Question

If x^4- x^3 - 4x^2 -x +1=0 then sum of all possible value of x+1/x

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: {x}^{4} - {x}^{3} - {4x}^{2} - x + 1 = 0$

Divide both sides by x², we get

$\rm :\longmapsto\: {x}^{2} - x - 4 - \dfrac{1}{x} + \dfrac{1}{ {x}^{2} } = 0$

can be rewritten as

$\rm :\longmapsto\:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - \bigg( x+ \dfrac{1}{x} \bigg) - 4 = 0$

$\red{\rm :\longmapsto\:Let \: \bigg(x + \dfrac{1}{x} \bigg) = y} \\ \red{ \rm\bigg(x+ \dfrac{1}{x} \bigg) ^{2} = {y}^{2}} \\ \red{ \rm \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = {y}^{2}} \\ \red{ \rm \: {x}^{2} + \dfrac{1}{ {x}^{2}} = {y}^{2} - 2}$

On substituting all these values, we get

$\rm :\longmapsto\: ({y}^{2} - 2) - y - 4 = 0$

$\rm :\longmapsto\: {y}^{2} - 2- y - 4 = 0$

$\rm :\longmapsto\: {y}^{2} - y - 6 = 0$

$\rm :\longmapsto\: {y}^{2} - 3y + 2y - 6 = 0$

$\rm :\longmapsto\:y(y - 3) + 2(y - 3) = 0$

$\rm :\longmapsto\:(y - 3)(y + 2) = 0$

$\rm :\implies\:y = 3 \: \: \: \: or \: \: \: y = - 2$

$\bf\implies \:x + \dfrac{1}{x} = 3 \: \: \: or \: \: \: x + \dfrac{1}{x} = - 2$

Hence,

$\purple{\bf :\longmapsto\:Possible \: values \: of \: x + \dfrac{1}{x} \: are \: 3, \: - \: 2 }$

Additional Information:-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac