Algebra theory class 3 solved examples
Q1. x 2 + ( 2 a − 1 ) x + a 2 = 0 x^2 + (2a - 1)x + a^2 = 0 α , β > 0 \alpha, \beta > 0 α \alpha & β \beta are real roots of above quadratic & a a is integer, Find value of ∣ α − β ∣ |\sqrt{\alpha} - \sqrt{\beta}| . Solution: S 1 = α + β = − ( 2 a − 1 ) 1 S_1 = \alpha + \beta = -\frac{(2a - 1)}{1} α + β = 1 − 2 a \alpha + \beta = 1 - 2a S 2 = α β = a 2 1 S_2 = \alpha \beta = \frac{a^2}{1} α β = a 2 \alpha \beta = a^2 ( α − β ) 2 = ( α ) 2 + ( β ) 2 − 2 α β (\sqrt{\alpha} - \sqrt{\beta})^2 = (\sqrt{\alpha})^2 + (\sqrt{\beta})^2 - 2\sqrt{\alpha\beta} = α + β − 2 α β = \alpha + \beta - 2\sqrt{\alpha\beta} = 1 − 2 a − 2 a 2 = 1 - 2a - 2\sqrt{a^2} = 1 − 2 a − 2 ∣ a ∣ = 1 - 2a - 2|a| D ≥ 0 D \geq 0 a x 2 + b x + c = 0 ax^2 + bx + c = 0 1 x 2 + ( 2 a − 1 ) x + a 2 = 0 1x^2 + (2a - 1)x + a^2 = 0 b 2 − 4 a c ≥ 0 b^2 - 4ac \geq 0 ( 2 a − 1 ) 2 − 4 ( 1 ) ( a 2 ) ≥ 0 (2a - 1)^2 - 4(1)(a^2) \geq 0 4 a 2 − 4 a + 1 − 4 a 2 ≥ 0 4a^2 - 4a + 1 - 4a^2 \geq 0 1 − 4 a ≥ ...
