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配方法應用題 (8條)
1．9(y-1)^2-144=0 2．7(x+4)^2+28=0 3．8(x+1)^2+25=0 4．4(x-5)^2-49=0 5．4y^2-15y+9=0 6．18y^2-25y-3=0 7．56(5x+7)+48x^2=0 8．92x(x+3)=161 就e 8條 唔該哂 更新: 第2和第3題好像錯了呢 答案好像錯了 更新 2: 我知道第2和第3題的答案 但我不知道全算式 第2條的答案是 沒有實根，(-4,2i) 第3條的答案是 沒有實根，(-1,5√2/4i)
最佳解答:
1．9(y-1)^2-144=0 [3(y-1)]^2 - 12^2 = 0 (3y-3-12)(3y-3+12) = 0 (3y - 15)(3y + 9) = 0 (y - 5)(y + 3) = 0 y = 5 or -3 2．7(x+4)^2+28=0 (x+4)^2 + 2^2 = 0 (x+4+2)(x+4-2) = 0 (x+6)(x+2) = 0 x = -6 or -2 3．4(x+1)^2+25=0 [2(x+1)]^2 + 5^2 = 0 (2x+2+5)(2x+2-5) = 0 (2x + 7)(2x - 3) = 0 x = -7/2 or 3/2 4．4(x-5)^2-49=0 [2(x-5)]^2 - 7^2 = 0 (2x-10-7)(2x-10+7) = 0 (2x-17)(2x-3) = 0 x = 17/2 or 3/2 5．4y^2-15y+9=0 (y - 3)(4y - 3) = 0 y = 3 or 3/4 6．18y^2-25y-3=0 (2y-3)(9y+1) = 0 y = 3/2 or -1/9 7．56(5x+7)+48x^2=0 7(5x+7) + 6x^2 = 0 6x^2 + 35x + 49 = 0 (2x+7)(3x+7) = 0 x = -7/2 or -7/3 8．92x(x+3)=161 4x(x+3) = 7 4x^2 + 12x - 7 = 0 (2x-1)(2x+7) = 0 x = 1/2 or -7/2 2010-09-11 20:47:25 補充： 2．7(x+4)^2-28=0 (x+4)^2 + 2^2 = 0 (x+4+2)(x+4-2) = 0 (x+6)(x+2) = 0 x = -6 or -2 3．4(x+1)^2-25=0 [2(x+1)]^2 + 5^2 = 0 (2x+2+5)(2x+2-5) = 0 (2x + 7)(2x - 3) = 0 x = -7/2 or 3/2 2010-09-11 20:49:54 補充： if it's not corrected, For Q2. 7(x+4)^2+28=0 x^2 + 8x + 16 + 4 = 0 x^2 + 8x + 20 = 0 △ = 64 - 80 = -16 < 0 For Q3, 8(x+1)^2+25=0 8x^2 + 8x + 89 = 0 △ = 61 - 4(89) <0 2010-09-11 20:57:55 補充： 第2和第3題好像錯了呢 答案好像錯了<<< 黃色部分已改, 數式改動了才能得出x, 答案於第1次補充 否則, 請參照第2次補充, Q2 和 Q3 皆不可求出x的實根 2010-09-11 21:06:53 補充： 既然你學了虛數i,那x就可解了,只是不是實根罷了 For Q2. 7(x+4)^2+28=0 x^2 + 8x + 16 + 4 = 0 x^2 + 8x + 20 = 0 x = [-8+/-√(64-80)]/2 x = -4 +/- 2i 2010-09-11 21:09:54 補充： For Q3, 8(x+1)^2+25=0 (x+1)^2 = -25/8 x+1 = +/-√(-25/8) x = -1 +/- 5/2 √(i/2)
其他解答:
There should be no real solution in Q2, Q3
配方法應用題 (8條)
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發問:1．9(y-1)^2-144=0 2．7(x+4)^2+28=0 3．8(x+1)^2+25=0 4．4(x-5)^2-49=0 5．4y^2-15y+9=0 6．18y^2-25y-3=0 7．56(5x+7)+48x^2=0 8．92x(x+3)=161 就e 8條 唔該哂 更新: 第2和第3題好像錯了呢 答案好像錯了 更新 2: 我知道第2和第3題的答案 但我不知道全算式 第2條的答案是 沒有實根，(-4,2i) 第3條的答案是 沒有實根，(-1,5√2/4i)
最佳解答:
1．9(y-1)^2-144=0 [3(y-1)]^2 - 12^2 = 0 (3y-3-12)(3y-3+12) = 0 (3y - 15)(3y + 9) = 0 (y - 5)(y + 3) = 0 y = 5 or -3 2．7(x+4)^2+28=0 (x+4)^2 + 2^2 = 0 (x+4+2)(x+4-2) = 0 (x+6)(x+2) = 0 x = -6 or -2 3．4(x+1)^2+25=0 [2(x+1)]^2 + 5^2 = 0 (2x+2+5)(2x+2-5) = 0 (2x + 7)(2x - 3) = 0 x = -7/2 or 3/2 4．4(x-5)^2-49=0 [2(x-5)]^2 - 7^2 = 0 (2x-10-7)(2x-10+7) = 0 (2x-17)(2x-3) = 0 x = 17/2 or 3/2 5．4y^2-15y+9=0 (y - 3)(4y - 3) = 0 y = 3 or 3/4 6．18y^2-25y-3=0 (2y-3)(9y+1) = 0 y = 3/2 or -1/9 7．56(5x+7)+48x^2=0 7(5x+7) + 6x^2 = 0 6x^2 + 35x + 49 = 0 (2x+7)(3x+7) = 0 x = -7/2 or -7/3 8．92x(x+3)=161 4x(x+3) = 7 4x^2 + 12x - 7 = 0 (2x-1)(2x+7) = 0 x = 1/2 or -7/2 2010-09-11 20:47:25 補充： 2．7(x+4)^2-28=0 (x+4)^2 + 2^2 = 0 (x+4+2)(x+4-2) = 0 (x+6)(x+2) = 0 x = -6 or -2 3．4(x+1)^2-25=0 [2(x+1)]^2 + 5^2 = 0 (2x+2+5)(2x+2-5) = 0 (2x + 7)(2x - 3) = 0 x = -7/2 or 3/2 2010-09-11 20:49:54 補充： if it's not corrected, For Q2. 7(x+4)^2+28=0 x^2 + 8x + 16 + 4 = 0 x^2 + 8x + 20 = 0 △ = 64 - 80 = -16 < 0 For Q3, 8(x+1)^2+25=0 8x^2 + 8x + 89 = 0 △ = 61 - 4(89) <0 2010-09-11 20:57:55 補充： 第2和第3題好像錯了呢 答案好像錯了<<< 黃色部分已改, 數式改動了才能得出x, 答案於第1次補充 否則, 請參照第2次補充, Q2 和 Q3 皆不可求出x的實根 2010-09-11 21:06:53 補充： 既然你學了虛數i,那x就可解了,只是不是實根罷了 For Q2. 7(x+4)^2+28=0 x^2 + 8x + 16 + 4 = 0 x^2 + 8x + 20 = 0 x = [-8+/-√(64-80)]/2 x = -4 +/- 2i 2010-09-11 21:09:54 補充： For Q3, 8(x+1)^2+25=0 (x+1)^2 = -25/8 x+1 = +/-√(-25/8) x = -1 +/- 5/2 √(i/2)
其他解答:
There should be no real solution in Q2, Q3
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