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| 1. |
Find value of K for which the equation kx^2+2x+1=0 has real and distinct roots. |
| Answer» We have, {tex}kx^2+2x+1=0{/tex}here, a=k, b=2, c=1{tex}\\therefore D=b^2-4c=(2)^2-4(k)(1)=4-4k{/tex}The given equation will have real and distinct roots,{tex}D>0\\implies4-4k>0\\implies k<1{/tex} | |