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1.sole the following simultaneous equations using the algebraic method x^2=y^2=25{ 3x-4y+25=02.the line y=mx intersects the curve y=x^2-2x+4 at only one point P.(a)find the values of m.(b)for each of the values of m obtained in (a),find the coordinates of P3.ABCD and DEFG are two squares.If... 顯示更多 1.sole the following simultaneous equations using the algebraic method x^2=y^2=25 { 3x-4y+25=0 2.the line y=mx intersects the curve y=x^2-2x+4 at only one point P. (a)find the values of m. (b)for each of the values of m obtained in (a),find the coordinates of P 3.ABCD and DEFG are two squares.If EC=4cm and the sum of their areas is 400cm^2,find the lengths of the sides of the two squares
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我估你打錯題目： x2+y2=25 ....(1) 3x﹣4y+25=0 ...(2) From(2)， x = (4y﹣25)/3 ...(3) sub (3) to (1), [(4y﹣25)/3]2+y2=25 [4y﹣25)2/32]+y2=25 [(16y2﹣200y+625)/9]+y2=25 (16y2﹣200y+625)+9y2=225 25y2﹣200y+400 = 0 y2﹣8y+16 = 0 (y﹣4)2 = 0 y = 4 ....(4) sub(4) to (3), x = (16﹣25)/3 x = -3 ∴the solution of equation is (-3,4) 2(a)y = mx ....(1) y=x2﹣2x+4 ....(2) sub (2) to (1), x2﹣2x+4 = mx x2﹣(2+m)x+4 = 0 ∵the line y=mx intersects the curve y=x2﹣2x+4 at only one point P ∴△ = 0 [﹣(2+m)]2﹣4(1)(4) = 0 4+4m+m2﹣16 = 0 m2+4m﹣12 = 0 (m﹣2)(m+6) = 0 m = 2 or m = -6 (b)x2﹣(2+m)x+4 = 0 sub m = 2 to "x2﹣(2+m)x+4 = 0" x2﹣(2+2)x+4 = 0 x2﹣4x+4 = 0 (x﹣2) = 0 x = 2 y = 2(2) = 4 sub m = -6 to "x2﹣(2+m)x+4 = 0" x2﹣(2﹣6)x+4 = 0 x2+4x+4 = 0 (x+2) = 0 x = ﹣2 y = (-2)(-6) = 12 ∴the coordinates of P is (2,4) and (-2,12) 3.Let the lengths of ABCD is x,the lengths of DEFG is y x2+ y2 = 400.....(1) x+ 4 = y .....(2) sub (2) to (1), x2+ (x+ 4)2 = 400 x2+x2+8x+16 = 400 2x2+8x﹣384 = 0 x2+4x﹣192 = 0 (x﹣12)(x+16) = 0 x = 12 or x = -16(rejected) ....(3) sub(3) to (2), y = 12+4 = 16 ∴ the lengths of ABCD is 12cm,the lengths of DEFG is 16cm 2007-11-12 16:23:56 補充： 2.when m = 2,the coordinates of P is (2,4)when m = -6,the coordinates of P is (-2,12)
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f.4 maths
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發問:1.sole the following simultaneous equations using the algebraic method x^2=y^2=25{ 3x-4y+25=02.the line y=mx intersects the curve y=x^2-2x+4 at only one point P.(a)find the values of m.(b)for each of the values of m obtained in (a),find the coordinates of P3.ABCD and DEFG are two squares.If... 顯示更多 1.sole the following simultaneous equations using the algebraic method x^2=y^2=25 { 3x-4y+25=0 2.the line y=mx intersects the curve y=x^2-2x+4 at only one point P. (a)find the values of m. (b)for each of the values of m obtained in (a),find the coordinates of P 3.ABCD and DEFG are two squares.If EC=4cm and the sum of their areas is 400cm^2,find the lengths of the sides of the two squares
最佳解答:
我估你打錯題目： x2+y2=25 ....(1) 3x﹣4y+25=0 ...(2) From(2)， x = (4y﹣25)/3 ...(3) sub (3) to (1), [(4y﹣25)/3]2+y2=25 [4y﹣25)2/32]+y2=25 [(16y2﹣200y+625)/9]+y2=25 (16y2﹣200y+625)+9y2=225 25y2﹣200y+400 = 0 y2﹣8y+16 = 0 (y﹣4)2 = 0 y = 4 ....(4) sub(4) to (3), x = (16﹣25)/3 x = -3 ∴the solution of equation is (-3,4) 2(a)y = mx ....(1) y=x2﹣2x+4 ....(2) sub (2) to (1), x2﹣2x+4 = mx x2﹣(2+m)x+4 = 0 ∵the line y=mx intersects the curve y=x2﹣2x+4 at only one point P ∴△ = 0 [﹣(2+m)]2﹣4(1)(4) = 0 4+4m+m2﹣16 = 0 m2+4m﹣12 = 0 (m﹣2)(m+6) = 0 m = 2 or m = -6 (b)x2﹣(2+m)x+4 = 0 sub m = 2 to "x2﹣(2+m)x+4 = 0" x2﹣(2+2)x+4 = 0 x2﹣4x+4 = 0 (x﹣2) = 0 x = 2 y = 2(2) = 4 sub m = -6 to "x2﹣(2+m)x+4 = 0" x2﹣(2﹣6)x+4 = 0 x2+4x+4 = 0 (x+2) = 0 x = ﹣2 y = (-2)(-6) = 12 ∴the coordinates of P is (2,4) and (-2,12) 3.Let the lengths of ABCD is x,the lengths of DEFG is y x2+ y2 = 400.....(1) x+ 4 = y .....(2) sub (2) to (1), x2+ (x+ 4)2 = 400 x2+x2+8x+16 = 400 2x2+8x﹣384 = 0 x2+4x﹣192 = 0 (x﹣12)(x+16) = 0 x = 12 or x = -16(rejected) ....(3) sub(3) to (2), y = 12+4 = 16 ∴ the lengths of ABCD is 12cm,the lengths of DEFG is 16cm 2007-11-12 16:23:56 補充： 2.when m = 2,the coordinates of P is (2,4)when m = -6,the coordinates of P is (-2,12)
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