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NaNa asked in 科學及數學數學 · 6 years ago

求方程的判別式

1.考慮二次方程x^2+6x-(k+3) =0 求方程的判別式, 答案以K 表示

2.求下列各情況K 值的可能範圍中K 值的可能範圍,

i.方程有兩程相異實根,

ii 有實根

3.若二次方程3x^2-2x-k=0, 沒有實根, 求K 值的可能範圍

3 Answers

Rating
  • 6 years ago
    Favorite Answer

    圖片參考:https://s.yimg.com/lo/api/res/1.2/ZQit8r6YM3UMER1b...

    1.

    x² + 6x - (k + 3) = 0

    判別式 Δ

    = 6² - 4(1)[-(k + 3)]

    = 36 + 4(k + 3)

    = 36 + 4k + 12

    = 4k + 48

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    2. [相信是接續 第 1 題]

    i. 方程有兩程相異實根

    判別式 Δ > 0

    4k + 48 > 0

    4k > -48

    k > -12

    ii. 有實根

    判別式 Δ ≥ 0

    4k + 48 ≥ 0

    4k ≥ -48

    k ≥ -12

    ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀

    3.

    3x² - 2x - k = 0 沒有實根

    判別式 Δ < 0

    (-2)² - 4(3)(-k) < 0

    4 + 12k < 0

    12k < -4

    k < -4/12

    k < -1/3

    圖片參考:https://s.yimg.com/rk/HA00430218/o/1588069491.jpg

    2015-07-27 21:25:16 補充:

    (-8k)² - 4(2)(3k) = 0

    64k² - 8(3k) = 0

    全式 ÷ 8

    [64k² - 8(3k)] ÷ 8 = 0 ÷ 8

    64k² ÷ 8 - 8(3k) ÷ 8 = 0

    8k² - 3k = 0

    明白嗎?

    2015-07-27 21:53:19 補充:

    Cheers!

    ﹝。◕‿◕。◕‿◠。﹞

  • pui
    Lv 4
    6 years ago

    1.

    判別式=b^2 -4ac

    判別式=6^2 -4[-(k+3)]

    判別式=36+4k+12

    判別式=4k+48

    判別式=4(k+12)

    2i

    Δ>0

    2i

    Δ>0 or Δ=0

    3

    3x^2 -2x-k=0

    判別式<0

    (-2)^2 -4(4)(3)(-k)<0

    4+4(3)(k)<0

    1+3k<0

    3k<-1

    k<-1/3

  • 6 years ago

    你好, 書本有一條例題

    y=2x^2-8kx+3k只有一個軸截距

    Δ=0

    (-8k)^2 -4(2)(3k)=0

    8k^2-3k=0 為什麼是8k^2-3k=0 ??

    k(8k-3)=0

    k=0 or k =3/8

    P.S 看來你很喜歡貓...

    2015-07-27 21:51:46 補充:

    原來是這樣....明白了

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