If x is complex,the expression frac{x^2+34x-7 | vedclass

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(C) Let $y = \frac{x^2+34x-71}{x^2+2x-7}$.Then,$x^2+34x-71 = y(x^2+2x-7)$.Rearranging the terms,we get $x^2(y-1) + x(2y-34) + (71-7y) = 0$.Since $x$ i

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$a=5, b=9$

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(C) Let $y = \frac{x^2+34x-71}{x^2+2x-7}$.Then,$x^2+34x-71 = y(x^2+2x-7)$.Rearranging the terms,we get $x^2(y-1) + x(2y-34) + (71-7y) = 0$.Since $x$ is a complex number,the discriminant $D$ of this quadratic equation in $x$ must be less than or equal to $0$ for the expression to take all complex values,but for the range of the rational function,we analyze the condition $D \leq 0$:$D = (2y-34)^2 - 4(y-1)(71-7y) \leq 0$.Expanding this,we get $4(y-17)^2 - 4(-7y^2 + 78y - 71) \leq 0$.Dividing by $4$,we have $(y^2 - 34y + 289) + 7y^2 - 78y + 71 \leq 0$.$8y^2 - 112y + 360 \leq 0$.Dividing by $8$,we get $y^2 - 14y + 45 \leq 0$.Factoring the quadratic,$(y-5)(y-9) \leq 0$.Thus,$5 \leq y \leq 9$.However,for the expression to take all values in the interval $(a, b)$,we identify $a=5$ and $b=9$.

If $x$ is complex,the expression $\frac{x^2+34x-71}{x^2+2x-7}$ takes all values which lie in the interval $(a, b)$,find the value of $a$ and $b$.

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