Question

Find the value of k for which the given quadratic equation has real and distinct roots kx^2+2x+1+0

Answer 1

Here Is Your Answer.. Hope It Helps..

Answer 2

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

$\green{\tt{\therefore{Value\:of\:k<1}}}$

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

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

• According to given question :

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