Question

Find the values of k for which the quadratic equation (k+4)x^2+(k+1)x+1=0 has equal roots

Answer 1

Given Equation is (k + 4)x^2 + (k + 1)x + 1 = 0

Where a = (k + 4), b = (k + 1) ,c = 1

Given that the quadratic equation has equal roots.

Hence  $b^{2} -4ac = 0$

 = $( k+1)^{2} - 4(k + 4)(1)$ = 0

We know that $(a+b)^2 = a^2 + b^2 + 2ab$

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

$k^{2} - 5k - 3k - 15 = 0$

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

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

k = -3 (or) k = 5.


Hope this helps! 

Answer 2

Hey friend,

( k + 4 ) × x² + ( k + 1 ) × x + 1 = 0

has equal roots

Therefore

${b}^{2} - 4ac = 0$

b = k + 1

a = k + 4

c = 1

thereforei

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

${k}^{2} + 1 + 2k - 4k - 16 = 0$

${k }^{2} - 2k - 15 = 0$

Therefore by applying quadratic formula...

$\frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} = x$

x = ( 2 ± 8 ) /2

= 5 , -3

Your answer is k = 5 or - 3

Hope it helps