Question

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

Answer 1

quadratic equation has equal root if discriminant D = b^2 - 4ac =0
=> ( k+1 )^2 = 4(k+4)(1)
=> k^2 + 2k + 1 = 4k + 16
=> k^2 -2k -15= 0
=> k = -3 or 5

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

To find roots :

equation is

(I) when k = 5

9x² + 6x + 1 = 0

9x² + 3x + 3x + 1 = 0

3x ( 3x + 1 ) + (3x + 1 ) = 0

( 3x + 1 ) ( 3x + 1 ) = 0

x = -1/3 , -1/3

(ii) when k = -3

equation is

x² - 2x + 1 = 0

x² - 1x - 1x + 1 = 0

x ( x - 1 ) - 1( x - 1 ) = 0

(x -1)(x-1) = 0

then x = 1 , 1

Hope it helps