Math, asked by Badboy330, 22 days ago

please solve this problem , It's argent​

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Answers

Answered by Anonymous
185

 \large \underline{ \underline{ \text{Question:}}} \\

  • \tt (x) \:  \frac{x}{x + 1}  +  \frac{x + 1}{x}  = 2 \frac{1}{12} ,x ≠0, - 1 \\

 \large \underline{ \underline{ \text{Solution:}}} \\

We have,

  •   \frac{x}{x + 1}  +  \frac{x + 1}{x}  = 2 \frac{1}{12}  \\

Can be written as,

  •   \frac{x}{x + 1}  +  \frac{x + 1}{x}  =  \frac{25}{12}  \\

Making a Quadratic Equation,

\implies   \frac{x}{x + 1}  +  \frac{x + 1}{x}  =  \frac{25}{12}  \\  \\ \implies   \frac{ {x}^{2}  + (x + 1)(x + 1)}{x(x + 1)}  =  \frac{25}{12}  \\  \\ \implies   \frac{ {x}^{2} +  {x}^{2} + 2x + 1  }{ {x}^{2}  + x}  =  \frac{25}{12}  \\  \\ \implies   \frac{2 {x}^{2} + 2x + 1 }{ {x}^{2}  + x}  =  \frac{25}{12}  \\  \\ \implies  12(2 {x}^{2}  + 2x + 1) = 25( {x}^{2}  + x) \\  \\\implies  24 {x}^{2}   + 24x + 12 = 25 {x}^{2}  + 25x \\  \\ \implies  25 {x}^{2}  + 25x - 24 {x}^{2}  - 24x - 12 = 0 \\  \\ \implies   {x}^{2}  + x - 12 = 0 \\

We have,

  • {x}^{2}  + x - 12 = 0 \\

Solving the Quadratic Equation by Quadratic Formula,

As we know,

  • \boxed{x  =  \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} } \\

Here,

  • a = 1 , \: b= 1 \: \text{and} \: c = -12 \\

Substituting the values and Finding the value of x,

\implies x  =  \frac{ - 1± \sqrt{ {(1)}^{2} - 4(1)( - 12) } }{2(1)}  \\  \\  \implies x  =   \frac{ - 1± \sqrt{1 + 48} }{2}  \\  \\ \implies x  =  \frac{ - 1± \sqrt{49} }{2}  \\  \\ \implies x  =  \frac{ - 1± 7}{2}  \\  \\ \implies x  =  \frac{ - 1 + 7}{2}  \: \text{and}  \:  \frac{ - 1 - 7}{2}  \\  \\ \implies x  =  \frac{6}{2}  \: \text{and}  \:  \frac{ -8}{2} \\  \\ \implies x  =  3  \: \text{and}  \:   - 4 \\

Therefore,

  • The values of x is 3 and -4.

 \\  \large \underline{ \underline{ \text{Required Answer:}}} \\

  • The values of x is 3 and -4.
Answered by ramsingh17950
5

Step-by-step explanation:

Question:

\begin{gathered}\tt (x) \: \frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{1}{12} ,x ≠0, - 1 \\ \end{gathered}

(x)

x+1

x

+

x

x+1

=2

12

1

,x

=0,−1

\begin{gathered} \large \underline{ \underline{ \text{Solution:}}} \\ \end{gathered}

Solution:

We have,

\begin{gathered} \frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{1}{12} \\ \end{gathered}

x+1

x

+

x

x+1

=2

12

1

Can be written as,

\begin{gathered} \frac{x}{x + 1} + \frac{x + 1}{x} = \frac{25}{12} \\ \end{gathered}

x+1

x

+

x

x+1

=

12

25

Making a Quadratic Equation,

\begin{gathered}\implies \frac{x}{x + 1} + \frac{x + 1}{x} = \frac{25}{12} \\ \\ \implies \frac{ {x}^{2} + (x + 1)(x + 1)}{x(x + 1)} = \frac{25}{12} \\ \\ \implies \frac{ {x}^{2} + {x}^{2} + 2x + 1 }{ {x}^{2} + x} = \frac{25}{12} \\ \\ \implies \frac{2 {x}^{2} + 2x + 1 }{ {x}^{2} + x} = \frac{25}{12} \\ \\ \implies 12(2 {x}^{2} + 2x + 1) = 25( {x}^{2} + x) \\ \\\implies 24 {x}^{2} + 24x + 12 = 25 {x}^{2} + 25x \\ \\ \implies 25 {x}^{2} + 25x - 24 {x}^{2} - 24x - 12 = 0 \\ \\ \implies {x}^{2} + x - 12 = 0 \\ \end{gathered}

x+1

x

+

x

x+1

=

12

25

x(x+1)

x

2

+(x+1)(x+1)

=

12

25

x

2

+x

x

2

+x

2

+2x+1

=

12

25

x

2

+x

2x

2

+2x+1

=

12

25

⟹12(2x

2

+2x+1)=25(x

2

+x)

⟹24x

2

+24x+12=25x

2

+25x

⟹25x

2

+25x−24x

2

−24x−12=0

⟹x

2

+x−12=0

We have,

\begin{gathered}{x}^{2} + x - 12 = 0 \\ \end{gathered}

x

2

+x−12=0

Solving the Quadratic Equation by Quadratic Formula,

As we know,

\begin{gathered}\boxed{x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} } \\ \end{gathered}

x=

2a

−b±

b

2

−4ac

Here,

\begin{gathered}a = 1 , \: b= 1 \: \text{and} \: c = -12 \\ \end{gathered}

a=1,b=1andc=−12

Substituting the values and Finding the value of x,

\begin{gathered}\implies x = \frac{ - 1± \sqrt{ {(1)}^{2} - 4(1)( - 12) } }{2(1)} \\ \\ \implies x = \frac{ - 1± \sqrt{1 + 48} }{2} \\ \\ \implies x = \frac{ - 1± \sqrt{49} }{2} \\ \\ \implies x = \frac{ - 1± 7}{2} \\ \\ \implies x = \frac{ - 1 + 7}{2} \: \text{and} \: \frac{ - 1 - 7}{2} \\ \\ \implies x = \frac{6}{2} \: \text{and} \: \frac{ -8}{2} \\ \\ \implies x = 3 \: \text{and} \: - 4 \\ \end{gathered}

⟹x=

2(1)

−1±

(1)

2

−4(1)(−12)

⟹x=

2

−1±

1+48

⟹x=

2

−1±

49

⟹x=

2

−1±7

⟹x=

2

−1+7

and

2

−1−7

⟹x=

2

6

and

2

−8

⟹x=3and−4

Therefore,

The values of x is 3 and -4.

\begin{gathered} \\ \large \underline{ \underline{ \text{Required Answer:}}} \\ \end{gathered}

Required Answer:

The values of x is 3 and -4.

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