1. It is given that -2 is a root of u^5+2u^4-(k+1)u^3-10u^2+ku+8=0,where k is constant.
a) Find the value of k.
b) Using the result of (a), solve the equation.
2. Express each of the following in the form of a(x-h)^2+k.
i) x^2+2x+3
ii) 4x^2+8x-3
b) It is given that x^2 - (b^2+2b+3)x + (4b^2+8b-3)=0 has one double real root,where b is a real number. Find the values of b.
唔識做功課62
2015-04-19 10:31 pm
回答 (3)
2015-04-19 11:26 pm
✔ 最佳答案
1.(a)
-2 is a root of u⁵ + 2u⁴ - (k + 1)u³ - 10u² + ku + 8 = 0
Therefore,
(-2)⁵ + 2(-2)⁴ - (k + 1)(-2)³ - 10(-2)² + k(-2) + 8 = 0
-32 + 32 + 8(k + 1) - 40 - 2k + 8 = 0
8k + 8 - 2k - 32 = 0
6k = 24
k = 4
(b)
The equation is
u⁵ + 2u⁴ - 5u³ - 10u² + 4u + 8 = 0
u⁵ + 2u⁴ - 5u³ - 10u² + 4u + 8 = 0
u⁴(u + 2) - 5u²(u + 2) + 4(u + 2) = 0
(u + 2)(u⁴ - 5u² + 4) = 0
(u + 2)(u² - 4)(u² - 1) = 0
(u + 2)(u + 2)(u - 2)(u + 1)(u - 1) = 0
u = -2 (repeated), 2, -1, 1
2.
(a)
(i) x² + 2x + 3
= x² + 2x + 1 + 2
= (x + 1)² + 2
(ii) 4x² + 8x - 3
= 4x² + 8x + 4 - 7
= 4(x² + 2x + 1) - 7
= 4(x + 1)² - 7
(b)
x² - (b² + 2b + 3)x + (4b² + 8b - 3) = 0 has double real roots.
Δ = 0
[-(b² + 2b + 3)]² - 4(1)(4b² + 8b - 3) = 0
(b² + 2b + 3)² - 4(4b² + 8b - 3) = 0
Using the results of (a),
[(b + 1)² + 2]² - 4[4(b + 1)² - 7] = 0
Let a = (b + 1)²
(a + 2)² - 4(4a - 7) = 0
a² + 4a + 4 - 16a + 28 = 0
a² - 12a + 32 = 0
(a - 4)(a - 8) = 0
a = 4 or a = 8
(b + 1)² = 4 or (b + 1)² = 8
b + 1 = ±2 or b + 1 = ±√8
b = -1 ± 2 or b = -1 ± √8
b = -3 or 1 or -1 ± 2√2
2015-04-19 15:59:59 補充：
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2015-04-20 1:54 am
1.(a)
-2 is a root of u⁵ + 2u⁴ - (k + 1)u³ - 10u² + ku + 8 = 0
Therefore,
(-2)⁵ + 2(-2)⁴ - (k + 1)(-2)³ - 10(-2)² + k(-2) + 8 = 0
-32 + 32 + 8(k + 1) - 40 - 2k + 8 = 0
8k + 8 - 2k - 32 = 0
6k = 24
k = 4
(b)
The equation is
u⁵ + 2u⁴ - 5u³ - 10u² + 4u + 8 = 0
u⁵ + 2u⁴ - 5u³ - 10u² + 4u + 8 = 0
u⁴(u + 2) - 5u²(u + 2) + 4(u + 2) = 0
(u + 2)(u⁴ - 5u² + 4) = 0
(u + 2)(u² - 4)(u² - 1) = 0
(u + 2)(u + 2)(u - 2)(u + 1)(u - 1) = 0
u = -2 (repeated), 2, -1, 1
2.(a)
(i) x² + 2x + 3
= x² + 2x + 1 + 2
= (x + 1)² + 2
(ii) 4x² + 8x - 3
= 4x² + 8x + 4 - 7
= 4(x² + 2x + 1) - 7
= 4(x + 1)² - 7
(b)
x² - (b² + 2b + 3)x + (4b² + 8b - 3) = 0 has double real roots.
Δ = 0
[-(b² + 2b + 3)]² - 4(1)(4b² + 8b - 3) = 0
(b² + 2b + 3)² - 4(4b² + 8b - 3) = 0
Using the results of (a),
[(b + 1)² + 2]² - 4[4(b + 1)² - 7] = 0
Let a = (b + 1)²
(a + 2)² - 4(4a - 7) = 0
a² + 4a + 4 - 16a + 28 = 0
a² - 12a + 32 = 0
(a - 4)(a - 8) = 0
a = 4 or a = 8
(b + 1)² = 4 or (b + 1)² = 8
b + 1 = ±2 or b + 1 = ±√8
b = -1 ± 2 or b = -1 ± √8
b = -3 or 1 or -1 ± 2√2
-2 is a root of u⁵ + 2u⁴ - (k + 1)u³ - 10u² + ku + 8 = 0
Therefore,
(-2)⁵ + 2(-2)⁴ - (k + 1)(-2)³ - 10(-2)² + k(-2) + 8 = 0
-32 + 32 + 8(k + 1) - 40 - 2k + 8 = 0
8k + 8 - 2k - 32 = 0
6k = 24
k = 4
(b)
The equation is
u⁵ + 2u⁴ - 5u³ - 10u² + 4u + 8 = 0
u⁵ + 2u⁴ - 5u³ - 10u² + 4u + 8 = 0
u⁴(u + 2) - 5u²(u + 2) + 4(u + 2) = 0
(u + 2)(u⁴ - 5u² + 4) = 0
(u + 2)(u² - 4)(u² - 1) = 0
(u + 2)(u + 2)(u - 2)(u + 1)(u - 1) = 0
u = -2 (repeated), 2, -1, 1
2.(a)
(i) x² + 2x + 3
= x² + 2x + 1 + 2
= (x + 1)² + 2
(ii) 4x² + 8x - 3
= 4x² + 8x + 4 - 7
= 4(x² + 2x + 1) - 7
= 4(x + 1)² - 7
(b)
x² - (b² + 2b + 3)x + (4b² + 8b - 3) = 0 has double real roots.
Δ = 0
[-(b² + 2b + 3)]² - 4(1)(4b² + 8b - 3) = 0
(b² + 2b + 3)² - 4(4b² + 8b - 3) = 0
Using the results of (a),
[(b + 1)² + 2]² - 4[4(b + 1)² - 7] = 0
Let a = (b + 1)²
(a + 2)² - 4(4a - 7) = 0
a² + 4a + 4 - 16a + 28 = 0
a² - 12a + 32 = 0
(a - 4)(a - 8) = 0
a = 4 or a = 8
(b + 1)² = 4 or (b + 1)² = 8
b + 1 = ±2 or b + 1 = ±√8
b = -1 ± 2 or b = -1 ± √8
b = -3 or 1 or -1 ± 2√2
2015-04-19 11:52 pm
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2015-04-21 19:29:02 補充：
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2015-04-21 19:29:02 補充：
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