Question

Solve the equation to find number of integral roots

$\frac{ {x}^{2} }{ {(x + 1)}^{2} } + {x}^{2} = 3$
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Answer 1

I left a screenshot of my answer since $\LaTeX$ doesn't work properly. Please view the attachment.

$\textbf{The first case.}$

$\rm{(x<-1)}$

$\rm{\dfrac{x}{x+1}>1}$

$\rm{\therefore \left(\dfrac{x}{x+1}\right)^{2}>1}$

$\rm{\left(\dfrac{x}{x+1}\right)^{2}+x^{2}=3}$

$0</p><p>[tex]\textrm{ \therefore No solution.}$

$\textbf{The second case.}$

$\rm{(x>-1)}$

$\rm{\dfrac{x}{x+1}<1}$

$\rm{\therefore 0<\left(\dfrac{x}{x+1}\right)^{2}<1}$

$\rm{\left(\dfrac{x}{x+1}\right)^{2}+x^{2}=3}$

$2-1$

$\textrm{ \therefore No solution.}$

Hence

$\rm{\cdots\longrightarrow\boxed{\rm\left(\dfrac{x}{x+1}\right)^{2}+x^{2}=3\ has\ no\ integral\ solution.}}$

Answer 2

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

Given equation is

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

can be rewritten as

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

We know,

$\boxed{ \rm{ \: {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \: \: }}$

So, using this identity, we get

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

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

$\rm \: {\bigg(\dfrac{ {x}^{2}}{x + 1} \bigg) }^{2} + \frac{ 2{x}^{2} }{x + 1} - 3 = 0$

Let assume that

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

So, above equation can be rewritten as

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

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

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

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

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

So, Consider

$\rm \: y = - 3$

$\rm \: \frac{ {x}^{2} }{x + 1} = - 3$

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

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

Now, Discriminant = 9 - 12 = - 3 < 0

$\rm\implies \: {x}^{2} + 3x + 3 = 0 \: have \: no \: real \: root.$

Now, Consider

$\rm \: y = 1$

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

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

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

So, using Quadratic formula, we get

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

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

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

So, it means given equation

$\rm \: \dfrac{ {x}^{2} }{ {(x + 1)}^{2} } + {x}^{2} = 3 \: have \: irrational \: roots$

So,

$\rm\implies \boxed{ \rm{ \dfrac{ {x}^{2} }{{(x + 1)}^{2} }+{x}^{2} = 3\: have \:no\:integeral\:roots\: }}$