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超急急急急急急急急急急急(2230前) F.4一元二次方程2
發問:
(42) Given that r is a root of the quadratic equation x^2 - 3x - 5 = 0, find the value of r^5 - 3r^4 - 5r^3 + 2r^2 - 6r(48) The fig. shows a piece of rectangle paper ABCD. The paper is folded such that 8 identical triangles of dimensions 3 cm X 4cm are obtained, It is given that the width and the length of paper... 顯示更多 (42) Given that r is a root of the quadratic equation x^2 - 3x - 5 = 0, find the value of r^5 - 3r^4 - 5r^3 + 2r^2 - 6r (48) The fig. shows a piece of rectangle paper ABCD. The paper is folded such that 8 identical triangles of dimensions 3 cm X 4cm are obtained, It is given that the width and the length of paper are (3x + 1) cm and (5x - 2) cm respectively 圖片參考：http://blog.yimg.com/3/IQ5sLCh7s5_2vQghH4qrxgCzDCp.eASGlCYD2L6aGzreM6vF2WCYMw--/70/l/8WYM3dxy2GEDY4MxatS3VA.jpg (48a) Express the area of thw shaded region in terms of x (48b i) If the area of the shaded region is 186 cm^2, show that 15x^2 - x - 236 = 0 (48b ii) Hence find tge lengths of JK and GH (54) The fig. shows the fraph of y = -2x^2 + 16x - 24 which cuts the x-axis at 2 points M & N, and passes through 2 points P & Q which are both above the x-axis. It is given that area of triangle PMN = area of triangle QMN = 12 sq. units 圖片參考：http://blog.yimg.com/3/IQ5sLCh7s5_2vQghH4qrxgCzDCp.eASGlCYD2L6aGzreM6vF2WCYMw--/71/l/7R8YzVL4KNUXI11oT5TIKw.jpg (54a) Find coordinates of M & N (54b) Find coordinates of P & Q (54c) If R is a point on the graph other than P & Q, it is possible that area of triangle RMN is 12 sq. units? Explain Answers: (Show equations and steps) (42) 10 (48a) (15x^2 - x -50) cm^2 (48b ii) JK: 5 cm, GH: 12 cm (54a) M(2,0), N(6,0) (54b) P(3,6), Q(5,6) (53c) Yes
42) x^2 - 3x - 5 = 0 r^2-3r=5 r^5 - 3r^4 - 5r^3 + 2r^2 - 6r =r^3(r^2-3r-5) + 2(r^2-3r) =r^3 * 0 + 2*5 =10 48a)Area=(5x-2)(3x+1)-8*(3*4/2) =15x^2-6x+5x-2-48 =(15x^2-x-50)cm^2 48bi)Area=15x^2-x-50=186 15x^2-x-236=0 bii)15x^2-x-236=0 x=4 OR-3.93(rejected) JK=3x+1-4*2 =5cm GH=5x-2-3*2 =12cm 54a) -2x^2+16x-24=0 x^2-8x+12 (x-6)(x-2)=0 x=6 OR x=2 M(2,0), N(6,0) 54b) Area=MN*height of P/2=12 height=6 The y-coordinate of P and Q are both 6 sub.y=6 -2x^2+16x-24=6 x^2-8x+15=0 (x-3)(x-5)=0 x=3 OR x=5 P(3,6), Q(5,6) 54c) Yes, R can be a point below x-axis, with y=-6
其他解答:
超急急急急急急急急急急急(2230前) F.4一元二次方程2
發問:
(42) Given that r is a root of the quadratic equation x^2 - 3x - 5 = 0, find the value of r^5 - 3r^4 - 5r^3 + 2r^2 - 6r(48) The fig. shows a piece of rectangle paper ABCD. The paper is folded such that 8 identical triangles of dimensions 3 cm X 4cm are obtained, It is given that the width and the length of paper... 顯示更多 (42) Given that r is a root of the quadratic equation x^2 - 3x - 5 = 0, find the value of r^5 - 3r^4 - 5r^3 + 2r^2 - 6r (48) The fig. shows a piece of rectangle paper ABCD. The paper is folded such that 8 identical triangles of dimensions 3 cm X 4cm are obtained, It is given that the width and the length of paper are (3x + 1) cm and (5x - 2) cm respectively 圖片參考：http://blog.yimg.com/3/IQ5sLCh7s5_2vQghH4qrxgCzDCp.eASGlCYD2L6aGzreM6vF2WCYMw--/70/l/8WYM3dxy2GEDY4MxatS3VA.jpg (48a) Express the area of thw shaded region in terms of x (48b i) If the area of the shaded region is 186 cm^2, show that 15x^2 - x - 236 = 0 (48b ii) Hence find tge lengths of JK and GH (54) The fig. shows the fraph of y = -2x^2 + 16x - 24 which cuts the x-axis at 2 points M & N, and passes through 2 points P & Q which are both above the x-axis. It is given that area of triangle PMN = area of triangle QMN = 12 sq. units 圖片參考：http://blog.yimg.com/3/IQ5sLCh7s5_2vQghH4qrxgCzDCp.eASGlCYD2L6aGzreM6vF2WCYMw--/71/l/7R8YzVL4KNUXI11oT5TIKw.jpg (54a) Find coordinates of M & N (54b) Find coordinates of P & Q (54c) If R is a point on the graph other than P & Q, it is possible that area of triangle RMN is 12 sq. units? Explain Answers: (Show equations and steps) (42) 10 (48a) (15x^2 - x -50) cm^2 (48b ii) JK: 5 cm, GH: 12 cm (54a) M(2,0), N(6,0) (54b) P(3,6), Q(5,6) (53c) Yes
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最佳解答:42) x^2 - 3x - 5 = 0 r^2-3r=5 r^5 - 3r^4 - 5r^3 + 2r^2 - 6r =r^3(r^2-3r-5) + 2(r^2-3r) =r^3 * 0 + 2*5 =10 48a)Area=(5x-2)(3x+1)-8*(3*4/2) =15x^2-6x+5x-2-48 =(15x^2-x-50)cm^2 48bi)Area=15x^2-x-50=186 15x^2-x-236=0 bii)15x^2-x-236=0 x=4 OR-3.93(rejected) JK=3x+1-4*2 =5cm GH=5x-2-3*2 =12cm 54a) -2x^2+16x-24=0 x^2-8x+12 (x-6)(x-2)=0 x=6 OR x=2 M(2,0), N(6,0) 54b) Area=MN*height of P/2=12 height=6 The y-coordinate of P and Q are both 6 sub.y=6 -2x^2+16x-24=6 x^2-8x+15=0 (x-3)(x-5)=0 x=3 OR x=5 P(3,6), Q(5,6) 54c) Yes, R can be a point below x-axis, with y=-6
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