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) \[{{b}^{2}}-4ac=0\] \[\Rightarrow \] \[{{(2k)}^{2}}-4.1.(k(k-1)+2)=0\] \[\Rightarrow \] \[4{{k}^{2}}-4({{k}^{2}}-k+2)=0\] \[\Rightarrow \] \[4{{k}^{2}}-4{{k}^{2}}+4k-8=0\] \[\Rightarrow \] \[4k=8\] \[\Rightarrow \] \[k=\frac{8}{4}=2\]
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