Question

For what integer value/s of k, the quadratic equation $(x-1)(x-k)=-4$ has real and equal roots. (k > 0)

• A. 5, -3
• B. 5
• C. 5, 2
• D. 2, -3

Answer 1

The correct option is B 5

$(x-1)(x-k)=-4$

$\text{Converting into standard from}$

$x^2-kx-x+k=-4$

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

$a=1,b=-(k+1),c=k+4$

Since the quadratic equation has real and equal roots,

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

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

$k^2+2k+1-4K-16=0$

$k^2-2k-15=0$

$k^2-5k+3k-15=0$

$(k-5)(k+3)=0$

$k=5,-3$

$\text{Since, we have to find the positive value of k.}$

$\text{Hence, k =5}$

$\text{(b) is correct}$