Find the value of k for …
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Posted by Smriti Sharma 4 years, 10 months ago
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Sia ? 4 years, 10 months ago
In quadratic equation, (k - 2)x2 + 2(2k - 3)x + 4(5k - 6) = 0
a = (k-2) , b = 2(2k-3) , c = 4(5k-6)
Since equation has equal roots,
{tex}\therefore {/tex} D = 0
{tex}b ^ { 2 } - 4 a c = 0{/tex}
{tex}\{ 2 ( 2 k - 3 ) \} ^ { 2 } - 4 ( k - 2 ) ( 5 k - 6 ) = 0{/tex}
{tex}4 \left( 4 k ^ { 2 } - 12 k + 9 \right) - 4 ( k - 2 ) ( 5 k - 6 ) = 0{/tex}
{tex}4 k ^ { 2 } - 12 k + 9 - 5 k ^ { 2 } + 6 k + 10 k - 12 = 0{/tex}
{tex}k ^ { 2 } - 4 k + 3 = 0{/tex}
{tex}k ^ { 2 } - 3 k - k + 3 = 0{/tex}
{tex}k(k-3)-1(k-3)=0{/tex}
{tex}(k-1)(k-3) =0{/tex}
{tex}\therefore k=1 \ or \ k=3{/tex}
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