# quadratic equation - nature of roots

For what values of a does the equation $$x^2-( 2^a-1)x-3(4^{a-1}2^{a-2})=0$$ possess real roots? Since the roots are to be real that means the discriminant should be $\geq 0$ $$\Rightarrow (2^a-1)^2+4\cdot 3\cdot (4^{a-1}2^{a-2}) \geq 0$$

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Also, if the discriminant is nonnegative, the roots are real (so any solution to the inequality truly does make the polynomial have real roots). Anyways, have you tried solving that inequality? –  Hurkyl Feb 18 '13 at 16:55

here by substituting $2^a=t$ I am getting an equation in the form : $3t^3+4t^2-8t+4=$ which is not giving any solution. Please suggest. –  Sachin Sharmaa Feb 19 '13 at 3:32
@SachinSharmaa: The first one is a square, so is positive. Powers of positive numbers are always positive, so the second is, too. This is general: given a quadratic $ax^2+bx+c=0$ with $ac \lt 0$, both roots are real. –  Ross Millikan Feb 19 '13 at 3:51