Question

Find the value of k for which the quadratic equation k^2x^2-2[k-1]x+4=0 has real and equal roots

Answer 1

Hope you will get the answer from the attachment

Answer 2

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

$\green{\tt{\therefore{Value\:of\:k=\frac{1}{3}\:and\:-1}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies k^{2}x^{2} -2(k-1)x + 4= 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies k^{2}x^{2} -2( k - 1)x + 4 = 0} \\ \\ \tt{\circ \: a = k^{2}} \\\\ \tt{\circ \: b = -2(k -1)}\\ \\\tt{\circ \:c = 4}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (-(2k -2))^{2} - 4 \times k^{2} \times 4= 0 } \\ \\ \tt{: \implies \: 4{k}^{2} + 4-8k - 16k^{2} = 0 } \\ \\ \tt{ : \implies \: 12{k}^{2} +8k -4 = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies 3{k}^{2} +2 -1 = 0} \\ \\ \tt{: \implies 3k^{2}+3k-k-1= 0} \\ \\ \tt{: \implies (3k-1)(k+1)=0} \\ \\ \green{\tt{: \therefore k = \frac{1}{3} \: and \: -1}}$