The values of frac{x^2-2x+1}{x^2+x-1} do not | vedclass

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(A) Let $y = \frac{x^2-2x+1}{x^2+x-1}$.$y(x^2+x-1) = x^2-2x+1$$yx^2 + yx - y = x^2 - 2x + 1$$(y-1)x^2 + (y+2)x - (y+1) = 0$.Since $x \in \mathbb{R}$,

Vedclass · Accepted Answer

$\left(-\frac{4}{5}, 0\right)$

Vedclass · Answer

(A) Let $y = \frac{x^2-2x+1}{x^2+x-1}$.$y(x^2+x-1) = x^2-2x+1$$yx^2 + yx - y = x^2 - 2x + 1$$(y-1)x^2 + (y+2)x - (y+1) = 0$.Since $x \in \mathbb{R}$, the discriminant $D \geq 0$.$D = (y+2)^2 - 4(y-1)(-(y+1)) \geq 0$$(y+2)^2 + 4(y-1)(y+1) \geq 0$$y^2 + 4y + 4 + 4(y^2-1) \geq 0$$y^2 + 4y + 4 + 4y^2 - 4 \geq 0$$5y^2 + 4y \geq 0$$y(5y+4) \geq 0$.The solution to this inequality is $y \in (-\infty, -\frac{4}{5}] \cup [0, \infty)$.Therefore, the values of $y$ do not lie in the interval $\left(-\frac{4}{5}, 0\right)$.

The values of $\frac{x^2-2x+1}{x^2+x-1}$ do not lie in the interval

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