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3 years ago

(4y^2-3y+2)+(5y+12y-4)-(13y^2-6y+20)

Mathematics

1 answer:

Andrews [41]3 years ago

4 0

Answer:

-(9 y^2 - 20 y + 22) or if you need y = 10/9 + (7 i sqrt(2))/9 or y = 10/9 - (7 i sqrt(2))/9

Step-by-step explanation:

Simplify the following:

-(13 y^2 - 6 y + 20) + 4 y^2 + 12 y + 5 y - 3 y - 4 + 2

Grouping like terms, -(13 y^2 - 6 y + 20) + 4 y^2 + 12 y + 5 y - 3 y - 4 + 2 = -(13 y^2 - 6 y + 20) + 4 y^2 + (-3 y + 5 y + 12 y) + (2 - 4):

-(13 y^2 - 6 y + 20) + 4 y^2 + (-3 y + 5 y + 12 y) + (2 - 4)

-3 y + 5 y + 12 y = 14 y:

-(13 y^2 - 6 y + 20) + 4 y^2 + 14 y + (2 - 4)

2 - 4 = -2:

-(13 y^2 - 6 y + 20) + 4 y^2 + 14 y + -2

-(13 y^2 - 6 y + 20) = -13 y^2 + 6 y - 20:

-13 y^2 + 6 y - 20 + 4 y^2 + 14 y - 2

Grouping like terms, 4 y^2 - 13 y^2 + 14 y + 6 y - 20 - 2 = (-13 y^2 + 4 y^2) + (6 y + 14 y) + (-20 - 2):

(-13 y^2 + 4 y^2) + (6 y + 14 y) + (-20 - 2)

4 y^2 - 13 y^2 = -9 y^2:

-9 y^2 + (6 y + 14 y) + (-20 - 2)

6 y + 14 y = 20 y:

-9 y^2 + 20 y + (-20 - 2)

-20 - 2 = -22:

-9 y^2 + 20 y + -22

Factor -1 out of -9 y^2 + 20 y - 22:

Answer: -(9 y^2 - 20 y + 22)

______________________________________________

Solve for y:

-9 y^2 + 20 y - 22 = 0

Divide both sides by -9:

y^2 - (20 y)/9 + 22/9 = 0

Subtract 22/9 from both sides:

y^2 - (20 y)/9 = -22/9

Add 100/81 to both sides:

y^2 - (20 y)/9 + 100/81 = -98/81

Write the left hand side as a square:

(y - 10/9)^2 = -98/81

Take the square root of both sides:

y - 10/9 = (7 i sqrt(2))/9 or y - 10/9 = -(7 i sqrt(2))/9

Add 10/9 to both sides:

y = 10/9 + (7 i sqrt(2))/9 or y - 10/9 = -(7 i sqrt(2))/9

Add 10/9 to both sides:

Answer: y = 10/9 + (7 i sqrt(2))/9 or y = 10/9 - (7 i sqrt(2))/9

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