One root of
3x^2-2(3b+1)x+4b = 0 is 8
Find the value of b
If a root of f(x) is 8, then f(8) = 0
3*64 - 8(6b + 2) + 4b = 0
192 - 48b - 16 + 4b = 0
b = 4
also
In the quadratics equation
(k-2)x^2 - 5x+ 2k + 3 = 0, the other roots are reciprocals of each other
find the value of k
the 'other' roots?
anyway let a and b be the roots of the quadratic
so ab = c/a = 1 (since they're reciprocals)
i.e. (2k + 3)/(k - 2) = 1
2k + 3 = k - 2
k = -5
also
In the quadratic equation
x^2+mx+2 = 0, the roots are consecutive. find the values of m
the roots are consecutive, so let the roots be b and b+1
therefore sum of roots i.e. b + (b + 1) = -b/a = -m
so 2b + 1 = -m --(1)
product of roots b(b+1) = b2 + b = c/a = 2 --(2)
solving simultaneously,
from (1), b = -(m + 1)/2
sub that in (2): (m2 + 2m + 1)/4 - (m + 1)/2 - 2 = 0
m2 + 2m + 1 - 2m - 2 - 8 = 0
m2 = 9
m = +/- 3