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(s) 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.
Answer : 1. -7/2 or 3
2. 2x^2 + 7x - 6=0
3. 9x^2 + 6x +1= 0
Math F.4 Quadratic Equation~~~
2012-10-09 2:42 am
回答 (2)
2012-10-09 3:34 am
1.x^2 + x = k(2x+3)
x^2 + x = 2xk+3k
x^2+(1-2k)x-3k=0
α + β= -(1-2k)
αβ=-3k
α^2 +β^2 = 43
(α + β)^2-2αβ=43
(2k-1)^2-2(-3k)=43
4k^2+2k-42=0
(2k-6)(2k+7)=0
k=-7/2 or k=3
2.α + β+4=1/2
α + β =-7/2
(α +2)(β+2)=αβ+2(α + β)+4=-6
αβ+2(-7/2)=-10
αβ=-3
The eqt is x^2-(-7/2)x+3=0
2x^2 + 7x - 6=0
3. 1/αβ=-3
αβ=-1/3
1/α+1/β=1
(α + β)/αβ=1
α + β==1/3
α + β+ αβ=-2/3
(α + β)(αβ)=1/9
The eqt is x^2-(-2/3)+1/9=0
9x^2 + 6x +1= 0
x^2 + x = 2xk+3k
x^2+(1-2k)x-3k=0
α + β= -(1-2k)
αβ=-3k
α^2 +β^2 = 43
(α + β)^2-2αβ=43
(2k-1)^2-2(-3k)=43
4k^2+2k-42=0
(2k-6)(2k+7)=0
k=-7/2 or k=3
2.α + β+4=1/2
α + β =-7/2
(α +2)(β+2)=αβ+2(α + β)+4=-6
αβ+2(-7/2)=-10
αβ=-3
The eqt is x^2-(-7/2)x+3=0
2x^2 + 7x - 6=0
3. 1/αβ=-3
αβ=-1/3
1/α+1/β=1
(α + β)/αβ=1
α + β==1/3
α + β+ αβ=-2/3
(α + β)(αβ)=1/9
The eqt is x^2-(-2/3)+1/9=0
9x^2 + 6x +1= 0
2012-10-09 3:20 am
1. x^2 + x = k(2x+3).
x^2 + x = 2kx + 3k
x^2 + (1-2k)x - 3k = 0
α + β = -(1-2k)/1 = 2k -1
αβ = -3k/1 = -3k
α^2 +β^2 = 43
(α+β)^2 - 2αβ = 43
(2k-1)^2 - 2(-3k) = 43
(4k^2-4k+1)+6k = 43
4k^2 + 2k + 42 = 0
2k^2 + k +21 = 0
(2k+7)(k-3) = 0
So k = -7/2 or 3
2. (α+2)+(β+2) = -(-1)/2
α + β + 4 = 1/2
α + β = -7/2
(α+2)(β+2) = -12/2
αβ + 2(α+β) + 4 = -6
αβ + 2 (-7/2) + 4 = -6
αβ - 7 + 4 = -6
αβ = -3
So the quadratic equation is 2x^2 + 7x - 6 = 0
3. (1/α)(1/β) = -3/1
1/(αβ) = -3
αβ = -1/3
(1/α)+(1/β) = -(-1)/1
(β+α) / (αβ) = 1/1
(β+α)/ (-1/3) = 1/1
-3( α + β ) = 1/1
α + β = -1/3
(α + β)+ (α β) = (-1/3)+(-1/3)
= -2/3
(α + β)(α β) = (-1/3)(-1/3)
= 1/9
So the quadratic equation is 9x^2 + 6x + 1 =0
x^2 + x = 2kx + 3k
x^2 + (1-2k)x - 3k = 0
α + β = -(1-2k)/1 = 2k -1
αβ = -3k/1 = -3k
α^2 +β^2 = 43
(α+β)^2 - 2αβ = 43
(2k-1)^2 - 2(-3k) = 43
(4k^2-4k+1)+6k = 43
4k^2 + 2k + 42 = 0
2k^2 + k +21 = 0
(2k+7)(k-3) = 0
So k = -7/2 or 3
2. (α+2)+(β+2) = -(-1)/2
α + β + 4 = 1/2
α + β = -7/2
(α+2)(β+2) = -12/2
αβ + 2(α+β) + 4 = -6
αβ + 2 (-7/2) + 4 = -6
αβ - 7 + 4 = -6
αβ = -3
So the quadratic equation is 2x^2 + 7x - 6 = 0
3. (1/α)(1/β) = -3/1
1/(αβ) = -3
αβ = -1/3
(1/α)+(1/β) = -(-1)/1
(β+α) / (αβ) = 1/1
(β+α)/ (-1/3) = 1/1
-3( α + β ) = 1/1
α + β = -1/3
(α + β)+ (α β) = (-1/3)+(-1/3)
= -2/3
(α + β)(α β) = (-1/3)(-1/3)
= 1/9
So the quadratic equation is 9x^2 + 6x + 1 =0
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