Question

Kx²+5x-12=0 find the value of K

Answer 1

Step-by-step explanation:

put value of x=1

=k(1)^2+5(1)-12

=1k+5-12

k=-7

Answer 2

Answer:

$k {x}^{2} - 5x + k = 0$

Here, a = k, b = - 5 c, = k

Root of quadrant equation will be real and equal, if :

$d = 0$

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

$( - 5 {)}^{2} - 4.k.k = 0$

$25 - 4 {k}^{2} = 0$

$4 {k}^{2} = 25$

${k}^{2} = \frac{25}{4}$

$k = { + }{ - } \frac{5}{2}$