Question

find the value of k for which of the quadratic equation kx^2-3kx+9=0 has real equal roots

Answer 1

here is your answer OK

Look...

We know that cos thita = sin(90-thita)

now look at the ques.

Cos 48 degree - sin 42 degree

= cos(90deg. -42deg.) - sin42 deg.

= sin42 deg. - sin 42 deg.

= 0

Answer 2

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

$\green{\tt{\therefore{Value\:of\:k=0\:and\:4}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies kx^{2} -3kx + 9 = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies kx^{2} -3kx + 9= 0} \\ \\ \tt{\circ \: a = k} \\ \\ \tt{\circ \: b = -3k}\\\\ \tt{\circ \:c = 9}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (-3k)^{2} - 4\times k\times 9= 0 } \\ \\ \tt{: \implies \: 9{k}^{2} -36k= 0 } \\ \\ \tt{ : \implies \: 9({k}^{2} - 4k) = 0 } \\\\ \tt{: \implies k(k-4)= 0} \\ \\ \tt{: \implies k= 0\:and\:4} \\ \\ \green{\tt{: \therefore k =0\:and\:4}}$