Question

If the roots of the equation x^2-bx+c=0 be two consecutiv

Answer 1

(x - a)(x - a - 1)

= x² - (a + 1)x - ax + a(a + 1)

= x² - (2a + 1)x + a(a + 1)

then

b² - 4ac = (2a + 1)² - 4a(a + 1)

= 4a² + 1 + 4a - 4a² - 4a

= 1

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