Question

The quadratic equation 2x^2+Ky+3=0 have equal roots then the value of K is​

Answer 1

2x2 + kx + 3 = 0

It’s of the form of

ax2 + bx + c = 0

Where, a = 2, b = k, c = 3

For the given quadratic equation to have real roots

D = b2 – 4ac ≥ 0 D

= (k)2 – 4(3)(2) ≥ 0

⇒ k2 – 24 ≥ 0

⇒ k2 ≥ 24

⇒ k ≥ 2√6 and k ≤ -2√6

[After taking square root on both sides]

The value of k can be represented as

(∞, 2√6] and [-2√6, -∞) .

Answer 2

Answer:

       a) √24

Step-by-step explanation:

2x² + kx + 3 = 0

if it has equal roots, then the determinant = 0

determinant of a quadratic equation = √(b² - 4ac) = 0

a = 2 , b = k , c = 3

√(b² - 4ac) = 0

√(k² - 4 × 2 × 3) = 0  

√(k² - 24) = 0   {square both sides}

k² - 24 = 0

k² = 24

k = √24

 ⇒ a) √24