Question

Find the values of k , for which the quadratic equation 2x^2+kx+3=0 has two real and equal roots

Answer 1

The condition for equal roots is-

b^2-4ac=0-----------(1)

Here,
a=2, b= k, and c= 3

Now putting all the values in equation (1) we get-

b^2-4ac=0
or, k^2-4(2)(3)=0
or, k^2-24=0
or, k^2=24
or, k= \sqrt{24} (ANS)

Answer 2

Answer:

given,

2x^2+Kx+3

here it is in the form ax^2+bc+C=0

here a=2, b=K, c=3

if it has two equal roots. then ∆=0

b^2-4ac=0

k^2-4(2)(3)=0

k^2-24=0

k^2=24

k=√24

k=2√6

or

we have to find the values of k for quadratic equations 2x² + kx + 3 = 0 so that they have two equal roots.

we know, quadratic equation will be equal only when

discriminant, D = b² - 4ac = 0

on comparing 2x² + kx + 3 = 0 with general form of quadratic equation , ax² + bx + c = 0 we get, a = 2, b = k and c = 3

so Discriminant , D = (k)² - 4(2)(3) = 0

or, k² - 24 = 0

or, k = ± √24 = ±2√6

hence, the value of k = 2√6 or -2√6