Question

solve for x
${x}^{2} + \frac{ {x}^{2} }{ {(x + 1)}^{2} } = 3$
find roots of X.. find zeroes of X​

Answer 1

Answer:

In real number system

$\qquad\qquad\boxed{ \sf{ \: \bf \: x = \dfrac{ 1 \: \pm \: \sqrt{ 5} }{2} \: }}$

Or

In complex number system,

$\qquad\boxed{ \sf{ \: \bf \: x = \dfrac{ 1 \: \pm \: \sqrt{ 5} }{2}, \: \: \dfrac{ - 3\: \pm \: i \sqrt{ 3} }{2} \: }}$

Step-by-step explanation:

Given equation is

$\sf \: {x}^{2} + \dfrac{ {x}^{2} }{ {(x + 1)}^{2} } = 3$

We know,

$\qquad\boxed{ \sf{ \: {a}^{2}+{b}^{2} = {(a - b)}^{2} + 2ab \: }}$

So, using this result, we get

$\sf \: {\bigg(x - \dfrac{x}{x + 1} \bigg) }^{2} + 2 \times x \times \dfrac{x}{x + 1} = 3$

$\sf \: {\bigg( \dfrac{ {x}^{2} + x - x}{x + 1} \bigg) }^{2} + \dfrac{2 {x}^{2} }{x + 1} = 3$

$\sf \: {\bigg( \dfrac{ {x}^{2}}{x + 1} \bigg) }^{2} + \dfrac{2 {x}^{2} }{x + 1} = 3$

Let assume that

$\qquad\sf \: \dfrac{ {x}^{2} }{x + 1} = y$

So, using this, the above expression can be rewritten as

$\sf \: {y}^{2} + 2y = 3$

$\sf \: {y}^{2} + 2y - 3= 0$

$\sf \: {y}^{2} + 3y - y - 3= 0$

$\sf \: y(y + 3) - 1(y + 3) = 0$

$\sf \: (y + 3)(y - 1) = 0$

$\bf\implies \: y = - 3 \: \: or \: \: y = 1$

Now, Consider

$\sf \: y = 1$

$\sf \: \dfrac{ {x}^{2} }{x + 1} = 1$

$\sf \: {x}^{2} = x + 1$

$\sf \: {x}^{2} - x - 1 = 0$

On comparing with a$x^2$ + bx + c = 0, we get

$\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: a=1\qquad \: \\ \\& \qquad \:\sf \: b= - 1 \\ \\& \qquad \:\sf \: c= - 1\end{aligned}} \qquad$

We know, Solution of quadratic equation is given by using quadratic formula, which is

$\qquad\boxed{ \sf{ \:\sf \: x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} \: }}$

So, using quadratic formula, we get

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

$\sf \: x = \dfrac{ 1 \: \pm \: \sqrt{ 1 + 4 } }{2}$

$\sf\implies \bf \: x = \dfrac{ 1 \: \pm \: \sqrt{ 5} }{2}$

Now, Consider

$\sf \: y = - 3$

$\sf \: \dfrac{ {x}^{2} }{x + 1} = - 3$

$\sf \: {x}^{2} = - 3x - 3$

$\sf \: {x}^{2} + 3x + 3 = 0$

On comparing with a$x^2$ + bx + c = 0, we get

$\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: a=1\qquad \: \\ \\& \qquad \:\sf \: b= 3\\ \\& \qquad \:\sf \: c=3\end{aligned}} \qquad$

So, Solution of equation using quadratic formula is given by

$\sf \: x = \dfrac{ - (3) \: \pm \: \sqrt{ {(3)}^{2} - 4(1)(3) } }{2(1)}$

$\sf \: x = \dfrac{ - 3\: \pm \: \sqrt{9 - 12 } }{2}$

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

$\bf\implies \: There \: is \: no \: solution \: in \: real \: numbers$

$\rule{190pt}{2pt}$

But for complex solutions

$\sf \: x = \dfrac{ - 3\: \pm \: \sqrt{ - 3} }{2} \:$

$\sf\implies \sf \: x = \dfrac{ - 3\: \pm \: i \sqrt{ 3} }{2} \:$

Hence,

$\sf\implies \bf \: x = \dfrac{ 1 \: \pm \: \sqrt{ 5} }{2}, \: \: \dfrac{ - 3\: \pm \: i \sqrt{ 3} }{2} \:$