Question

find the value of k for which the quadratic equation( K - 2) X square + 2 (2 k - 3 ) X +(5 K - 6 )equal to zero has equal roots​

Answer 1

here is the answer

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Value\:of\:k=1\:and\:3}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k - 2)x^{2} +2(2k -3)x + (5k-6) = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies (k - 2)x^{2} +2(2k - 3)x + (5k-6) = 0} \\ \\ \tt{\circ \: a = (k -2)} \\ \\ \tt{\circ \: b = 2(2k -3)}\\ \\\tt{\circ \:c = (5k-6)}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (4k -6)^{2} - 4 \times( k - 2) \times (5k-6) = 0 } \\ \\ \tt{: \implies \: 16{k}^{2} + 36-48k - 20k^{2} +64k -48= 0 } \\ \\ \tt{ : \implies \: 4{k}^{2} -16k +12 = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} - 4k + 3= 0} \\ \\ \tt{: \implies k(k - 3) - 1(k - 3) = 0} \\ \\ \tt{: \implies (k - 1)(k - 3) = 0} \\ \\ \green{\tt{: \implies k = 1 \: and \: 3}}$