How do you solve y^2+3/2y=10y2+32y=10?

2 Answers
Jul 6, 2018

color(violet)(y = -4, 5/2y=4,52

Explanation:

y^2 + (3/2) y = 10y2+(32)y=10

2y^2 + 3x = 202y2+3x=20

2y^2 + 3x - 20 = 02y2+3x20=0

2y^2 + 8y - 5y - 20 = 02y2+8y5y20=0

2y (y + 4) - 5(y +4) = 02y(y+4)5(y+4)=0

(y+4) * (2y-5) = 0(y+4)(2y5)=0

y = -4, 5/2y=4,52

Jul 6, 2018

y=-4" or "y=5/2y=4 or y=52

Explanation:

"subtract 10 from both sides"subtract 10 from both sides

y^2+3/2y-10=0y2+32y10=0

"multiply through by 2"multiply through by 2

2y^2+3y-20=02y2+3y20=0

"factor the quadratic using the a-c method"factor the quadratic using the a-c method

"the factors of the product "2xx-20=-40the factors of the product 2×20=40

"which sum to "+3" are "+8" and "-5which sum to +3 are +8 and 5

"split the middle term using these factors"split the middle term using these factors

2y^2+8y-5y-20=0larrcolor(blue)"factor by grouping"2y2+8y5y20=0factor by grouping

color(red)(2y)(y+4)color(red)(-5)(y+4)=02y(y+4)5(y+4)=0

"take out the "color(blue)"common factor "(y+4)take out the common factor (y+4)

(y+4)(color(red)(2y-5))=0(y+4)(2y5)=0

"equate each factor to zero and solve for y"equate each factor to zero and solve for y

y+4=0rArry=-4y+4=0y=4

2y-5=0rArry=5/22y5=0y=52