Question

Find the value of k for which the quadratic equation (k-12)x^2+2(k-12)x+2=0 has equal roots

Answer 1

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

$\green{\tt{\therefore{Value\:of\:k=12\:and\:14}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k-12)x^{2} +2(k-12)x + 2 = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies ( k-12)x^{2} +2( k - 12)x + 2 = 0} \\ \\ \tt{\circ \: a = (k-12)} \\\\ \tt{\circ \: b = 2(k -12 )}\\ \\\tt{\circ \:c = 2}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (2k -24)^{2} - 4 \times( k -12) \times 2 = 0 } \\ \\ \tt{: \implies \: 4{k}^{2} + 576-96k - 8k + 96= 0 } \\ \\ \tt{ : \implies \: 4 {k}^{2} -104k+672 = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} - 26k +168 = 0} \\ \\ \tt{: \implies (k-14)(k-12)= 0} \\ \\ \tt{: \implies k =14\:and\:12} \\ \\ \green{\tt{: \therefore k = 12 \: and \: 14}}$