Question

find the values of k, for which the given quadratic equation has equal roots: 4x2+ kx + 6=0

Answer 1

Since the equation has equal roots

Discriminent = 0

=> b^2 - 4ac = 0

=> k^2 - 4(4)(6)= 0

=> k^2 - 96 = 0

=> k^2 = 96

=> k = + - 4 root6

$k = + 4 \sqrt{6} \\ \\ k = - 4 \sqrt{6}$

Answer 2

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

$\green{\tt{\therefore{Value\:of\:k=\pm 4\sqrt{6}}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies 4x^{2} +kx + 6= 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies 4x^{2} +kx + 6= 0} \\ \\ \tt{\circ \: a = 4} \\ \\ \tt{\circ \: b = k}\\\\ \tt{\circ \:c = 6}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies k^{2} - 4\times4\times 6= 0 } \\ \\ \tt{: \implies \: {k}^{2} -96 = 0 } \\ \\ \tt{ : \implies \: ({k}^{2} - (4\sqrt{6})^{2}) = 0 } \\\\ \tt{: \implies (k-4\sqrt{6})(k+4\sqrt{6})= 0} \\ \\ \tt{: \implies k=4\sqrt{6}\:and\:-4\sqrt{6}} \\ \\ \green{\tt{: \implies k = \pm4\sqrt{6} }}$