Question

Solve for x :
$4(x - \dfrac{1}{x} ) {}^{2} + 8(1 + \dfrac{1}{x} ) = 29. \: x≠0.$
​

Answer 1

Given:–

• $4( \text{x}- \dfrac{1}{ \text{x}} ) {}^{2} + 8(1 + \dfrac{1}{ \text{x}} ) = 29$

To find:–

• Value of x

Topic:–

• Quadratic equations

Formula used:–

The roots of the quadratic equation ax² + bx + c = 0; where a≠0 will be obtained by this formula.

• $\pink{ \bold{ \boxed{ \text{x} = \dfrac{ \text - b± \sqrt{b{}^{2} - 4ac} }{2a}}}}$

Step by step explaination:-

$\underline \bold{ \mathfrak{★let \: us \: assume \: the \: value \: of \: \text{x} + \dfrac{{1}}{ \text{x}} \: be \: y}}$

$\underline \bold{ \mathfrak{★Here \: we \: have \: to \: use \: this \: identity \: in \: terms \: of \: \text{x} + \dfrac{1}{ \text{x}} = y \: :- }( \text{x} + \dfrac{1}{ \text{x}} ) {}^{2} - (x - \dfrac{1}{x}) {}^{2} = 4 }$

$\underline \bold{ \mathfrak{★Therefore,}}$

$\: \: \: \: \: :\implies \text{y} {}^{2} - ({ \text{x}- \dfrac{1}{ \text{x}} {}^{2} ) = 4},( \text{x} - \dfrac{1}{ \text{x}} ) {}^{2} = \text{y} {}^{2} - 4$

$\underline \bold{ \mathfrak{★Now, }}$

$\: \: \: \: \: :\implies4( \text{y} - 4) + 8 \text{y} = 29 \\ \: \: \: \: \: : \implies4 \text{y} {}^{2} - 16 + 8y = 29$

$\underline \bold{\mathfrak{★Splitting \: middle \: term.}}$

$\: \: \: \: \: :\implies4 \text{y} {}^{2} + \text{18y} - \text{10y} - 45 = 0$

$\underline \bold {\mathfrak{★Factorising}}$

→ 2y (2y + 9 ) - 5 (2y + 9) = 0

→ (2y + 9) (2y - 5) = 0

$\underline \bold {\mathfrak{★Therefore, \: value of \: y \: will \: be \: -9/2 \: or \: 5/2 }}$

_________

$\underline \bold {\mathfrak{★Evaluating \:the \:value\: of \:y \:of \:-9/2 \:in \:x+1/x }}$

$\: \: \: \: \: :\implies \text{x} + \dfrac{1}{ \text{x}} = - \dfrac{ 9}{2}$

$\underline \bold{ \mathfrak{★Evaluating\: values\: in\:the \: given \: formula}}$

$\: \: \: \: \: :\implies \text{x} = \dfrac{ - 9± \sqrt{(9) {}^{2} - 4 \times 2 \times 2 } }{2 \times 2}$

$\: \: \: \: \: :\implies \dfrac{ - 9± \sqrt{65} }{4}$

$\bold{ \underline{ \mathfrak{★Again \: evaluating \: the \: second \: value \: of \: y \: that \: is \: \frac{5}{2} \:in \: \text{x} + \dfrac{ \text{1}}{ \text{x}} }}}$

$\: \: \: \: \: :\implies2x² - 5x + 2 = 0$

$\: \: \: \: \: :\implies 2x² - 4x - x + 2 = 0$

$\bold{ \underline{\mathfrak{★Factorising..}}}$

$\: \: \: \: \: :\implies 2x(x-2) -1 (x-2)=0$

$\: \: \: \: \: :\implies (x-2) (2x-1) = 0$

Thus, x = 1/2 , x = 2

Conclusion:–

$\dfrac{ - 9± \sqrt{65} }{4} \: \text{or} \: \dfrac{1}{2} ,\: 2$

More to know:–

★ An equation which has only one variable in which highest power of variable is two, is known as quadratic equation.

★ The standard form of a quadratic equation is ax²+bx+c = 0, where a, b and c are all real numbers.

★ How to solve a quadratic equation by using factorisation method?

