Question

Friends. Please help me!!!
Find the value of k for which the quadratic equation
2kx^2-2(1+2k)x+(3+2k)
Has equal roots

Answer 1

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

$\green{\therefore{\tt{Value\:of\:k=\frac{1}{2}}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies 2kx^{2} -2(1 +2k)x + (3+2k) = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies 2kx^{2} - 2(1+2k)x + (3+2k)= 0} \\ \\ \tt{\circ \: a = (2k)} \\ \tt{\circ \: b = -2(1+2k)}\\ \tt{\circ \:c = (3+2k)}\\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (-(2+4k))^{2} - 4 \times2k\times (3+2k)= 0 } \\ \\ \tt{: \implies \: \cancel{16{k}^{2}} + 4+16k - 24k \cancel{-16k^{2} }= 0 } \\ \\ \tt{ : \implies \: 8k=4} \\ \\ \tt{: \implies k = \frac{4}{8}} \\ \\ \green{\tt{: \implies k = \frac{1}{2}}}$