Question

find the values of k for which the quardatic equation 3x^2+kx-3 =0 has real and equal roots

Answer 1

given
a=3 ,b=k ,c=-3
for real and equal root
D=0
D = b^2 -4ac= 0.
k^2-4(3)(-3)=0
k^2+36=0
k^2 =-36
k=√-36
k =6
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Answer 2

using discriminant formula

b^2-4ac=0

k^2+36=0

k^2= -36

square of any value cannot be -ve.

therefoe,

k^2= 36

k=+or- 36

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