Question

find the value of k for which the quadratic equation (k+9)x^2-2(k-3)x+1=0 has equal roots

Answer 1

This is your answer:

Answer 2

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

$\green{\tt{\therefore{Value\:of\:k=0\:and\:7}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k+9)x^{2} -2(k-3)x + 1 = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies ( k+9)x^{2} -2( k - 3)x + 1 = 0} \\ \\ \tt{\circ \: a = (k+9)} \\\\ \tt{\circ \: b = -2(k -3 )}\\ \\\tt{\circ \:c = 1}\\ \\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (-(2k -6))^{2} - 4 \times( k +9) \times 1 = 0 } \\ \\ \tt{: \implies \: 4{k}^{2} + 36-24k - 4k - 36 = 0 } \\ \\ \tt{ : \implies \: 4{k}^{2} - 28k = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} - 7k = 0} \\ \\ \tt{: \implies k(k - 7)= 0} \\ \\ \tt{: \implies k = 0\:and\:7} \\ \\ \green{\tt{: \therefore k = 0 \: and \: 7}}$