Question

solve the following quadratic equation :
1) 5t^2-4t-2

2)x^2-4x+1

3)x^2-6x+2
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Answer 1

Answer:

1) 5t²-4t-2=0

comparing the above equation with

ax²+bx+c=0..... (Standard form of the quadratic equation)

therefore,

a=5 , b=-4 and c=-2

b²-4ac=(-4) ²×5×-2

=16×-10

=-160

therefore, by formula method,

t=-b²±√b²-4ac/2a

=16±√-160/10

therefore, t=16+√-160/10 OR

t=16-√-160/10

2) x²-4x+1=0

heeet also comparing the above equation we get, a =1 , b=-4 and c=1

b²-4ac=(-4) ²×1× 1

=16

therefore, by formula method,

x=-b±√b²-4ac/2a

=16±4/2

x=16+4/2 or x=16-4/2

=20/2 =12/2

=10 =6

3) x²-6x+2=0

here,

a=1 , b=-6 and c=2

b²-4ac=(-6) ²×1×2

=72

x=-b±√b²-4ac/2a

=6+6√2/2 or 6-6√2/2