急!急!急! Quadratic Equation F4

1.
x = k(x - 1)(x - 4)
x = k(x² - 5x + 4)
kx² - (5k + 1)x - 4k = 0

Since the equation has two identical roots, then determinant Δ = 0
[-(5k + 1)]² - 4(k)(-4k) = 0
25k² + 10k + 1 - 16k² = 0
9k² + 10k + 1= 0
(9k + 1)(k + 1) = 0
k = -1/9 or k = -1


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2.(a)
The graph : y = x² + (k + 1)x + 4...... (1)
x - axis : y = 0 ...... (2)

(1) = (2) :
x² + (k + 1)x + 4 =0

Since the graph touches the x-axis, thus determinant Δ = 0
(k + 1)² - 4(1)(4) = 0
k² + 2k + 1 - 16 = 0
k² + 2k - 15 = 0
(k + 5)(k - 3) = 0
k = -5 or k = 3


2.(b)
When k = -5 :
x² + (-5 + 1)x + 4 =0
x² - 4x + 4 =0
(x - 2)² = 0
x = 2 (double roots)
Hence, the x-intercept = 2

When k = 3 :
x² + (3 + 1)x + 4 =0
x² + 4x + 4 =0
(x + 2)² = 0
x = -2 (double roots)
Hence, the x-intercept = -2


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3.(a)
x² + px - 1 = 0

(i)
Determinant Δ
= p² - 4(1)(-1)
= p² + 4

(ii)
For all real numbers p, p² ≥ 0
Since Δ = p² + 4 ≥ 4, the roots are real.


3.(b)
x² + ax + a² = 0

(i)
Determinant Δ
= a² - 4(1)(a²)
= -3a²

(ii)
For all real numbers a, a² ≥ 0
If a = 0, Δ = 0 and the roots are realand equal.
If a ≠ 0, Δ < 0 and the roots are notreal.


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4.
(a)
px² + 24x + 9 = 0 has equal roots, then determinant Δ = 0
(24)² - 4(p)(9) = 0
36p = 576
p = 16


(b)
2x² + (p - 1)x + 28 = 0
2x² + (16 - 1)x + 28 = 0
2x² + 15x + 28 = 0
(2x + 7)(x + 4) = 0
x = -7/2 or x = -4

2011-07-23 17:26:02 補充：
To myisland8132:

In Q.1, it is (5k + 1)x, but NOT (k + 5)x.

In Q.3(b)(ii), when a = 1, determinant = 0, and the two roots are real and equal (0).