1.5 \( 2 \sqrt{2-7 x}=\sqrt{-36 x} \) \( \sqrt{2-x}-x=-2 \) .7 \( 26-5^{2 x}=\left(5^{x}-6\right)^{2} \) Show that the equation \( 2^{2 x+1}+7.2^{x}-4=0 \) has only ONE solution. SIMPLIFICATION

To show that the equation \[ 2^{2x+1} + 7 \cdot 2^{x} - 4 = 0 \] has only **one solution**, let's proceed with a step-by-step simplification and analysis. ### Step 1: Simplify the Equation First, rewrite \(2^{2x + 1}\) to make the equation easier to handle: \[ 2^{2x + 1} = 2 \cdot 2^{2x} = 2 \cdot (2^{x})^2 \] So, the original equation becomes: \[ 2 \cdot (2^{x})^2 + 7 \cdot 2^{x} - 4 = 0 \] ### Step 2: Substitute to Form a Quadratic Equation Let’s make a substitution to simplify the equation further. Let: \[ t = 2^{x} \] Since \(2^{x} > 0\) for all real \(x\), \(t > 0\). Substituting \(t\) into the equation: \[ 2t^2 + 7t - 4 = 0 \] ### Step 3: Solve the Quadratic Equation Now, solve for \(t\) using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 2\), \(b = 7\), and \(c = -4\). Plugging these values in: \[ t = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4} \] This gives two potential solutions for \(t\): 1. \(t = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}\) 2. \(t = \frac{-7 - 9}{4} = \frac{-16}{4} = -4\) ### Step 4: Consider the Valid Solutions Recall that \(t = 2^{x} > 0\). Therefore, \(t = -4\) is **not valid**. Only \(t = \frac{1}{2}\) is acceptable. ### Step 5: Find the Corresponding \(x\) Solve for \(x\) using the valid \(t\): \[ 2^{x} = \frac{1}{2} \implies x = \log_{2}\left(\frac{1}{2}\right) = -1 \] ### Conclusion The equation \[ 2^{2x+1} + 7 \cdot 2^{x} - 4 = 0 \] has only **one valid solution**, which is: \[ x = -1 \] This single solution arises because the quadratic equation derived from the substitution yields only one positive value for \(t\), corresponding to a unique real value for \(x\).