Question

For what values of k, the roots of the quadratic equation (k + 4)x2 + (K + 1)x + 1 = 0
are equal?​

Answer 1

Answer:

k=5

Step-by-step explanation:

(k+4)x²+(k+1)x+1=0

a=k+4, b=k+1, c=1

Discriminant=0

b²-4ac=0

(k+1)²-4(k+4)(1)=0

k²+2k+1-4k-16=0

k²+2k-4k+1-16=0

k²-2k-15=0

k²-5k+3k-15=0

(k²-5k)+(3k-15)=0

k(k-5)+3(k-5)=0

(k-5)(k+3)=0

k=5 and k=(-3)

k=5

Therefore, the value of k is 5

Answer 2

$\huge\mathcal\colorbox{blue}{{\color{white}{AnSwEr~↓~↓~}}}$

$\bf \color{navy} Given \: that \: the \: quadratic \: equation \\ \bf \color{navy} has \: equal \: roots \\ \bf \therefore D , \: \: {b}^{2} - 4ac = 0 \\ \bf here \: a = (k + 4) \: , \: \: b = (k + 1) \: , \: \: c = 1 \\ \bf \implies (k + 1)^{2} - 4(k + 4)(1) = 0 \\ \bf ({k}^{2} + 2k + 1) - 4k - 16 = 0 \\ \bf {k}^{2} + 2k + 1 - 4k - 16 = 0 \\ \bf {k }^{2} - 2k - 15 = 0 \\ \bf k^{2} - 5k + 3k - 15 = 0 \\ \bf [k(k - 5)~]+[3(k - 5)~] = 0 \\ \bf (k + 3)(k - 5) = 0 \\ \bf k + 3 = 0~~~or~~~k - 5 = 0 \\ \bf k = -3~~~or~~~k = 5$

$\huge\color{blue}{\textbf{\textsf{Hope~it~helps..!!}}}$