Question

2x^2 - 3x - 1 = 0 by completing the square

Answer 1

Given 2x² - 3x - 1 = 0

x² - 3/2 x - 1/2 = 0

x² - 3/2 x = 1/2

x² - 3/2 x + (3/4)² = 1/2 + (3/4)²

\({{(x-3)}\over2}^2\) = 1/2 + 9/16

\({{(x-3)}\over2}^2\) = (8+9)/16

x - 3/2 = ±\(\sqrt{17}\)/4

x = +\(\sqrt{17}\)/4 + 3/2

⇒ \(\frac{(\sqrt{17}+ 6)}4\)

x = - \(\sqrt{17}\)/4 + 3/2

⇒ \(\frac{(-\sqrt{17}+ 6)}4\)