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2x^2 - 3x - 1 = 0 by completing the square |
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Answer» 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 ⇒ \(\frac{(-\sqrt{17}+ 6)}4\) |
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