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F.5 M2 about Determinants~~
發問:
*只需教我做(b)題便可以了Consider the system of linear equations (E) ax + y + z = 4x + ay + z = 37x - y + 2z = 11,where a is a real number.(a) If a = 2, solve (E). (ans. x=t , y=t-1 , z=5-3t)(b) Suppose all solutions (x,y,z) of 2x + y + z = 4x + 2y + z = 37x - y + 2z = 11satisfy (p-1)(x^2) ≥ (p+1)yz.Find... 顯示更多 *只需教我做(b)題便可以了 Consider the system of linear equations (E) ax + y + z = 4 x + ay + z = 3 7x - y + 2z = 11, where a is a real number. (a) If a = 2, solve (E). (ans. x=t , y=t-1 , z=5-3t) (b) Suppose all solutions (x,y,z) of 2x + y + z = 4 x + 2y + z = 3 7x - y + 2z = 11 satisfy (p-1)(x^2) ≥ (p+1)yz. Find the range of values of p. 更新: To 螞蟻雄兵 : 但答案係 -1/2 < p <= 3/2 不等號之類既野有冇搞錯? 同埋我唔係好明 b^2 - 4ac 果舊野代表咩...點解係<=0 ?
最佳解答:
(b)Suppose all solutions (x，y，z) of 2x + y + z = 4 x + 2y + z = 3 7x - y + 2z = 11 satisfy (p-1)(x^2) ≥ (p+1)yz. Find the range of values of p. Sol ｜２１１｜ ｜１２１｜＝（１／２）｜３１｜＝０ ｜７－１２｜｜－９－３｜ ｜４１１｜ ｜３２１｜＝（１／４）｜５１｜ ｜１１－１２｜｜－１５－３｜＝０ So 為相依有無限組解 (2x+y+z)－(x+2y+z)=4－3 x－y=1 y=x－1 2x+(x－1)+z=4 z=5－3x (p－1)x^2>=(p+1)(x－1)(5－3x) (p－1)x^2>=(p+1)(－3x^2+8x－5) (p－1)x^2+(p+1)(3x^2－8x+5)>=0 (p－1+3p+3)x^2－8(p+1)x+5(p+1)>=0 (4p+2)x^2－8(p+1)x+5(p+1)>=0 4p+2>=0 p>=－1/2------------------------(1) 64(p+1)^2－4*(4p+2)*5(p+1)<=0 64(p^2+2p+1)－20(4p^2+6p+2)<=0 －16p^2+8p+24<=0 2p^2－p－3>=0 (2p－3)(p+1)>=0 (p-3/2)(p+1)>=0 p<=－1 orp>=3/2-------------(2) 綜合(1)(2) p>=3/2 2011-07-10 05:30:24 補充： 修改如下 綜合(1)(2) －１－１／２３／２ －－－－－＊－－－－－－－＊－－－－－－－＊－－－－－－－ ＠－－－－－－－－－－－－＞ ＠－－－－－－－－－－－－－－－＠ ＠－－－－－－－＠result －1/2<=p<=3/2 2011-07-10 05:33:59 補充： ax^2+bx+c，a>0，恆>0 一定有2個條件 (1) 圖形開口向上 a>0 (2) 圖形與x軸最多只有一個交點 b^2-4ac<=0
F.5 M2 about Determinants~~
發問:
*只需教我做(b)題便可以了Consider the system of linear equations (E) ax + y + z = 4x + ay + z = 37x - y + 2z = 11,where a is a real number.(a) If a = 2, solve (E). (ans. x=t , y=t-1 , z=5-3t)(b) Suppose all solutions (x,y,z) of 2x + y + z = 4x + 2y + z = 37x - y + 2z = 11satisfy (p-1)(x^2) ≥ (p+1)yz.Find... 顯示更多 *只需教我做(b)題便可以了 Consider the system of linear equations (E) ax + y + z = 4 x + ay + z = 3 7x - y + 2z = 11, where a is a real number. (a) If a = 2, solve (E). (ans. x=t , y=t-1 , z=5-3t) (b) Suppose all solutions (x,y,z) of 2x + y + z = 4 x + 2y + z = 3 7x - y + 2z = 11 satisfy (p-1)(x^2) ≥ (p+1)yz. Find the range of values of p. 更新: To 螞蟻雄兵 : 但答案係 -1/2 < p <= 3/2 不等號之類既野有冇搞錯? 同埋我唔係好明 b^2 - 4ac 果舊野代表咩...點解係<=0 ?
最佳解答:
(b)Suppose all solutions (x，y，z) of 2x + y + z = 4 x + 2y + z = 3 7x - y + 2z = 11 satisfy (p-1)(x^2) ≥ (p+1)yz. Find the range of values of p. Sol ｜２１１｜ ｜１２１｜＝（１／２）｜３１｜＝０ ｜７－１２｜｜－９－３｜ ｜４１１｜ ｜３２１｜＝（１／４）｜５１｜ ｜１１－１２｜｜－１５－３｜＝０ So 為相依有無限組解 (2x+y+z)－(x+2y+z)=4－3 x－y=1 y=x－1 2x+(x－1)+z=4 z=5－3x (p－1)x^2>=(p+1)(x－1)(5－3x) (p－1)x^2>=(p+1)(－3x^2+8x－5) (p－1)x^2+(p+1)(3x^2－8x+5)>=0 (p－1+3p+3)x^2－8(p+1)x+5(p+1)>=0 (4p+2)x^2－8(p+1)x+5(p+1)>=0 4p+2>=0 p>=－1/2------------------------(1) 64(p+1)^2－4*(4p+2)*5(p+1)<=0 64(p^2+2p+1)－20(4p^2+6p+2)<=0 －16p^2+8p+24<=0 2p^2－p－3>=0 (2p－3)(p+1)>=0 (p-3/2)(p+1)>=0 p<=－1 orp>=3/2-------------(2) 綜合(1)(2) p>=3/2 2011-07-10 05:30:24 補充： 修改如下 綜合(1)(2) －１－１／２３／２ －－－－－＊－－－－－－－＊－－－－－－－＊－－－－－－－ ＠－－－－－－－－－－－－＞ ＠－－－－－－－－－－－－－－－＠ ＠－－－－－－－＠result －1/2<=p<=3/2 2011-07-10 05:33:59 補充： ax^2+bx+c，a>0，恆>0 一定有2個條件 (1) 圖形開口向上 a>0 (2) 圖形與x軸最多只有一個交點 b^2-4ac<=0
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