If $y = x + \frac{1}{x}$ $y = x + 1/x$ , then write the equation $x^{4} + x^{3} - 4 x^{2} + x + 1 = 0$ $x^4 + x^3 -4x^2 +x +1 = 0$ in terms of $y$ $y$ ?

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If $y = x + \frac{1}{x}$ $y = x + 1/x$ , then write the equation $x^{4} + x^{3} - 4 x^{2} + x + 1 = 0$ $x^4 + x^3 -4x^2 +x +1 = 0$ in terms of $y$ $y$ ?

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Answer 1 · 2017-06-06T23:03:04

$y + y^{2} - 6 = 0$ $y+y^2-6=0$

Explanation:

$x^{4} + x^{3} - 4 x^{2} + x + 1 = x^{2} (x^{2} + x - 4 + \frac{1}{x} + \frac{1}{x^{2}}) = x^{2} (y + y^{2} - 6) = 0$ $x^4 + x^3 -4x^2 +x +1=x^2(x^2+x-4+1/x+1/x^2) = x^2(y+y^2-6)=0$

but $x \neq 0$ $x ne 0$ so

$x^{4} + x^{3} - 4 x^{2} + x + 1 = 0 \Leftrightarrow y + y^{2} - 6 = 0$ $x^4 + x^3 -4x^2 +x +1=0 hArr y+y^2-6=0$

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If $y = x + \frac{1}{x}$ $y = x + 1/x$ , then write the equation $x^{4} + x^{3} - 4 x^{2} + x + 1 = 0$ $x^4 + x^3 -4x^2 +x +1 = 0$ in terms of $y$ $y$ ?

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If y=x+1xy = x + 1/x, then write the equation x4+x3−4x2+x+1=0x^4 + x^3 -4x^2 +x +1 = 0 in terms of yy?

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If $y = x + \frac{1}{x}$ $y = x + 1/x$ , then write the equation $x^{4} + x^{3} - 4 x^{2} + x + 1 = 0$ $x^4 + x^3 -4x^2 +x +1 = 0$ in terms of $y$ $y$ ?