From given question, we observe that α occurs at x=‌
−b
2a
i.e. x=‌
−b
2
and β occurs at x=‌
−b
2a
, i.e. x=‌
+a
2
So, ‌‌α=(‌
−b
2
)2+b⋅(‌
−b
2
)+5 ⇒‌‌α=5−‌
b2
4
‌‌... (i) and ‌‌β=−(‌
+a
2
)2+a(‌
+a
2
)+5 =−‌
a2
4
+‌
a2
2
+5 ⇒‌‌β=5−‌
a2
4
. . . (ii) Given, x2−10x+24≤0 ⇒‌‌(x−4)(x−6)≤0 ⇒‌‌4≤x≤6 ∴‌‌α=4 and β=6 From Eqs. (i) and (ii), we get ∴b2=4‌ and ‌a2=4 ∴‌‌a2b2=16