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ashokhari8782
23.06.2018
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Y²+1/3y=2 Solve using formula.
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abhi178
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y² + 1/3 y = 2
first of all resolve it in general form ay² + by + c = 0 .
e.g., y² + 1/3 y = 2
=> 3y² + y = 2 × 3
=> 3y² + y - 6 = 0
if ay² + by + c = 0
then, y = {-b ± √(b² - 4ac)}/2a use it here,
3y² + y - 6 = 0
so, y = {-1 ± √(1² - 4.3(-6))}/2(3)
= {-1 ± √(1 + 72)}/6
= (-1 ± √73)/6
hence, roots are (-1 + √73)/6 and (-1 - √73)/6
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rohitkumargupta
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given equation is ,
Y² + 1/3y = 2
now change in its general form ax² + bx + c = 0
so, it will become "3y² + y - 6 = 0
Where, a = 3 , b = 1 , c = -6
now we know the quadratic foumula for finding the value of roots.
Y = [ -b ]/2a
Y = [ -1 ]/2(3)
Y = [-1 ]/6
hence, the roots are, Y = (-1 + )/6
Y = (1 - )/6
Step-by-step explanation:
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abhi178
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y² + 1/3 y = 2
first of all resolve it in general form ay² + by + c = 0 .
e.g., y² + 1/3 y = 2
=> 3y² + y = 2 × 3
=> 3y² + y - 6 = 0
if ay² + by + c = 0
then, y = {-b ± √(b² - 4ac)}/2a use it here,
3y² + y - 6 = 0
so, y = {-1 ± √(1² - 4.3(-6))}/2(3)
= {-1 ± √(1 + 72)}/6
= (-1 ± √73)/6
hence, roots are (-1 + √73)/6 and (-1 - √73)/6
Answer:
Step-by-step explanation:
y²+(1/3)*y=2
Multipying by 3 we get
3y²+y-6=0
Here a=3,b=1 and c=-6
Thus x={ -1±√(1-4*(3)(-6) } 2*3
=1/6 ( -1±√(1+72)
x=(-1±√(73) )/6