Question

pls solve this no.10

Answer 1

kx²+kx+1= -4x²-x
=kx²+kx+4x²+x+1=0
=x²(k+4)+x(k+1)+1=0

Answer 2

$k {x}^{2} + kx + 1 = - 4 {x}^{2} - x \\ = > 4 {x}^{2} + k {x}^{2} + x + kx + 1 = 0 \\ = > (k + 4) {x}^{2} + x(k + 1) + 1 = 0$
So
$b = k + 1 \\ a = k + 4\\ c = 1$
So if roots are real and equal,
${b}^{2} - 4ac = 0 \\ = > {(k + 1)}^{2} - 4(k + 4) = 0 \\ = > {k}^{2} + 2k + 1 - 4k - 16 = 0 \\ = > {k}^{2} - 2k - 15 = 0 \\ = > {k }^{2} + 3k - 5k - 15 = 0 \\ = > k(k + 3) - 5(k + 3) = 0 \\ = > (k + 3)(k - 5) = 0 \\ or \\ k = - 3 \: or \: + 5$
Solved!
Hope you like the answer!