Solve each of the following equations using any method.(Leave the answers in surd form if necessary, and identify those equations that have no real roots.)
1. √2 x^2 -5x+2√2=0
2. 6(x+3)^2 -25 = 2(x+3)^2
3. 3√3 x^2 +(√3)/4=3√2 x
4. 2(5-x)=(5-x)(3x +4)
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5. It is given that ax^2 +(a+b)x +b=0 is a quadratic equatiion, where a and b are real numbers.
(a) Find the discriminant of the equation in terms of a and b.
(b) Show that the equation must have real roots.
(c)Does the equation 998x^2 +1000x +2 = 0 have real roots?
Quadratic Equations (F.4)
2009-04-30 5:33 am
回答 (2)
2009-04-30 8:28 am
✔ 最佳答案
1.√2x2 - 5x + 2√2 = 0
(√2x - 1)(x - 2√2) = 0
x = 1/√2 or x = 2√2
2.
6(x + 3)2 - 25 = 2(x+3)2
4(x + 3)2 - 25 = 0
(2x + 6)2 - 52 = 0
[(2x + 6) + 5] [(2x + 6) - 5] = 0
(2x + 11)(2x + 1) = 0
x = -11/2 or x = -1/2
3.
3√3x2 + √3/4 = 3√2x
3√3x2 - 3√2x + √3/4 = 0
12√3x2 - 12√2x + √3 = 0
x = {-(-12√2) √(-12√2)2 - 4(12√3)(√3)}/(2 x 12√3)
x = (12√2 12)/(24√3)
x = (√2 1)/(2√3)
x = (√6 + √3)/6 or x = (√6 - √3)/6
4.
2(5 - x) = (5 - x)(3x + 4)
2(5 - x) - (5 - x)(3x + 4) = 0
(5 - x)[2 - (3x + 4)] = 0
(5 - x)(-2 - 3x) = 0
(3x + 2)(x - 5) = 0
x = -2/3 or x = 5
5.
(a)
ax2 + (a + b)x + b = 0
D = (a + b)2 - 4ab
D = (a2 + 2ab + b2) - 4ab
D = a2 - 2ab + b2
D = (a - b)2
(b)
D = (a - b)2 ≥ 0
When D = 0, the equation has double real roots.
When D > 0, the equation has two real roots.
Hence, the equation must have real roots.
(c)
When a = 998 and b = 2, the equation becomes:
998x2 + 1000x + 2 = 0
D = (998 - 2)2 > 0
Hence, the equation has two real roots.
=
2009-04-30 11:44 am
sorry...
樓上第2條解答我有少少問題
4(x+3)^2 - 25 = 0
你expend完變左
(2x+6)^2 - 5^2 = 0
係咪乘錯? 定我short左? 勿插
樓上第2條解答我有少少問題
4(x+3)^2 - 25 = 0
你expend完變左
(2x+6)^2 - 5^2 = 0
係咪乘錯? 定我short左? 勿插
收錄日期: 2021-04-28 15:17:40
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