A. maths(F.4)

我諗你搞錯左少少

f(x)=0 唔係話 D=0
f(x)=0
f(x)=2x^2-kx+3=0
2x^2-kx+3=0

而佢話呢條equation有2個distinct real roots
即係 D>0

a=2 b=-k c=3
D=(-k)^2-(4)(2)(3)>0

Solve左佢就計到k個range

Let f(x)=2x²-kx+3, where k is a constant .

Find the range of values of k for which the equation f(x)=0 has two distinct real roots.
Δ＞0
(-k)²-4(2)(3)＞0
k²＞24
k＞2√6

f(x)=0應該只得一個 real root , 但我想知點解會有 two distinct real roots ,
由於一元二次方程於多數情況下都有兩舍solution,所以題目話the equation has two distinct root 係絕對合理

Discriminant is greater than 0 because the question said it has two distinct root

two distinct real roots 即係 Discriminant > 0

(-k)^2 - 4(2)(3) > 0