Answers to 2 decimal places where necessary. Proof, by solving for \( x \) that the roots of \( x^{2}-2 x=7 \) are irrational.

To prove that the roots of the quadratic equation \( x^{2} - 2x = 7 \) are irrational, we'll solve for \( x \) and analyze the nature of the solutions. ### Step 1: Rewrite the Equation First, rearrange the equation to standard quadratic form: \[ x^{2} - 2x - 7 = 0 \] ### Step 2: Apply the Quadratic Formula The quadratic formula to find the roots of \( ax^{2} + bx + c = 0 \) is: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For our equation, \( a = 1 \), \( b = -2 \), and \( c = -7 \). Plugging these values into the formula gives: \[ x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-7)}}{2(1)} = \frac{2 \pm \sqrt{4 + 28}}{2} = \frac{2 \pm \sqrt{32}}{2} \] ### Step 3: Simplify the Square Root Simplify \( \sqrt{32} \): \[ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \] So, the solutions become: \[ x = \frac{2 \pm 4\sqrt{2}}{2} = 1 \pm 2\sqrt{2} \] ### Step 4: Analyze the Solutions The solutions are: \[ x = 1 + 2\sqrt{2} \quad \text{and} \quad x = 1 - 2\sqrt{2} \] Since \( \sqrt{2} \) is an irrational number, multiplying it by 2 and adding or subtracting 1 will still result in an irrational number. Therefore, both roots \( 1 + 2\sqrt{2} \) and \( 1 - 2\sqrt{2} \) are irrational. ### Step 5: Decimal Approximations (to Two Decimal Places) For practical purposes, we can approximate the roots as follows: \[ \sqrt{2} \approx 1.4142 \] \[ 2\sqrt{2} \approx 2 \times 1.4142 = 2.8284 \] Thus, \[ x \approx 1 + 2.8284 = 3.8284 \approx 3.83 \] \[ x \approx 1 - 2.8284 = -1.8284 \approx -1.83 \] ### Conclusion The roots of the equation \( x^{2} - 2x = 7 \) are: \[ x = 1 + 2\sqrt{2} \approx 3.83 \quad \text{and} \quad x = 1 - 2\sqrt{2} \approx -1.83 \] Both roots are **irrational**.