15marks quadratic equations
2017/06/24 18:43
瀏覽72
迴響0
推薦0
引用0
標題:
15marks quadratic equations
1.It is given that α and β are the roots of the quadratic equation x^2+x=k(2x+3). IF α^2+β^2=43,find the possible value of k.2. If α+2 and β+2 are the roots of the quadratic equation 2x^2-x-12=0,form a quadratic equation in x whose roots are α... 顯示更多 1.It is given that α and β are the roots of the quadratic equation x^2+x=k(2x+3). IF α^2+β^2=43,find the possible value of k. 2. If α+2 and β+2 are the roots of the quadratic equation 2x^2-x-12=0,form a quadratic equation in x whose roots are α and β. 3.If 1/α and 1/β are the roots of the quadratic equation x^2-x-3=0,form a quadratic equation in x whose roots are α+β and αβ. 4.It is given that 3α and 3β are the roots of the quadratic equation x^2+px+q=0. Form a quadratic equation in x whose roots are α+2β and β+2α, in terms of p and q. Thank you, please.
最佳解答:
1. x^2 + x = k(2x + 3) x^2 + (1 - 2k)x - 3k = 0 Sum of roots α + β = 2k - 1 Product of roots αβ = -3k α^2 + β^2 = 43 α^2 + 2αβ + β^2 = 43 + 2αβ (α + β)^2 = 43 + 2αβ (2k - 1)^2 = 43 - 6k 4k^2 - 4k + 1 - 43 + 6k = 0 4k^2 + 2k - 42 = 0 2k^2 + k - 21 = 0 (k - 3)(2k + 7) = 0 k = 3 or k = -7/2 2. 2x^2 - x - 12 = 0 Sum of roots α + 2 + β + 2 = 1/2 => α + β = -7/2 Product of roots = (α + 2)(β + 2) = -6 αβ + 2α + 2β + 4 = -6 αβ + 2(-7/2) + 4 = -6 αβ = -3 The new equation is x^2 + 7x/2 - 3 = 0 OR 2x^2 + 7x - 6 = 0 3. x^2 - x - 3 = 0 Sum of roots = 1/α + 1/β = 1 ... (1) Product of roots = (1/α)(1/β) = -3 => αβ = -1/3 (1) => (α + β)/αβ = 1 α + β = -1/3 The equation is (x + 1/3)(x + 1/3) = 0 <=> x^2 + 2x/3 + 1/9 = 0 <=> 9x^2 + 6x + 1 = 0 4. x^2 + px + q = 0 Sum of roots = 3α + 3β = -p => α + β = -p/3 Product of roots = (3α)(3β) = 9αβ = q => αβ = q/9 New sum of roots = α + 2β + β + 2α = 3(α + β) = -p New product of roots = (α + 2β)(β + 2α) = αβ + 2β^2 + 2α^2 + 4αβ = αβ + 2(α + β)^2 = q/9 + 2(-p/3)^2 = q/9 + 2p^2/9 New equation is x^2 + px + q/9 + 2p^2/9 = 0 <=> 9x^2 + 9px + q + 2p^2 = 0
其他解答:
15marks quadratic equations
此文章來自奇摩知識+如有不便請留言告知
發問:1.It is given that α and β are the roots of the quadratic equation x^2+x=k(2x+3). IF α^2+β^2=43,find the possible value of k.2. If α+2 and β+2 are the roots of the quadratic equation 2x^2-x-12=0,form a quadratic equation in x whose roots are α... 顯示更多 1.It is given that α and β are the roots of the quadratic equation x^2+x=k(2x+3). IF α^2+β^2=43,find the possible value of k. 2. If α+2 and β+2 are the roots of the quadratic equation 2x^2-x-12=0,form a quadratic equation in x whose roots are α and β. 3.If 1/α and 1/β are the roots of the quadratic equation x^2-x-3=0,form a quadratic equation in x whose roots are α+β and αβ. 4.It is given that 3α and 3β are the roots of the quadratic equation x^2+px+q=0. Form a quadratic equation in x whose roots are α+2β and β+2α, in terms of p and q. Thank you, please.
最佳解答:
1. x^2 + x = k(2x + 3) x^2 + (1 - 2k)x - 3k = 0 Sum of roots α + β = 2k - 1 Product of roots αβ = -3k α^2 + β^2 = 43 α^2 + 2αβ + β^2 = 43 + 2αβ (α + β)^2 = 43 + 2αβ (2k - 1)^2 = 43 - 6k 4k^2 - 4k + 1 - 43 + 6k = 0 4k^2 + 2k - 42 = 0 2k^2 + k - 21 = 0 (k - 3)(2k + 7) = 0 k = 3 or k = -7/2 2. 2x^2 - x - 12 = 0 Sum of roots α + 2 + β + 2 = 1/2 => α + β = -7/2 Product of roots = (α + 2)(β + 2) = -6 αβ + 2α + 2β + 4 = -6 αβ + 2(-7/2) + 4 = -6 αβ = -3 The new equation is x^2 + 7x/2 - 3 = 0 OR 2x^2 + 7x - 6 = 0 3. x^2 - x - 3 = 0 Sum of roots = 1/α + 1/β = 1 ... (1) Product of roots = (1/α)(1/β) = -3 => αβ = -1/3 (1) => (α + β)/αβ = 1 α + β = -1/3 The equation is (x + 1/3)(x + 1/3) = 0 <=> x^2 + 2x/3 + 1/9 = 0 <=> 9x^2 + 6x + 1 = 0 4. x^2 + px + q = 0 Sum of roots = 3α + 3β = -p => α + β = -p/3 Product of roots = (3α)(3β) = 9αβ = q => αβ = q/9 New sum of roots = α + 2β + β + 2α = 3(α + β) = -p New product of roots = (α + 2β)(β + 2α) = αβ + 2β^2 + 2α^2 + 4αβ = αβ + 2(α + β)^2 = q/9 + 2(-p/3)^2 = q/9 + 2p^2/9 New equation is x^2 + px + q/9 + 2p^2/9 = 0 <=> 9x^2 + 9px + q + 2p^2 = 0
其他解答:
你可能會有興趣的文章：
限會員,要發表迴響,請先登入
