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

To find the roots of the equation \( x^{2} - 2x = 7 \), we can rewrite it in standard form as \( x^{2} - 2x - 7 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -7 \), we first calculate the discriminant: \[ b^{2} - 4ac = (-2)^{2} - 4(1)(-7) = 4 + 28 = 32. \] Since the discriminant is positive, we have real roots. However, we need to check if they are rational or irrational. The square root of 32 can be simplified to \( 4\sqrt{2} \), which is not a perfect square. Therefore, the roots are: \[ x = \frac{2 \pm 4\sqrt{2}}{2} = 1 \pm 2\sqrt{2}. \] Since \( 2\sqrt{2} \) is irrational, the roots \( 1 + 2\sqrt{2} \) and \( 1 - 2\sqrt{2} \) are also irrational, proving our claim!