Question

Solve it.........


Thanks ❤️

Answer 1

( k + 4)x^2 + (k+1)x + 1 = 0

On comparing the given equation with general equation, we get

a = (k +4)

b = (k +1)

c = 1

For equal roots,

b^2 - 4ac = 0

=> (k+ 1)^2 - 4(k + 4)(1) = 0

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

k = 5 and - 3
______________________

For k = 5

Equation is 9x^2 + 6x + 1 = 0

=> 9x^2 + 3x + 3x + 1 = 0

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

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

Roots are - 1/3 and - 1/3
______________________

For k = - 3

Equation is x^2 - 2x + 1 = 0

=> x^2 - x - x + 1 = 0

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

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

Roots are 1 and 1

Answer 2

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

Now for equal roots D = 0

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

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

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

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

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

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

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

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

$k = - 3 \: \: \: \: and \: \: \: \: k = 5$


$thanks$