If x is real,then the value of frac{x^2 + 34x | vedclass

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(D) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.Rearranging the equation,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$,which simplifies to $x^2(y - 1) +

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$5$ and $9$

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(D) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.Rearranging the equation,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$,which simplifies to $x^2(y - 1) + 2x(y - 17) + (71 - 7y) = 0$.For $x$ to be real,the discriminant $D$ must be greater than or equal to $0$ $(D \ge 0)$.$D = [2(y - 17)]^2 - 4(y - 1)(71 - 7y) \ge 0$.Dividing by $4$,we get $(y - 17)^2 - (y - 1)(71 - 7y) \ge 0$.$(y^2 - 34y + 289) - (71y - 7y^2 - 71 + 7y) \ge 0$.$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$.$8y^2 - 112y + 360 \ge 0$.Dividing by $8$,we get $y^2 - 14y + 45 \ge 0$.Factoring the quadratic,we get $(y - 5)(y - 9) \ge 0$.This inequality holds when $y \le 5$ or $y \ge 9$.Therefore,the value of the expression does not lie between $5$ and $9$.

If $x$ is real,then the value of $\frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ does not lie between

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