Question

Find all values of $a$ for which the equation $4^x-{a2}^x-a+3=0$ has at least one solution.

• A. $a\in (-\infty ,2)$
• B. $a\in (2,\infty )$
• C. $a\in (-\infty ,-2)$
• D. $a\in (-2,\infty )$

Answer 1

The correct option is B $a\in (2,\infty )$

Given equation $4^x-{a2}^x-a+3=0$
Putting $2^x=t>0$
$\Rightarrow t^2-at-a+3=0.......\left(1\right)$
Equation $\left(1\right)$ should have at least one positive root $(t>0)$, then
Discriminant $D={(-a)}^2-4.1(-a+3)\ge 0$
$\Rightarrow a^2+4a-12\ge 0$
$\Rightarrow (a+6)(a-2)\ge 0$
$\therefore a\in (-\infty ,-6)\cup (2,\infty )$
If roots of the equation $\left(1\right)$ are $t_1,t_2$,
Then, $t_1+t_2=a$
$t_1{t}_2=3-a$
For $a\in (-\infty ,-6)$
$t_1+t_2\le 0$ and $t_1{t}_2\ge 0$.
Therefore, both roots are negative and consequently, the original equation has no solutions.
For $a\in (2,\infty )$
$t_1+t_2\ge 0$
and $t_1{t}_2\le 0$
Consequently, at least one roots $t_1$ or $t_2$, is greater than zero.
Thus, for $a\in (2,\infty )$ the given equation has at least one solution.