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Q → The Solution of set of $\frac{ {x}^{2} - \: 3x \: + \: 4}{x \: + \: 1} > 1$ , x belongs to R is

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Answer 1

Step-by-step explanation:

Dividing by x^2, 3(x^2 + 1/x^2) + 4(x + 1/x) - 14

= 3(x + 1/x)^2 -6 + 4(x + 1/x) - 14 = 3(x + 1/x)^2 + 4(x + 1/x) - 20 = 0.

Let y = x+1/x. Then 3y^2 + 4y -20 = (3y^2 + 10y) - (6y + 20)

= y(3y +10) - 2(3y +10) = (y - 2)(3y + 10)= 0.

==> (x + 1/x -2)(3x + 3/x + 10) = 0.

Multiplying by x^2, (x^2 - 2x + 1)(3x^2 + 10x + 3)= 0.

3x^2 + 10x +3 = (3x^2 + 9x) + (x + 3) = (3x+1)(x+3).

(x-1)^2*(x+3)(3x+1)=0.

x =1, 1,-3, -1/3.

Answer 2

Answer:

Solution−

Given inequality is

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

can be rewritten as

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

$\rm :\longmapsto\:\dfrac{ {x}^{2} - \: 3x \: + \: 4 - \: (x + 1)}{x \: + \: 1}$

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

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

$\rm :\longmapsto\:\dfrac{ {x}^{2} - \: 3x - x \: + \: 3 }{x \: + \: 1}$

$\rm :\longmapsto\:\dfrac{ x(x - 3) - 1(x - 3)}{x \: + \: 1}$

$\rm :\longmapsto\:\dfrac{ (x - 3) (x - 1)}{x \: + \: 1}$

So, breaking points are - 1, 1 and 3.

So, intervals along with their respective signs of inequality are as follow :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( - \infty , - 1) & \sf - \\ \\ \sf ( - 1,1) & \sf + \\ \\ \sf (1,3) & \sf - \\ \\ \sf (3, \infty ) & \sf + \end{array}} \\ \end{gathered}\end{gathered}$

So, Solution set is

$\bf\implies \:\boxed{ \bf{x \in \: ( - 1, \: 1) \: \cup \: (3, \: \infty ) \: }}$