question_answer

Find the value of k for which the equation \[{{x}^{2}}+k(2x+k-1)+2=0\] has real and equal roots.

Answer:

Given equation is,

\[{{x}^{2}}+k(2x+k-1)+2=0\]

\[\Rightarrow \] \[{{x}^{2}}+2kx+(k-1)+2=0\]

Hear \[a=1,\,b=2k\] and \[c=k(k-1)+2\]

For real and equal roots

(multiply)                       \[{{b}^{2}}-4ac=0\]

\[\Rightarrow \]       \[{{(2k)}^{2}}-4.1.(k(k-1)+2)=0\]

\[\Rightarrow \]              \[4{{k}^{2}}-4({{k}^{2}}-k+2)=0\]

\[\Rightarrow \]             \[4{{k}^{2}}-4{{k}^{2}}+4k-8=0\]

\[\Rightarrow \]                                   \[4k=8\]

\[\Rightarrow \]                              \[k=\frac{8}{4}=2\]