Question

find the roots of 2x^2+kx+3=0 are real and equal then find value of k

Answer 1

2X² + KX + 3 = 0



Here,


A = Coefficient of X² = 2


B = Coefficient of X = K



And,

C = Constant term = 3







Discriminant ( D ) = 0



B² - 4AC = 0



(K)² - 4 × 2 × 3 = 0



K² - 24 = 0


K² = 24


K = √24 = √2 × 2 × 2 × 3




K = 2√6

Answer 2

Answer:2

Step-by-step explanation:

2x  

2

+kx+3=0   ---(1)

if eq (1)has equal roots then

Discriminant (1)=  

b  

2

−4ac

​  

=0

b  

2

=4ac

k  

2

=4×2×3

k  

2

=24

k=  

24

​  

=±2  

6

​  

 

∴ for k = ±2  

16

​  

 given equation has equal roots.