How to solve : $\frac{p x^{2} - 4 + 5 x}{p^{3} x - x^{2}} = 1$ $(px^2-4+5x)/(p^3x-x^2) = 1$ ?

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How to solve : $\frac{p x^{2} - 4 + 5 x}{p^{3} x - x^{2}} = 1$ $(px^2-4+5x)/(p^3x-x^2) = 1$ ?

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Answer 1 · 2018-04-06T15:31:57

$\frac{p x^{2} - 4 + 5 x}{p^{3} x - x^{2}} = 1$ $(px^2-4+5x)/(p^3x-x^2)=1$

$\Rightarrow p x^{2} - 4 + 5 x = p^{3} x - x^{2}$ $=>px^2-4+5x=p^3x-x^2$

$\Rightarrow p x^{2} - 4 + 5 x - p^{3} x + x^{2} = 0$ $=>px^2-4+5x-p^3x+x^2=0$

$\Rightarrow (p + 1) \cdot x^{2} - (p^{3} - 5) \cdot x - 4 = 0$ $=>(p+1)cdot x^2-(p^3-5)cdot x-4=0$

Applying Shreedhar Acharya's Formula,

$x = \frac{(p^{3} - 5) \pm \sqrt{{(p^{3} - 5)}^{2} + 4 \times 4 \times (p + 1)}}{2 \cdot (p + 1)}$ $x=((p^3-5)+-sqrt((p^3-5)^2+4xx4xx(p+1)))/(2 cdot (p+1)$

$x = \frac{(p^{3} - 5) \pm \sqrt{p^{6} - 10 p^{3} + 25 + 16 p + 16}}{2 \cdot (p + 1)}$ $x=((p^3-5)+-sqrt(p^6-10p^3+25+16p+16))/(2 cdot (p+1)$

$x = \frac{(p^{3} - 5) \pm \sqrt{p^{6} - 10 p^{3} + 16 p + 41}}{2 \cdot (p + 1)}$ $x=((p^3-5)+-sqrt(p^6-10p^3+16p+41))/(2 cdot (p+1)$

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How to solve : $\frac{p x^{2} - 4 + 5 x}{p^{3} x - x^{2}} = 1$ $(px^2-4+5x)/(p^3x-x^2) = 1$ ?

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