Question

(x²+1/x²)-5(x+1/x)= 4 (solve it)​

Answer 1

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

Given that

$\rm :\longmapsto\:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 5\bigg(x + \dfrac{1}{x} \bigg) = 4$

can be rewritten as

$\rm :\longmapsto\:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 5\bigg(x + \dfrac{1}{x} \bigg) - 4 = 0 - - - (1)$

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

On squaring both sides, we get

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

$\purple{\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = {y}^{2}}$

$\purple{\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} }= {y}^{2} - 2} - - (3)$

On substituting equation (2) and (3) in equation (1), we get .

$\rm :\longmapsto\: ({y}^{2} - 2) - 5y - 4 = 0$

$\rm :\longmapsto\: {y}^{2} - 2- 5y - 4 = 0$

$\rm :\longmapsto\: {y}^{2} - 5y - 6 = 0$

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

$\rm :\longmapsto\:y(y - 6) + 1(y - 6) = 0$

$\rm :\longmapsto\:(y - 6)(y + 1) = 0$

$\bf\implies \:y = 6 \: \: \: \: \: or \: \: \: \: y = - 1$

Case :- 1

$\rm :\longmapsto\:y \: = \: 6$

$\rm :\longmapsto\:x + \dfrac{1}{x} = 6$

$\rm :\longmapsto\: \dfrac{ {x}^{2} + 1}{x} = 6$

$\rm :\longmapsto\: {x}^{2} + 1 = 6x$

$\rm :\longmapsto\: {x}^{2} - 6x + 1 = 0$

We know,

$\rm :\longmapsto\:The \: solution \: of \: {ax}^{2} + bx + c = 0 \: is \: given \: by$

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

Thus,

$\rm :\longmapsto\:x = \dfrac{ - ( - 6) \: \pm \: \sqrt{ {( - 6)}^{2} - 4(1)(1)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: \sqrt{ 36 - 4} }{2}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: \sqrt{ 32} }{2}$

$\rm :\longmapsto\:x = \dfrac{6 \: \pm \: 4\sqrt{ 2} }{2}$

$\bf :\longmapsto\:x = 3 \: \pm \: 2 \sqrt{2}$

Case :- 2

$\rm :\longmapsto\:y = - \: 1$

$\rm :\longmapsto\:x + \dfrac{1}{x} = - 1$

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

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

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

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4(1)(1) } }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1} - 4} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ - 3} }{2} \: \cancel \in \: R$

Hence,

Solution of

$\bf :\longmapsto\:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) - 5\bigg(x + \dfrac{1}{x} \bigg) = 4$

is given by

$\bf :\longmapsto\:x = 3 \: \pm \: 2 \sqrt{2}$

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