,

• Since,

$\rm :\longmapsto\: \alpha , \: \beta , \: \gamma \: and \: \delta \: are \: the \: roots \: of \: given \: equation$

$\rm :\implies\: \alpha + \beta + \gamma + \delta = - \dfrac{b}{a} = - \dfrac{7}{1} = - 7$

and

$\rm :\longmapsto\: \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \dfrac{c}{a} = 0$

Now,

• Consider,

$\rm :\longmapsto\: {( \alpha + \beta + \gamma + \delta)}^{2}$

$\tt \: = {(\alpha+\beta )}^{2}+ {( \gamma + \delta)}^{2} + 2( \alpha + \beta )( \gamma + \delta)$

$\tt = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta + { \gamma }^{2} + {\delta}^{2} + 2 \gamma \delta + 2( \alpha + \beta )( \gamma + \delta)$

$\tt = { \alpha }^{2} + { \beta }^{2} + { \gamma }^{2} + {\delta}^{2} + 2( \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta )$

On substituting the values, we get

$\rm :\longmapsto\: {( - 7)}^{2} = { \alpha }^{2} + { \beta }^{2} + { \gamma }^{2} + {\delta}^{2} + 2 \times 0$

$\rm :\implies\:{ \alpha }^{2} + { \beta }^{2} + { \gamma }^{2} + {\delta}^{2} = 49$