Question

j140b

1. given that (x^2 + 2mx + m^2 - 2m + 3 = 0) has 2 roots (alpha) and (\beta) (with (alpha < \beta)).

( 1 ) find (alpha + \beta) and (alpha\beta) in terms of (m).
( 2 ) find ((alpha - \beta)^2) in terms of (m).
( 3 ) if (\beta - alpha = 2), find (m).

Explanation:

Step1: Identify coefficients of the quadratic equation

$$x^2 + 2mx + (m^2 - 2m + 3) = 0$$

Step2: Apply Vieta's formulas for roots

$$\alpha + \beta = -2m, \quad \alpha\beta = m^2 - 2m + 3$$

Step3: Express the squared difference of roots

$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$

Step4: Substitute Vieta's relations into expression

$$(\alpha - \beta)^2 = (-2m)^2 - 4(m^2 - 2m + 3)$$

Step5: Simplify the algebraic expression

$$(\alpha - \beta)^2 = 4m^2 - 4m^2 + 8m - 12 = 8m - 12$$

Step6: Relate given difference to squared difference

$$\beta - \alpha = 2 \implies (\alpha - \beta)^2 = 2^2 = 4$$

Step7: Set up equation to find m

$$8m - 12 = 4$$

Step8: Solve the linear equation for m

$$8m = 16 \implies m = 2$$

Answer:

(1) \(\alpha + \beta = -2m\) and \(\alpha\beta = m^2 - 2m + 3\)
(2) \((\alpha - \beta)^2 = 8m - 12\)
(3) \(m = 2\)