IB Mathematics SL 2.7 Use of the discriminant AA SL Paper 1- Exam Style Questions- New Syllabus

(a)
For two distinct real roots, the discriminant must be positive: \( \Delta > 0 \).
\( \Delta = k^2 – 4(1)(15 – k) = k^2 + 4k – 60 \)
Solve \( k^2 + 4k – 60 > 0 \):
\( k^2 + 4k – 60 = 0 \) has solutions \( k = 6 \) and \( k = -10 \).
Since the quadratic opens upwards, \( \Delta > 0 \) when \( k < -10 \) or \( k > 6 \).
\( \boxed{k < -10 \quad \text{or} \quad k > 6} \)

(b)
Let the roots be \( \alpha \) and \( \beta \). For both roots positive or both negative, the product \( \alpha\beta > 0 \).
By Vieta’s formulas: \( \alpha\beta = 15 – k \).
So \( 15 – k > 0 \Rightarrow k < 15 \).
Combine with part (a) result \( k < -10 \) or \( k > 6 \):
\( k < -10 \) or \( 6 < k < 15 \).
\( \boxed{k < -10 \quad \text{or} \quad 6 < k < 15} \)