Question

find the nature of roots of the quadratic equation
5x²-4x+1=0
please help me friend​

Answer 1

Hii friend..

Your answer is below :

Equation : $5x^2-4x+1=0$

$D=b^2-4ac\\a=5\\b=-4\\c=1\\D=(-4)^2-4*5*1\\D=16-20\\D=-4$

As, discriminant is negative which is smaller than zero, hence, it has no real roots.

Hope this helps you..  :D

Answer 2

Answer:

Given quadratic equation is

5x²-4x+1=0

comparing the above equation with ax² + bx+c=0 we get,

a=5, b=-4 and c=1

therefore, b²-4ac=(-4) ²×-4×5×1

=-4

therefore by using formula,

x=-b±√b²-4ac/2a

=4±√-4/10

=4±2/10

therefore, 4+2/10 or 4-2/10

6/10 or 2/10

3/5 or 1/5

3/5=0.6 and 1/5=0.2

here the roots are greater than 0

therefore the roots of the given quadratic equation are real and unequal.