Question

Find the value of k in the following quadratic equations for which roots are real and equal :-2x^2+kx+3=0

Answer 1

Answer:

Step-by-step explanation:

2x^2+kx+3=0

α,β are the roots

α=β

therefore

sum of roots = -b/a

α+α = -k/2

2α= -k/2

4α= -k

product of the roots

αβ = α² = c/a = 3/2

α=√1.5

4×√1.5 = -k

k= -4√1.5

Answer 2

as we know equal roots have discriminant =0

so,b^2-4ac=0

here a=2 b=k c=3

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

k^2-24=0

k=√24

We split it into the following:

=√4⋅6

Now, we use the radical rule which states that, √ab=√a⋅√b,a,b>0.

So, we get,

=√4⋅√6

=2√6

here k=2√6...