Distinct Solutions 868Df9

1. **State the problem:** We want to find the number of integer values of $k$ in the range $-6 \leq k \leq 6$ such that the equation $$4^x - k2^x + 4 = 0$$ has two distinct real solutions. 2. **Rewrite the equation:** Let $c = 2^x$. Since $2^x > 0$ for all real $x$, $c > 0$. The equation becomes: $$4^x - k2^x + 4 = 0 \implies c^2 - kc + 4 = 0$$ because $4^x = (2^x)^2 = c^2$. 3. **Analyze the quadratic in $c$:** The quadratic equation is: $$c^2 - kc + 4 = 0$$ We want two distinct real solutions for $x$, which means two distinct positive real solutions for $c$ (since $c=2^x > 0$). 4. **Conditions for two distinct real solutions in $c$:** - The discriminant must be positive: $$\Delta = k^2 - 4 \cdot 1 \cdot 4 = k^2 - 16 > 0$$ - The roots must be positive. For a quadratic $c^2 - kc + 4 = 0$, sum of roots $= k$ and product of roots $= 4$. Since product $=4 > 0$, both roots have the same sign. If sum $= k > 0$, both roots are positive. If sum $= k < 0$, both roots are negative. We want positive roots, so $k > 0$. 5. **Summarize conditions:** $$k^2 - 16 > 0 \implies k^2 > 16 \implies |k| > 4$$ and $$k > 0$$ So $k$ must be an integer greater than 4. 6. **Check integer values of $k$ in $-6 \leq k \leq 6$ satisfying these:** Possible $k$ values with $|k| > 4$ are $k = -6, -5, 5, 6$. From these, only $k=5$ and $k=6$ satisfy $k > 0$. 7. **Conclusion:** There are 2 integer values of $k$ in the given range for which the equation has two distinct real solutions. **Final answer:** 2