If exactly one root of the equation x^2 + (a | vedclass

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(B) Let $f(x) = x^2 + (a - 1)x + 2a$. For exactly one root to lie in $(0, 3)$,we consider the following cases:Case $1$: $f(0) \cdot f(3) < 0$$f(0) = 2

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$(-\infty, 0] \cup (6, \infty)$

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(B) Let $f(x) = x^2 + (a - 1)x + 2a$. For exactly one root to lie in $(0, 3)$,we consider the following cases:Case $1$: $f(0) \cdot f(3) $f(0) = 2a$$f(3) = 9 + 3(a - 1) + 2a = 9 + 3a - 3 + 2a = 5a + 6$So,$2a(5a + 6) Case $2$: One root is at the boundary.If $x = 0$ is a root,$f(0) = 0 \Rightarrow 2a = 0 \Rightarrow a = 0$. The equation becomes $x^2 - x = 0$,roots are $0, 1$. Since $1 \in (0, 3)$,$a = 0$ is a solution.If $x = 3$ is a root,$f(3) = 0 \Rightarrow 5a + 6 = 0 \Rightarrow a = -1.2$. The equation becomes $x^2 - 2.2x - 2.4 = 0$. Roots are $3$ and $-0.8$. Since only one root $3$ is not in $(0, 3)$ (it is on the boundary),we check if the other root is in $(0, 3)$. $-0.8$ is not in $(0, 3)$. Thus $a = -1.2$ is not included.Combining these,$a \in (-1.2, 0]$.Wait,re-evaluating the original equation $x^2 + (a-1)x + 2a = 0$:$f(0) = 2a$,$f(3) = 9 + 3a - 3 + 2a = 5a + 6$.$f(0)f(3) Checking $a=0$: $x^2-x=0 \Rightarrow x=0, 1$. Root $1 \in (0,3)$. Correct.Checking $a=-1.2$: $x^2-2.2x-2.4=0 \Rightarrow (x-3)(x+0.8)=0$. Roots $3, -0.8$. Neither is in $(0,3)$.Thus,the set is $(-1.2, 0]$.Given the options provided in the prompt,there is a discrepancy. Based on the standard interpretation of such problems,the correct set is $(-1.2, 0]$.

If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

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