Question

For what integer value/s, of k, the quadratic equation $(1-k)x^2-4kx+2=0$ has real and equal roots?

• A. $-1$
• B. $\frac{1}{2},-1$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A $-1$

$(1-k)x^2-4kx+2=0$

$a=1-k,b=-4k,c=2$

Since the quadratic equation has real and equal roots,

$D=b^2-4ac=0$

$(-4k{)}^2-4\times (1-k)\times 2=0$

$16k^2-8(1-k)=0$

$16k^2-8+8k=0$

$2k^2+k-1=0$

$2k^2+2k-k-1=0$

$(k+1)(2k-1)=0$

$k=-1,\frac{1}{2}$

$\text{Since, we have to find the inetger value}$

$\text{Hence, k=-1}$.

$\text{(a) is correct}$