Question

find the value of k for which the quadratic equation( K + 4)x square+(k+1) x+1=0 has equal roots​

Answer 1

Answer:

Equal roots means

D = 0

D = b^2 - 4ac =0

$= (k + 1) {}^{2} - 4(k + 4)(1) \\ k {}^{2} + 2k + 1 - 4k - 16 = 0 \\ k {}^{2} - 2k - 15 = 0$

by splitting the middle term

$k {}^{2} - 5k + 3k - 15 = 0 \\ k(k - 5) + 3(k - 5) = 0 \\ k - 5 = 0 \: \: \: \: \: k + 3 = 0 \\ k = 5 \: and \: k = - 3$

Answer 2

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

$\green{\therefore{\text{Value\:of\:k=5\:and\:-3}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k +4)x^{2} +( k+1)x + 1 = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies ( k+4)x^{2} +( k +1)x + 1= 0} \\ \\ \tt{\circ \: a = (k +4)} \\\\ \tt{\circ \: b = (k +1)}\\\\ \tt{\circ \:c = 1}\\ \\\bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (k +1)^{2} - 4 \times( k+4) \times 1 = 0 } \\ \\ \tt{: \implies \: {k}^{2} + 1+2k - 4k -16 = 0 } \\ \\ \tt{ : \implies \: {k}^{2} - 2k -15= 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} - 5k +3k -15 = 0} \\ \\ \tt{: \implies k(k - 5) +3(k - 5) = 0} \\ \\ \tt{: \implies (k +3)(k -5) = 0} \\ \\ \green{\tt{: \implies k = 5 \: and \: -3}}$