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(40) maths questions. quick
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2. Solve the equation (x+2)^2 = (2x+1)^29. Solve the equation (4x+3)^2 - (5x+1)^2 = 010. Solve the equation (5x+2)^2 + (3x-1)^2 = 40, give your answer correct to 3 significant figures.12b. Hence solve the equation 4/(x^2) + 7/x - 2 = 0.15.Set up a quadraic equation in x whose roots are reciprocals of the... 顯示更多 2. Solve the equation (x+2)^2 = (2x+1)^2 9. Solve the equation (4x+3)^2 - (5x+1)^2 = 0 10. Solve the equation (5x+2)^2 + (3x-1)^2 = 40, give your answer correct to 3 significant figures. 12b. Hence solve the equation 4/(x^2) + 7/x - 2 = 0. 15.Set up a quadraic equation in x whose roots are reciprocals of the roots of x^2 -5x +4 = 0. 17.Set up a quadraic equation in x whose roots are one-third of the roots of 5x^2 - x - 6 = 0.
圖片參考：http://hk.yimg.com/i/icon/16/3.gif 2. Solve the equation (x+2)^2 = (2x+1)^2 (x+2)^2 = (2x+1)^2 x^2+4x+4 = 4x^2+4x+1 3x^2 = 3 x = -1 or x = 1 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 9. Solve the equation (4x+3)^2 - (5x+1)^2 = 0 (4x+3)^2 - (5x+1)^2 = 0 [(4x+3)+(5x+1)][(4x+3)-(5x+1)] = 0 (9x+4)(-x+2) = 0 x = -4/9 or x = 2 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 10. Solve the equation (5x+2)^2 + (3x-1)^2 = 40, give your answer correct to 3 significant figures. (5x+2)^2 + (3x-1)^2 = 40 25x^2+20x+4+9x^2-6x+1-40 = 0 34x^2+14x-35 = 0 x = [-14 + - sqrt(14^2+4(34)(35))]/(2)(34) = 0.829 or -1.24 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 12b. Hence solve the equation 4/(x^2) + 7/x - 2 = 0. 圖片參考：http://hk.yimg.com/i/icon/16/17.gif 4/(x^2) + 7/x - 2 = 0 4 + 7x - 2x^2 = 0 2x^2-7x-4 = 0 (x-4)(2x+1) = 0 x = 4 or x = -1/2 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 15.Set up a quadraic equation in x whose roots are reciprocals of the roots of x^2 -5x +4 = 0. Let α and β be the two roots of x^2-5x+4 = 0 α+β = 5 αβ = 4 To find (1/α) + (1/β) and (1/α)(1/β) (1/α) + (1/β) = (α+β)/(αβ) = 5/4 (1/α)(1/β) = 1/4 So, the required equation be x^2 - 5/4 x + 1/4 = 0 i.e. 4x^2 - 5x + 1 = 0 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 17.Set up a quadraic equation in x whose roots are one-third of the roots of 5x^2 - x - 6 = 0. Let α and β be the two roots of 5x^2-x-6 = 0 α+β = 1/5 αβ = -6/5 To find (α/3)+(β/3) and (α/3)(β/3) (α/3)+(β/3) = (α+β)/3 = 1/15 (α/3)(β/3) = (αβ)/9 = -6/45 = -3/15 So, the required equation be x^2 - 1/15 x - 3/15 = 0 i.e. 15x^2 - x - 3 = 0 圖片參考：http://hk.yimg.com/i/icon/16/3.gif
其他解答:
(40) maths questions. quick
發問:
2. Solve the equation (x+2)^2 = (2x+1)^29. Solve the equation (4x+3)^2 - (5x+1)^2 = 010. Solve the equation (5x+2)^2 + (3x-1)^2 = 40, give your answer correct to 3 significant figures.12b. Hence solve the equation 4/(x^2) + 7/x - 2 = 0.15.Set up a quadraic equation in x whose roots are reciprocals of the... 顯示更多 2. Solve the equation (x+2)^2 = (2x+1)^2 9. Solve the equation (4x+3)^2 - (5x+1)^2 = 0 10. Solve the equation (5x+2)^2 + (3x-1)^2 = 40, give your answer correct to 3 significant figures. 12b. Hence solve the equation 4/(x^2) + 7/x - 2 = 0. 15.Set up a quadraic equation in x whose roots are reciprocals of the roots of x^2 -5x +4 = 0. 17.Set up a quadraic equation in x whose roots are one-third of the roots of 5x^2 - x - 6 = 0.
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最佳解答:圖片參考：http://hk.yimg.com/i/icon/16/3.gif 2. Solve the equation (x+2)^2 = (2x+1)^2 (x+2)^2 = (2x+1)^2 x^2+4x+4 = 4x^2+4x+1 3x^2 = 3 x = -1 or x = 1 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 9. Solve the equation (4x+3)^2 - (5x+1)^2 = 0 (4x+3)^2 - (5x+1)^2 = 0 [(4x+3)+(5x+1)][(4x+3)-(5x+1)] = 0 (9x+4)(-x+2) = 0 x = -4/9 or x = 2 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 10. Solve the equation (5x+2)^2 + (3x-1)^2 = 40, give your answer correct to 3 significant figures. (5x+2)^2 + (3x-1)^2 = 40 25x^2+20x+4+9x^2-6x+1-40 = 0 34x^2+14x-35 = 0 x = [-14 + - sqrt(14^2+4(34)(35))]/(2)(34) = 0.829 or -1.24 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 12b. Hence solve the equation 4/(x^2) + 7/x - 2 = 0. 圖片參考：http://hk.yimg.com/i/icon/16/17.gif 4/(x^2) + 7/x - 2 = 0 4 + 7x - 2x^2 = 0 2x^2-7x-4 = 0 (x-4)(2x+1) = 0 x = 4 or x = -1/2 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 15.Set up a quadraic equation in x whose roots are reciprocals of the roots of x^2 -5x +4 = 0. Let α and β be the two roots of x^2-5x+4 = 0 α+β = 5 αβ = 4 To find (1/α) + (1/β) and (1/α)(1/β) (1/α) + (1/β) = (α+β)/(αβ) = 5/4 (1/α)(1/β) = 1/4 So, the required equation be x^2 - 5/4 x + 1/4 = 0 i.e. 4x^2 - 5x + 1 = 0 圖片參考：http://hk.yimg.com/i/icon/16/3.gif 17.Set up a quadraic equation in x whose roots are one-third of the roots of 5x^2 - x - 6 = 0. Let α and β be the two roots of 5x^2-x-6 = 0 α+β = 1/5 αβ = -6/5 To find (α/3)+(β/3) and (α/3)(β/3) (α/3)+(β/3) = (α+β)/3 = 1/15 (α/3)(β/3) = (αβ)/9 = -6/45 = -3/15 So, the required equation be x^2 - 1/15 x - 3/15 = 0 i.e. 15x^2 - x - 3 = 0 圖片參考：http://hk.yimg.com/i/icon/16/3.gif
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