Question

for what value of k will the following quadratic equations k²x²-(k+1)x+9=0 have real and equal root​

Answer 1

Answer:

${k}^{2} {x}^{2} - (k + 1)x + 9$

if

$a = {k}^{2} \\ b = - (k + 1) \\ c = 9$

According to question

real and equal root

so,

$D = 0$

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

$({ - (k + 1))}^{2} - 4 \times {k}^{2} \times 9$

${k}^{2} + 2 k + 1 - 36 {k}^{2} = 0$

$- 35 {k}^{2} + 2k + 1 = 0 \\ 35 {k}^{2} - 2k - 1 = 0$

$7 \times 5 \times {k}^{2} - 7k + 5k - 1 = 0$

$7k(5k - 1) + 1(5k - 1) = 0$

$(7k + 1)(5k - 1) = 0$

$7k + 1 = 0 \\ k = \frac{ - 1}{7}$

and

$5k - 1 = 0 \\ k = \frac{1}{5}$