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answers:
cidyah say: Compare with ax^2+bx+c
a = k
b = (1-3k)
c = k-6
sum of the roots = -b/a = (3k-1) / k
product of the roots = c/a = (k-6) / k
Let m and n be the roots
n = -1/m
mn=-1
(k-6) / k = -1
k-6 = -k
2k=6
k= 3
The roots are:
(3k-1)/k = ((3)(3)-1) /3 = 9/3= 3
and
(k-6)/k = (3-6)/3 = -1
k= 3
The roots are 3 and -1
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Ian H say: k(x – r)(x + 1/r) = kx^2 + k[(1- r)/r]x – k = 0
compare with …… kx^2 + (1-3k)x + (k - 6) = 0
k = 3, so, 3x^2 - 8x - 3 = (3x + 1)(x – 3) = 0
The roots are x = 1/3 and x = -3
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Philip say: Let roots of kx^2 + (1-3k)x +(k-6) = 0..[1] be p and (-1/p). Then (x-p)[x+(1/p)] = 0, ie., (x-p)(px+1) = 0. Then px^2 +(1-p^2)x -p = 0. Then p = k, (1-p^2) =(1-3k) and (k-6)= -p. So (1-k^2) =1-3k, ie., k(k-3) = 0,
ie., k = 3. {k =/= 0, otherwise [1] becomes x-1=0, a linear eqn} Then p = 3 & -(1/p) = -(1/3) and [1]
becomes 3x^2 -8x -3 = 0, ie., 6x = 8(+/-)D, where D^2 = (8)^2 - 4*3*(-3) = 64 + 36 = (10)^2 and
x = (1/6)[8(+/-)10] = (1/6)[-2 or 18] = -(1/3) or 3. This proves that values for k and the 2 roots of [1] are
correct.
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Pope say: kx² + (1 - 3k)x + (k - 6) = 0
It is a quadratic equation. Let the roots be α and β.
αβ = (k - 6)/k
If α = -1/β, then αβ = -1.
(k - 6)/k = -1
k - 6 = -k
2k = 6
k = 3
kx² + (1 - 3k)x + (k - 6) = 0
(3)x² + [1 - 3(3)]x + [(3) - 6] = 0
3x² - 8x - 3 = 0
(3x + 1)(x - 3) = 0
x = -1/3 or x = 3
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david say: kx^2+(1-3k)x+(k-6)=0 ist roox = a, 2nd root = -1/a
x^2 + [(1-3k)/k]x + (k-6)/k = 0
sum of roots = -b = a + (-1/a) = [(1-3k)/k]
product of roots = c = a(-1/a) = (k-6)/k
-k = k - 6
-2k = -6 .... k = 3
a + (-1/a) = [(1-3k)/k] <<< sub k=3
(a^2 - 1)/a = -8/3
3a^2 - 3 = -8a
3a^2 + 8a - 3 = 0
(3a - 1)(a + 3) = 0
a = 1/3 or a = -3 ...
1st root x = -3 ... 2nd root x = 1/3
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