Question

Pls help me out.....​

Answer 1

Step-by-step explanation:

here is your answer

please try to solve it yourself

Answer 2

$k=-3/4$ or $k=3$

Step-by-step explanation:

Given,

$\left ( k+1 \right )x^{2}-4kx+9=0$

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

$a=k+1,b=-4k,c=9$

find the value of k

Solution,

D=0 if both roots are real and equal

$\left ( -4k \right )^{2}-4\left ( k+1 \right )9=0$

$16k^{2}-36\left ( k+1 \right )=0$

$4\left ( 4k^{2} -9\left ( k+1 \right )\right )=0$

$4k^{2}-9k-9=0$

$4k^{2}-12k+3k-9=0$     (middle term split)

$4k\left ( k-3 \right )+3\left ( k-3 \right )=0$

$\left ( 4k+3 \right )\left ( k-3 \right )=0$

$4k+3=0$

$k=-3/4$

or

$k-3=0$

$k=3$