Question

1. given that (x^2 - 2x - m + 1 = 0) has 2 positive real number roots, (alpha) and (\beta), determine the range of values of (m).

consider this!
i) given that (x^2 - 6x + 3k = 0) has 2 real number roots, (alpha) and (\beta), with opposite signs (one root is positive and the other root is negative), determine the range of values of (k).
sol (alpha + \beta = 6), (alpha\beta = 3k)
from (alpha\beta = 3k < \box),
(k < \box) ... (1)
and, from (\frac{d}{4} = 9 - 3k > 0),
(k < 3) ... (2)
from (1) and (2),
(k < \box)

ii) in order for (ax^2 + bx + c = 0) to have 2 different real number roots, (alpha) and (\beta), with opposite signs,
\\\

$$\begin{cases} \\alpha\\beta = \\frac{c}{a} < 0 \\\\ d = b^2 - 4ac > 0 \\end{cases}$$

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• if (\frac{c}{a} < 0), then (ac < 0), so the discriminant is always positive.
• therefore we only need the condition (alpha\beta < 0).

Explanation:

Step1: Set discriminant condition

For real roots, the discriminant must be non-negative:

$$D = (-2)^2 - 4(1)(-m + 1) \ge 0$$

Step2: Solve discriminant inequality

Simplify the inequality to find the range for \(m\):

$$4 + 4m - 4 \ge 0 \implies m \ge 0$$

Step3: Set sum of roots condition

Since both roots are positive, their sum must be positive:

$$\alpha + \beta = 2 > 0$$

Step4: Set product of roots condition

Since both roots are positive, their product must be positive:

$$\alpha\beta = -m + 1 > 0 \implies m < 1$$

Step5: Combine all conditions

Find the intersection of \(m \ge 0\) and \(m < 1\):

$$0 \le m < 1$$

Answer:

\(-1 < m \le 0\)