Find the values of k when the equation kx^2+k=8x-2xk has two real distinct roots?
Thanks
$\displaystyle kx^2+k=8x-2xk$
$\displaystyle kx^2+k-8x+2xk=0$
$\displaystyle kx^2-8x+2xk+k=0$
$\displaystyle kx^2+(-8+2k)x+k=0$
A quadratic equation $\displaystyle ax^2 +bx+c=0$ will have two distinct real roots when $\displaystyle b^2-4ac>0$
In your case solve $\displaystyle (-8+2k)^2-a\times k\times k>0 $
Thx for ur ans.....I cudnt find d way of thinking d ans 4 dis qn.
Suppose for fixed real numbers a and b, f suffix c(x)=x cube+ax square+bx+c has 3 distinct roots for c=0.Then
(1)f suffix c(x) has 3 distinct roots for all real c
(2)f suffix c(x) has 3 distinct roots for all real c>0 or for all real c<0, but not for both
(3)f suffix c(x) has 3 distinct roots for all real c in (p,q) for some p<0 and q>0
(4)f suffix c(x)need not have 3 distinct roots for all real c other than 0
kx^2+k=8x-2xk,
Discriminant = D = (2k - 8)^2 - 4(k)(k) > 0, two real distict roots
k = -4, D = 256 - 64 = 192
k = - 3, D = 196 - 36 = 160
k = -2, D = 144 - 16 = 128
k = -1, D = 100 - 4 = 96
k = 0, D = 64 - 0 = 64
k = 1, D = 36 - 4 = 32
k = 2, D = 0,
k = 3, D = 4 - 36 = -34
k = 4, D = 0 - 64 = - 64
k = 5 D = 4 - 100 = - 96
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