Question

For what values of $a$ does the equation $x^2-(2^a-1)x-3\left(4^{a-1}{2}^{a-2}\right)=0$ possess real roots?

Answer 1

$x^2-(2^a-1)x-3\left(4^{a-1}{2}^{a-2}\right)=0$

Roots are real

$\therefore △\ge 0$

$(2^a-1{)}^2+12(4^{a-1}{2}^{a-2})\ge 0$

$2^{2a}-{2.2}^a+1+{3.2}^2{.2}^{2a-2}{.2}^{a-2}\ge 0$

Let $2^a=t$

$t^2-2t+1+\frac{3}{4}{t}^3\ge 0$

$3t^3+4t^2-8t+1\ge 0$

$(t-1)(3t^2+7t-1)\ge 0$

$\therefore t\ge 1$

$2^a\ge 1$

$a\in (0,\infty )$