Number of integral values of a for which both | vedclass

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(A) Let $f(x) = x^2 - (2a + 3)x + a^2 + 3a$. For both roots to lie in $(0, 4)$,the following conditions must be satisfied:$1$. Discriminant $D \ge 0$:

Vedclass · Accepted Answer

$0$

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(A) Let $f(x) = x^2 - (2a + 3)x + a^2 + 3a$. For both roots to lie in $(0, 4)$,the following conditions must be satisfied:$1$. Discriminant $D \ge 0$: $D = (2a + 3)^2 - 4(a^2 + 3a) = 4a^2 + 12a + 9 - 4a^2 - 12a = 9$. Since $9 > 0$,the roots are always real.$2$. $0 $3$. $f(0) > 0$: $a^2 + 3a > 0 \implies a(a + 3) > 0 \implies a \in (-\infty, -3) \cup (0, \infty)$.$4$. $f(4) > 0$: $16 - 4(2a + 3) + a^2 + 3a > 0 \implies 16 - 8a - 12 + a^2 + 3a > 0 \implies a^2 - 5a + 4 > 0 \implies (a - 1)(a - 4) > 0 \implies a \in (-\infty, 1) \cup (4, \infty)$.Taking the intersection of all conditions: $a \in (0, 1)$.The only integer value in this interval is none. However,checking the boundary conditions for roots to be in $(0, 4)$,we find no integer $a$ satisfies the condition. Thus,the number of integral values is $0$.

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

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