Show that the roots of equation \( 2 x^{2}+3=7 x \) are rational

To show that the roots of the equation \( 2x^{2} + 3 = 7x \) are rational, we first rewrite it in standard form: \( 2x^{2} - 7x + 3 = 0 \). Next, we will use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, \( a = 2 \), \( b = -7 \), and \( c = 3 \). Now, we calculate the discriminant \( b^2 - 4ac \): \[ (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25. \] Since the discriminant is a perfect square (\( 25 = 5^2 \)), the roots are: \[ x = \frac{7 \pm 5}{2 \cdot 2} = \frac{7 \pm 5}{4}. \] Calculating the two possible values gives: 1. \( x = \frac{12}{4} = 3 \) 2. \( x = \frac{2}{4} = \frac{1}{2} \) Both roots \( 3 \) and \( \frac{1}{2} \) are rational numbers. Hence, the roots of the given equation are rational!