If x is real,then the value of the expression | vedclass

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(C) Let $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.Multiplying both sides by the denominator,we get $x^2 + 14x + 9 = y(x^2 + 2x + 3)$.Rearranging the te

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$-5$ and $4$

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(C) Let $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.Multiplying both sides by the denominator,we get $x^2 + 14x + 9 = y(x^2 + 2x + 3)$.Rearranging the terms to form a quadratic equation in $x$: $x^2(1 - y) + x(14 - 2y) + (9 - 3y) = 0$.Since $x$ is real,the discriminant $D$ must be greater than or equal to $0$ $(D \ge 0)$.$D = (14 - 2y)^2 - 4(1 - y)(9 - 3y) \ge 0$.Expanding the terms: $(196 + 4y^2 - 56y) - 4(9 - 3y - 9y + 3y^2) \ge 0$.$196 + 4y^2 - 56y - 36 + 48y - 12y^2 \ge 0$.$-8y^2 - 8y + 160 \ge 0$.Dividing by $-8$ (and reversing the inequality sign): $y^2 + y - 20 \le 0$.Factoring the quadratic: $(y + 5)(y - 4) \le 0$.Thus,the value of $y$ lies between $-5$ and $4$ inclusive,i.e.,$-5 \le y \le 4$.

If $x$ is real,then the value of the expression $\frac{x^2 + 14x + 9}{x^2 + 2x + 3}$ lies between

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