唔識做功課75

2x² ‒ (k + 3)x + k = ‒1
2x² ‒ (k + 3)x + (k + 1)= 0

Discriminant of the quadratic equation, Δ
= (k + 3)² ‒ 4 × 2 × (k + 1)
= k² + 6k + 9 ‒ 8k ‒ 8
= k² ‒ 2k + 1
= (k ‒ 1)²

For all real value of k, Δ = (k ‒ 1)² ≥ 0
Hence, for all real value of k, the equation must have 2 real roots.

I agree what is claimed.

2015-07-11 16:53:16 補充：
2 real roots means 2 different real roots or double real roots.

2015-07-11 16:59:57 補充：
0² = 0, (negative real number)² > 0, (positive real number)² > 0
Hence, (any real number)² ≥ 0

For any real value of k, (k ‒ 1) is a real number.
Hence, (k ‒ 1)² ≥ 0

2015-07-11 17:01:05 補充：
Δ is the the discriminant of a quadratic equation.

Δ = 0 : double real roots (two identical real roots)
Δ > 0 : two different real roots
Δ < 0 : no real roots

As Δ = (k ‒ 1)² ≥ 0, there are two real roots.

2015-07-11 17:04:42 補充：
Δ = k² ‒ 4k + 8
= (k² ‒ 4k + 4) + 4
= (k ‒ 2)² + 4

For any real value of k, (k ‒ 2)² ≥ 0
Hence, Δ = (k ‒ 2)² + 4 ≥ 4

As Δ > 0, there are two different real roots.

我係求到(k-1)^2但我唔知点解會 ≥ 0同埋点解 must have 2 real roots.

2015-07-10 23:18:23 補充：
但若果我求到 Δ = k^2-4k+8咁喺咪no real roots.

2015-07-13 14:11:52 補充：
thanks for your help!