The Nature of Roots (F.4)

小雪

2009-05-01 1:24 am

In each of the following equations [1-2],
(a) put y = 0 and find the discriminant,
(b) hence determine the number of x-intercepts in the graph of the given equation.
1. y = 3x^2 + 4x - 2
2. y = 2x^2 + 4x + 5

3. If the equation 14x^2 - 7x + k = 0 has two distinct real roots, find the range of values of k.
4. If the equation x^2 + 4x - k = 0 has real roots, find the possible values of k.

In the following quadratic equations, find the value of the discriminant and state whether the equation has two distinct real roots, two equal real roots or no real roots. [5]
5. 2x^2 = 0.25x(1-x)

6.If 4x^2 -(1+m)x + 1 = 0 has two equal real roots, find the calues of m.
7. If 4(x^2 -3x) + k = 3 has no real roots, find the range of values of k
8. It is given that the quadratic equation kx^2 + (k+3)x - 1 = 0 has two equal real roots.
(a) Find the two possible values of k.
(b) Hence find the roots of the equation corresppnding to each value of k.
9. It is given that the quadratic equation (m-2)x^2 + mx + 2 = 0 has two equal real roots.
(a) Find the value(s) of m.
(b) Hence find the roots of the equation.
10. If the graph of y = x^2 + 2px - p + 2 touches the x-axis at only one point, find the two possible values of p.

回答 (1)

老爺子

2009-05-06 4:49 am

✔ 最佳答案

1.
(a)
Put y = 0: 3x2 + 4x - 2 = 0

Determinant, Δ
= (4)2 - 4(3)(-2)
= 40

(b)
Since Δ = 40 > 0
thus 3x2 + 4x - 2 = 0 has two real roots.
Therefore, no. of x-intercepts in the graph of the given equation = 2

2.
(a)
Put y = 0: 2x2 + 4x + 5 = 0

Determinant, Δ
= (4)2 - 4(2)(5)
= -24

(b)
Since Δ = -24 < 0
thus 2x2 + 4x + 5 = 0 has no real root.
Therefore, no. of x-intercepts in the graph of the given equation = 0

3.
Determinant Δ > 0
(-7)2 - 4(14)(k) > 0
49 - 56k > 0
56k < 49
k < 7/8

4.
Determinant Δ ≥ 0
(4)2 - 4(1)(-k) ≥ 0
16 + 4k ≥ 0
4k ≥ -16
k ≥ -4

5.
2x2 = 0.25x(1 - x)
8x2 = x(1 - x)
8x2 = x - 8x2
9x2 - x = 0

Determinant, Δ
= (-1)2 - 4(9)(0)
= 1

Since Δ = 1 > 0, the equation has two distinct real roots.

6.
Determinant Δ = 0
[-(1+m)]2 - 4(4)(1) = 0
m2 + 2m - 15 = 0
(m + 5)(m - 3) = 0
m = -5 or m = 3

7.
4(x2 - 3x) + k = 3
4x2 - 12x + (k - 3) = 0

Determinant Δ < 0
(`-12)2 - 4(4)(k - 3) < 0
144 - 16k + 48 < 0
16k > 192
k > 12

8.
(a)
Determinant Δ = 0
(k + 3)2 - 4(k)(-1) = 0
k2 + 10k + 9 = 0
(k + 1)(k + 9) = 0
k = -1 or k = -9

(b)
When k = -1, the equation is:
(-1)x2 + [(-1) + 3]x - 1 = 0
x2 - 2x + 1 = 0
(x - 1)2 = 0
x = 1 (double roots)

When k = -9, the equation is:
(-9)x2 + [(-9) + 3]x - 1 = 0
9x2 + 6x + 1 = 0
(3x + 1)2 = 0
x = -1/3 (double roots)

9.
(a)
Determinant Δ = 0
m2 - 4(m - 2)(2) = 0
m2 - 8m + 16 = 0
(m - 4)2 = 0
m = 4

(b)
(4 - 2)x2 + 4x + 2 = 0
2x2 + 4x + 2 = 0
2(x + 1)2 = 0
x = -1 (double roots)

10.
y = x2 + 2px - p + 2

Put y = 0: x2 + 2px + (-p + 2) = 0
Determinant Δ = 0
(2p)2 - 4(1)(-p + 2) = 0
4p2 + 4p - 8 = 0
4(p + 2)(p - 1) = 0
p = -2 or p = 1
=

👍 0 👎 0