Answer: The roots of the given equation are non-real.
Step-by-step explanation: We are given to use the discriminant and select whether the roots of the following quadratic equation are real or non-real :
[tex]5x^2-4x+3=0~~~~~~~~~~~~~~~~~~~~~~~`(i)[/tex]
We know that
for the quadratic equation [tex]ax^2+bx+c,~a\neq 0,[/tex] the discriminat is given by
[tex]D=b^2-4ac.[/tex]
And, the roots will be
(i) real, if D is greater than or equal to zero,
(ii) non-real, if D < 0.
For the given equation (i), we have
a = 5, b = -4 and c = 3.
So, the discriminant is given by
[tex]D=b^2-4ac=(-4)^2-4\times5\times3=16-60=-44<0.[/tex]
Therefore, the roots of the given equation are non-real.
