f4 maths mc X1

mx² - 2(m+2)x + m+5 = 0 has no real root.△ = 2²(m+2)² - 4m(m+5) < 0m² + 4m + 4 - m² - 5m < 0m > 4;(m -5)x² - 2(m+2)x + m = 0For m ≠ 5 , △ = 2²(m+2)² - 4(m -5)m= 4 (m² + 4m + 4 - m² + 5m)= 4 (9m + 4) > 0 for m > 4 , So it has 2 real roots for m ≠ 5 ,For m = 5 , it becomes - 14x + 5 = 0 , 1 real root.So the answer is undetermined , choose (D) .

A quadratic equation has no real roots if its determinant b^2 - 4ac < 0

that means

(-2(m+2))^2 - 4m(m+5) < 0
4m^2 + 8m + 8 - 4m^2 - 20m < 0
8 < 12m
m < 2/3

The determinant D of the 2nd equation is

(-2(m+2))^2 - 4(m-5)m
= 4m^2 + 8m + 8 -4m^2 + 20m
= 28m + 8

so D < 56/3 + 8 = 80/3

so the value of D depends on the value of m but will always less than 80/3,
so we cannot determine if the 2nd equation has 2, 1 or no real roots.