, are zeros of x² - 4x - 3, find a quadratic polynomial whose zeros are 3α and 3β
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Given α and β are zeroes of the polynomial f(x) = x2- 4x + 3
α+ β = 4 αβ = 3
1) (3α + 3β) = 3x 4 = 12
3α x 3β = 9 x 3 = 27.
If 3α, 3β are zeros of the quadratic polynomial then the equation is
x2 -(3α + 3β)x + 9αβ = 0 then
x2 - 12x + 27 = 0.
2)
(1/2α + 1/2β) = (α + β) / 2αβ = 4 / 6 = 2 / 3.
1/4αβ = 1 /12
If 1 / 2α, 1 / 2β are zeros of the quadratic polynomial then the equation is
x2 -(1 / 2α + 1 / 2β)x + 1 / 4αβ = 0 then
x2 -(2 / 3)x + 1 / 12 = 0
12x2 - 8x + 1 = 0.
α+ β = 4 αβ = 3
1) (3α + 3β) = 3x 4 = 12
3α x 3β = 9 x 3 = 27.
If 3α, 3β are zeros of the quadratic polynomial then the equation is
x2 -(3α + 3β)x + 9αβ = 0 then
x2 - 12x + 27 = 0.
2)
(1/2α + 1/2β) = (α + β) / 2αβ = 4 / 6 = 2 / 3.
1/4αβ = 1 /12
If 1 / 2α, 1 / 2β are zeros of the quadratic polynomial then the equation is
x2 -(1 / 2α + 1 / 2β)x + 1 / 4αβ = 0 then
x2 -(2 / 3)x + 1 / 12 = 0
12x2 - 8x + 1 = 0.
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