Find the value of K for which the roots are real and equal1) (2K+1)x^2

1.	Find the value of K for which the roots are real and equal1) (2K+1)x^2+2(K+3)x+(K+5)=0
Answer» We have, {tex}(2k+1)x^2+2(k+3)x+(k+5)=0{/tex}Here, a=(2k+1),b=2(k+3), c=(k+5){tex}\\therefore D=b^2-4ac=4(k+3)^2-4(2k+1)(k+5){/tex}{tex}=4k^2+36+24k-8k^2-40k-4k-20=-4k^2-20k+16{/tex}{tex}=-4(k^2+5k-4){/tex}For the equation having equal roots,{tex}D=0 \\implies k^2+5k-4=0{/tex}{tex}k = {-5 \\pm \\sqrt{5^2-4(1)(-4)} \\over 2(1)}={-5 \\pm \\sqrt{41} \\over 2}{/tex}

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