The number of integral values of a for which | vedclass

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(D) For $x^2 - (a - 1)x + 3 = 0$ to have both roots positive:$1$. Discriminant $D_1 \ge 0 \Rightarrow (a - 1)^2 - 12 \ge 0 \Rightarrow (a - 1)^2 \ge 1

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$1$

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(D) For $x^2 - (a - 1)x + 3 = 0$ to have both roots positive:$1$. Discriminant $D_1 \ge 0 \Rightarrow (a - 1)^2 - 12 \ge 0 \Rightarrow (a - 1)^2 \ge 12 \Rightarrow a - 1 \ge 2\sqrt{3}$ or $a - 1 \le -2\sqrt{3}$. Since $2\sqrt{3} \approx 3.46$,$a \ge 4.46$ or $a \le -2.46$.$2$. Sum of roots $> 0 \Rightarrow a - 1 > 0 \Rightarrow a > 1$.$3$. Product of roots $> 0 \Rightarrow 3 > 0$ (always true).Combining these,$a \ge 4.46$.For $x^2 + 3x + 6 - a = 0$ to have both roots negative:$1$. Discriminant $D_2 \ge 0 \Rightarrow 3^2 - 4(6 - a) \ge 0 \Rightarrow 9 - 24 + 4a \ge 0 \Rightarrow 4a \ge 15 \Rightarrow a \ge 3.75$.$2$. Sum of roots $$3$. Product of roots $> 0 \Rightarrow 6 - a > 0 \Rightarrow a Combining these,$3.75 \le a Intersection of both conditions: $a \ge 4.46$ and $3.75 \le a The integral values of $a$ are $5$. Thus,there is only $1$ such value.

The number of integral values of $a$ for which $x^2 - (a - 1)x + 3 = 0$ has both roots positive and $x^2 + 3x + 6 - a = 0$ has both roots negative is

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