Find the complete set of values of k for whic | vedclass

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(B) Let $2^x = t$. Since $x > 0$,we have $t > 1$. The equation becomes $t^2 - (k + 2)t + 2k = 0$.Factoring the quadratic: $(t - k)(t - 2) = 0$.So,the

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$( - \infty, 1 ]$

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(B) Let $2^x = t$. Since $x > 0$,we have $t > 1$. The equation becomes $t^2 - (k + 2)t + 2k = 0$.Factoring the quadratic: $(t - k)(t - 2) = 0$.So,the roots are $t_1 = k$ and $t_2 = 2$.For the original equation to have exactly one positive root,we analyze the roots of $t$:$1$. One root $t_2 = 2$ is already greater than $1$,which corresponds to $x = \log_2(2) = 1$,which is a positive root.$2$. For the equation to have exactly one positive root,the other root $t_1 = k$ must not yield a positive $x$. Since $t = 2^x$,$t > 0$ is required. If $t_1 = k \le 1$,then $x = \log_2(k) \le 0$.$3$. If $k = 2$,the roots are $t = 2, 2$. Then $x = 1$ (only one distinct positive root). Thus $k=2$ is a solution.$4$. If $k \le 1$,the roots are $t_1 \le 1$ and $t_2 = 2$. $t_1 \le 1$ gives $x \le 0$,and $t_2 = 2$ gives $x = 1$. This satisfies the condition.Combining these,the set of values is $(-\infty, 1] \cup \{2\}$.

Find the complete set of values of $k$ for which the equation $4^x - (k + 2)2^x + 2k = 0$ has exactly one positive root.

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