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 $t = 2^x > 2^0 = 1$. Substituting $t$ into the equation,we get $t^2 - (k + 2)t + 2k = 0$. Factoring the quadr

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$k \in (-\infty, 0] \cup \{2\}$

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(B) Let $2^x = t$. Since $x > 0$,we have $t = 2^x > 2^0 = 1$. Substituting $t$ into the equation,we get $t^2 - (k + 2)t + 2k = 0$. Factoring the quadratic equation: $(t - k)(t - 2) = 0$. So,the roots are $t_1 = k$ and $t_2 = 2$. We require exactly one positive root for $x$,which means exactly one root for $t$ must be greater than $1$. We already have one root $t_2 = 2$,which is greater than $1$. For the equation to have exactly one root $t > 1$,the other root $t_1 = k$ must satisfy $k \le 1$. Also,if $k = 2$,the roots are $t_1 = 2, t_2 = 2$,which gives only one distinct root $t = 2 > 1$,which is valid. Thus,the condition is $k \le 1$ or $k = 2$.

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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