• First solve all the fractions and the given brackets
• Transpose all the given terms from LHS to get an equation which is in the form of ax² + bx + c = 0
• Factorise every expression on L.H.S.
• Now put the given each factor equal to zero and solve it.

Answer 2

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

Given equation is

$\rm :\longmapsto\:4{\bigg(x - \dfrac{1}{x} \bigg) }^{2} + 8\bigg(x + \dfrac{1}{x} \bigg) = 29$

We know that,

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

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

On substituting these values in given equation, we get

$\rm :\longmapsto\:4{\bigg(x + \dfrac{1}{x} \bigg) }^{2} - 16 + 8{\bigg(x + \dfrac{1}{x} \bigg) } = 29$

$\rm :\longmapsto\:4{\bigg(x + \dfrac{1}{x} \bigg) }^{2} + 8{\bigg(x + \dfrac{1}{x} \bigg) } - 16 - 29 = 0$

$\rm :\longmapsto\:4{\bigg(x + \dfrac{1}{x} \bigg) }^{2} + 8{\bigg(x + \dfrac{1}{x} \bigg) } - 45 = 0$

$\red{ \boxed{ \sf{Let \:{\bigg(x + \dfrac{1}{x} \bigg) } = y - - - (1) }}}$

So above equation can be reduced to

$\rm :\longmapsto\:4 {y}^{2} + 8y - 45 = 0$

$\rm :\longmapsto\:4 {y}^{2} + 18y - 10y - 45 = 0$

$\rm :\longmapsto\:2y(2y + 9) - 5(2y + 9) = 0$

$\rm :\longmapsto\:(2y - 5)(2y + 9) = 0$

$\rm :\implies\:y = \dfrac{5}{2} \: \: \: or \: \: \: y = - \: \dfrac{9}{2}$

Consider,

$\rm :\longmapsto\:\:y = \dfrac{5}{2}$

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

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

$\rm :\longmapsto\: {2x}^{2} + 2 = 5x$

$\rm :\longmapsto\: {2x}^{2} - 5x + 2 = 0$

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

$\rm :\longmapsto\:2x(x - 2) - 1(x - 2) = 0$

$\rm :\longmapsto\:(2x - 1)(x - 2) = 0$

$\rm :\implies\:x = 2 \: \: \: \: or \: \: \: \: x = \dfrac{1}{2}$

Consider,

$\rm :\longmapsto\: \: y = - \: \dfrac{9}{2}$

$\rm :\longmapsto\: \:{\bigg(x + \dfrac{1}{x} \bigg) } = - \: \dfrac{9}{2}$

$\rm :\longmapsto\: \:{\bigg( \dfrac{ {x}^{2} + 1}{x} \bigg) } = - \: \dfrac{9}{2}$

$\rm :\longmapsto\: {2x}^{2} + 2 = - 9x$

$\rm :\longmapsto\: {2x}^{2} + 9x + 2 = 0$

Now, we are not able to split the middle terms here.

So, to find the value of x, we use Quadratic Formula, given by

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

On comparing

$\rm :\longmapsto\: {2x}^{2} + 9x + 2 = 0 \: with \: {ax}^{2} + bx + c = 0 \: we \: get$

$\rm :\longmapsto\:a = 2$

$\rm :\longmapsto\:b = 9$

$\rm :\longmapsto\:c = 2$

So, using Quadratic formula,

$\rm :\longmapsto\:x = \dfrac{ - 9 \: \pm \: \sqrt{ {9}^{2} - 4(2)(2)} }{2(2)}$

$\rm :\longmapsto\:x = \dfrac{ - 9 \: \pm \: \sqrt{81 - 16} }{4}$

$\rm :\longmapsto\:x = \dfrac{ - 9 \: \pm \: \sqrt{65} }{4}$

Hence,

Solution of

$\bf :\longmapsto\:4{\bigg(x - \dfrac{1}{x} \bigg) }^{2} + 8\bigg(x + \dfrac{1}{x} \bigg) = 29$

is

$\bf :\longmapsto\:x = 2, \: \dfrac{1}{2}, \: \dfrac{ - 9 \: \pm \: \sqrt{65} }{4}$

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)