Question

Find the value of k for which the equation 4x^+kn+8=0 has real and equal root.​

Answer 1

Answer: k = 8√2

Step-by-step explanation:

for real and equal roots the D = b^2-4ac (i.e discriminant ) should be equal to zero.

using above : D = b^2 -4ac =0          -(1)

here equation is 4 x^2 + kx + 8 =0 comparing it with general equation

ax^2 + bx +c =0

we get a=4 , b=k , c=8

using  (1) we get  k^2 -4*4*8 =0

solving k^2 = 4*4*8

taking square root

ans  k = 8√2