Solve the quadratic equations by factorization method only x2 – x + 1 = 0

Given: x² – x + 1 = 0

x² – x + \( \dfrac{1}{4}\) + \( \dfrac{3}{4}\) = 0

x² – 2 (x) (\( \dfrac{1}{2}\)) + (\( \dfrac{1}{2}\))² + \( \dfrac{3}{4}\) = 0

(x – \( \dfrac{1}{2}\))² + \( \dfrac{3}{4}\) = 0 [Since, (a + b)² = a² + 2ab + b²]

(x – \( \dfrac{1}{2}\))² + \( \dfrac{3}{4}\) × 1 = 0

We know, i² = –1 = 1 = –i²

By substituting 1 = –i² in the above equation, we get

(x – \( \dfrac{1}{2}\))² + \( \dfrac{3}{4}\) (-1)² = 0

(x – \( \dfrac{1}{2}\))² + \( \dfrac{3}{4}\) (-i)² = 0

(x – \( \dfrac{1}{2}\))² – \( \sqrt{(\dfrac{3i}{2})^2}\)= 0

[By using the formula, a² – b² = (a + b) (a – b)]

(x – \( \dfrac{1}{2}\) + \( \sqrt{\dfrac{3i}{2}}\)) (x – \( \dfrac{1}{2}\) – \( \sqrt{\dfrac{3i}{2}}\)) = 0

(x – \( \dfrac{1}{2}\) + \( \sqrt{\dfrac{3i}{2}}\)) = 0 or (x – \( \dfrac{1}{2}\) –\( \sqrt{\dfrac{3i}{2}}\)) = 0

x = \( \dfrac{1}{2}\) – \( \sqrt{\dfrac{3i}{2}}\) or x = \( \dfrac{1}{2}\) + \( \sqrt{\dfrac{3i}{2}}\)

The roots of the given equation are \( \dfrac{1}{2}\) + \( \sqrt{\dfrac{3i}{2}}\), \( \dfrac{1}{2}\) – \( \sqrt{\dfrac{3i}{2}}\)