Y= x^2 +x -1 / x^2 -x-1

1.	Y= x^2 +x -1 / x^2 -x-1
Answer» ANSWER : To find the VALUE of the given EQUATION – $y= \frac{ x ^ { 2 } +x-1 }{ x ^ { 2 } -x-1 }$ , FOLLOW the STEPS... $\Rightarrow y\left(x-\left(-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)=x^{2}+x-1$ $\Rightarrow y\left(x+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(x-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)=x^{2}+x-1$ $\Rightarrow y\left(x+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(x-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)=x^{2}+x-1$ $\Rightarrow y\left(x+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(x-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)=x^{2}+x-1$ $\Rightarrow \left(yx+\frac{1}{2}y\sqrt{5}-\frac{1}{2}y\right)\left(x-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)=x^{2}+x-1$ $\Rightarrow yx^{2}-yx-\frac{1}{4}y\left(\sqrt{5}\right)^{2}+\frac{1}{4}y=x^{2}+x-1$ $\Rightarrow yx^{2}-yx-\frac{1}{4}y\times 5+\frac{1}{4}y=x^{2}+x-1$ $\Rightarrow yx^{2}-yx-\frac{5}{4}y+\frac{1}{4}y=x^{2}+x-1$ $\Rightarrow yx^{2}-yx-y=x^{2}+x-1$ $\Rightarrow yx^{2}-yx-y-x^{2}-x=-1$ $\Rightarrow yx^{2}-yx-y-x^{2}-x+1=0$ $\Rightarrow \left(y-1\right)x^{2}+\left(-y-1\right)x-y+1=0$ $\Rightarrow \left(y-1\right)x^{2}+\left(-y-1\right)x+1-y=0$ $\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(-y-1\right)^{2}-4\left(y-1\right)\left(1-y\right)}}{2\left(y-1\right)}$ $\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(y+1\right)^{2}-4\left(y-1\right)\left(1-y\right)}}{2\left(y-1\right)}$ $\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(y+1\right)^{2}+\left(4-4y\right)\left(1-y\right)}}{2\left(y-1\right)}$ $\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(y+1\right)^{2}+4\left(1-y\right)^{2}}}{2\left(y-1\right)}$ $\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{5y^{2}-6y+5}}{2\left(y-1\right)}$ $\Rightarrow x=\frac{y+1±\sqrt{5y^{2}-6y+5}}{2y-2}$ $\Rightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2y-2}$ $\Rightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$ $\Rightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2y-2}$ $\Rightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$ $\Rightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)} \\ \Rightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$ $\\$ This, we can see that the value for $y= \frac{ x ^ { 2 } +x-1 }{ x ^ { 2 } -x-1 }$ will be $\bf \hookrightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)} \\ \bf \hookrightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$ .

Answer»

ANSWER :

To find the VALUE of the given EQUATION – $y= \frac{ x ^ { 2 } +x-1 }{ x ^ { 2 } -x-1 }$ ,

$\Rightarrow y\left(x-\left(-\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)=x^{2}+x-1$

$\Rightarrow y\left(x+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(x-\left(\frac{1}{2}\sqrt{5}+\frac{1}{2}\right)\right)=x^{2}+x-1$

$\Rightarrow y\left(x+\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)\left(x-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)=x^{2}+x-1$

$\Rightarrow \left(yx+\frac{1}{2}y\sqrt{5}-\frac{1}{2}y\right)\left(x-\frac{1}{2}\sqrt{5}-\frac{1}{2}\right)=x^{2}+x-1$

$\Rightarrow yx^{2}-yx-\frac{1}{4}y\left(\sqrt{5}\right)^{2}+\frac{1}{4}y=x^{2}+x-1$

$\Rightarrow yx^{2}-yx-\frac{1}{4}y\times 5+\frac{1}{4}y=x^{2}+x-1$

$\Rightarrow yx^{2}-yx-\frac{5}{4}y+\frac{1}{4}y=x^{2}+x-1$

$\Rightarrow yx^{2}-yx-y=x^{2}+x-1$

$\Rightarrow yx^{2}-yx-y-x^{2}-x=-1$

$\Rightarrow yx^{2}-yx-y-x^{2}-x+1=0$

$\Rightarrow \left(y-1\right)x^{2}+\left(-y-1\right)x-y+1=0$

$\Rightarrow \left(y-1\right)x^{2}+\left(-y-1\right)x+1-y=0$

$\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(-y-1\right)^{2}-4\left(y-1\right)\left(1-y\right)}}{2\left(y-1\right)}$

$\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(y+1\right)^{2}-4\left(y-1\right)\left(1-y\right)}}{2\left(y-1\right)}$

$\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(y+1\right)^{2}+\left(4-4y\right)\left(1-y\right)}}{2\left(y-1\right)}$

$\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{\left(y+1\right)^{2}+4\left(1-y\right)^{2}}}{2\left(y-1\right)}$

$\Rightarrow x=\frac{-\left(-y-1\right)±\sqrt{5y^{2}-6y+5}}{2\left(y-1\right)}$

$\Rightarrow x=\frac{y+1±\sqrt{5y^{2}-6y+5}}{2y-2}$

$\Rightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2y-2}$

$\Rightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$

$\Rightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2y-2}$

$\Rightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$

$\Rightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)} \\ \Rightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$

$\\$

This, we can see that the value for $y= \frac{ x ^ { 2 } +x-1 }{ x ^ { 2 } -x-1 }$ will be $\bf \hookrightarrow x=\frac{\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)} \\ \bf \hookrightarrow x=\frac{-\sqrt{5y^{2}-6y+5}+y+1}{2\left(y-1\right)}$ .

Y= x^2 +x -1 / x^2 -x-1

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Y= x^2 +x -1 / x^2 -x-1​

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Y= x^2 +x -1 / x^2 -x-1