Question

For what value of $k$, the equation $(2k-1)x^3-2kx+1=0$ will have equal roots?

Answer 1

The equation $a{x}^3+b{x}^2+cx+d=0$ has equal roots if the discriminant$=0$

Discriminant$△=18abcd-4b^{3}d+b^2{c}^2-4a{c}^3-27a^2{d}^2=0$

where $a=2k-1,b=0,c=-2k$ and $d=1$

$△=18(2k-1)\times 0\times -2k\times 1-4\times 0\times 1+0-4(2k-1){(-2k)}^3-27(2k-1)1=0$

$\Rightarrow △=-4(2k-1){(-2k)}^3-27{(2k-1)}^{2}1=0$

$\Rightarrow 32k^3(2k-1)-27(4k^2+1-4k)=0$

$\Rightarrow (2k-1)(32k^3-27(2k-1))=0$

$\Rightarrow 2k-1=0$ and $32k^3-54k+27=0$

$\Rightarrow k=\frac{1}{2}$ and $32k^3-54k+27=0$

The value of $k=\frac{1}{2}$