Find the value of k for which the quadratic equation (k-2)x + 2(2k-3) + (5k-6) =0 have equal roots

Find the value of k for which the quadratic equation (k-2)x + 2(2k-3)

1.	Find the value of k for which the quadratic equation (k-2)x + 2(2k-3) + (5k-6) =0 have equal roots
Answer» In quadratic equation, (k - 2)x2 + 2(2k - 3)x +\xa04(5k - 6) = 0a = (k-2) , b = 2(2k-3) , c = 4(5k-6)Since equation has equal roots,{tex}\\therefore {/tex}\xa0D = 0{tex}b ^ { 2 } - 4 a c = 0{/tex}{tex}\\{ 2 ( 2 k - 3 ) \\} ^ { 2 } - 4 ( k - 2 ) ( 5 k - 6 ) = 0{/tex}\xa0{tex}4 \\left( 4 k ^ { 2 } - 12 k + 9 \\right) - 4 ( k - 2 ) ( 5 k - 6 ) = 0{/tex}\xa0{tex}4 k ^ { 2 } - 12 k + 9 - 5 k ^ { 2 } + 6 k + 10 k - 12 = 0{/tex}\xa0{tex}k ^ { 2 } - 4 k + 3 = 0{/tex}\xa0{tex}k ^ { 2 } - 3 k - k + 3 = 0{/tex}{tex}k(k-3)-1(k-3)=0{/tex}\xa0{tex}(k-1)(k-3) =0{/tex}{tex}\\therefore k=1 \\ or \\ k=3{/tex}