Question

find the zeros of the polynomial
$3y { } ^{2} - 3 \sqrt{2 } y + 1$
​

Answer 1

Answer:

Step-by-step explanation:

3y2−3(2

1

2

)y+1=0

3y2−4.242641y+1=0

For this equation: a=3, b=-4.242640687119286, c=1

3y2+−4.242641y+1=0

Step 1: Use quadratic formula with a=3, b=-4.242640687119286, c=1.

y=

−b±√b2−4ac

2a

y=

−(−4.242641)±√(−4.242641)2−4(3)(1)

2(3)

y=

4.242640687119286±√6

6

y=1.1153550716504108,0.2988584907226845

Answer:

y=1.1153550716504108,0.2988584907226845

Answer 2

Answer:

answer for the given problem is given