Question

find the value of K in which the quadratic equation (k-4)x²+2(k-4)x + 2 =0 has equal roots​

Answer 1

Answer:

b2-4ac= 0

Step-by-step explanation:

by this formula you find the value of k

Answer 2

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

$\green{\therefore{\text{Value\:of\:k=6\:and\:4}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k - 4)x^{2} +2( k - 4)x + 2 = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies ( k - 4)x^{2} +2( k - 4)x + 2 = 0} \\ \\ \tt{\circ \: a = (k - 4)} \\ \tt{\circ \: b = (2k -8)}\\ \tt{\circ \:c = 2}\\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (2k - 8)^{2} - 4 \times( k - 4) \times 2 = 0 } \\ \\ \tt{: \implies \: 4{k}^{2} + 64-32k - 8k + 32 = 0 } \\ \\ \tt{ : \implies \: 4{k}^{2} - 40k + 96 = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} - 6k - 4k + 24 = 0} \\ \\ \tt{: \implies k(k - 6) - 4(k - 6) = 0} \\ \\ \tt{: \implies (k - 6)(k -4) = 0} \\ \\ \green{\tt{: \implies k = 6 \: and \: 4}}$