Question

find the non zero values of K,for which the quadratic equation Kxsquare +(K-1)x+xsquare =0 has equal roots.find the roots of equation.

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

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

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

$\green{\therefore{\text{x=0}}}$

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

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

• According to given question :

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

$\text{Putting \: the \: values\:of\:k\:in\:given\:eqn} \\ \tt{: \implies (k+1)x^{2} +(k-1)x = 0 } \\ \\ \tt{: \implies \: ( k+1)x^{2}-0= 0 } \\ \\ \tt{ : \implies \: 2x^{2}-0= 0 } \\ \\ \tt{: \implies 2x^{2}= 0} \\ \\ \green{\tt{: \implies x= 0}}$