Question

Quarditic equation...
Solve the following equation by factorization method ​

Answer 1

If we replace $\dfrac{x}{x+1}=t$, we have $5t^2-4t-1$

$5t^2-4t-1$

5t            1

t              -1

$=(5t+1)(t-1)$

The zeros are at $5t+1=0$ or $t-1=0$

$t=-\dfrac{1}{5}$ or $t=1$ are the solutions.

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The solution for $\dfrac{x}{x+1}=-\dfrac{1}{5}$ is $x=-\dfrac{1}{6}$

The solution for $\dfrac{x}{x+1}=1$ does not exist.

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Therefore, a solution is $x=-\dfrac{1}{6}$

Answer 2

Answer:

x= -1/6

Step-by-step explanation:

5(x/x+1)^2 -4(x/x+1) -1 =0

(5x^2/x^2+2x+1) -(4x/x+1 )=1     {after multiplication}

(5x^3+5x^2-4x^3-8x^2-4x)/(x^3+3x^2+3x+1)=1     {after taking Lcm}

x^3-3x^2-4x=x^3+3x^2+3x+1      {taking the denominator to the right side}

x^3-3x^2-4x-x^3-3x^2-3x-1 =0

-6x^2-7x-1=0

6x^2+7x+1=0   {multiplying both sides with -1}

6x^2+6x+x+1=0

6x(x+1)+1(x+1)=0 after simplification u will get the answer

(x+1)(6x+1)=0 SO,x+1=0 or 6x+1=0 X= -1 OR -1/6 but since it is given that x is not equal to -1, x=-1/6

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