Question: 6. Solve $$2^{2x+2} - 3 \times 2^x - 10 = 0.$$ (3) Math expression: $$2^{2x+2} - 3 \times 2^x - 10 = 0,$$ positioned center.solve this step by step
1. **State the problem:** Solve the equation $$2^{2x+2} - 3 \times 2^x - 10 = 0.$$
2. **Rewrite the equation using properties of exponents:** Recall that $$2^{2x+2} = 2^{2x} \times 2^2 = 4 \times (2^{2x}).$$ Also, $$2^{2x} = (2^x)^2.$$ So the equation becomes:
$$4 \times (2^x)^2 - 3 \times 2^x - 10 = 0.$$
3. **Substitute:** Let $$y = 2^x.$$ Then the equation is:
$$4y^2 - 3y - 10 = 0.$$
4. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=4,$$ $$b=-3,$$ and $$c=-10.$$
Calculate the discriminant:
$$\Delta = (-3)^2 - 4 \times 4 \times (-10) = 9 + 160 = 169.$$
5. **Find the roots:**
$$y = \frac{3 \pm \sqrt{169}}{2 \times 4} = \frac{3 \pm 13}{8}.$$
6. **Calculate each root:**
- $$y_1 = \frac{3 + 13}{8} = \frac{16}{8} = 2.$$
- $$y_2 = \frac{3 - 13}{8} = \frac{-10}{8} = -\frac{5}{4}.$$
7. **Recall substitution:** Since $$y = 2^x,$$ and $$2^x > 0$$ for all real $$x,$$ the negative root $$y_2 = -\frac{5}{4}$$ is invalid.
8. **Solve for $$x$$:**
$$2^x = 2 \implies x = 1.$$
**Final answer:** $$x = 1.$$