How do you solve #5y^{2}+10y=-2y^{2}-8y+64#?

1 Answer
Pallu
Oct 21, 2017

#=> (7y+32)(y-2)=0#

#therefore y= -32/7 # or #y=2#

Explanation:

#5y^{2}+10y=-2y^{2}-8y+64#

Group like terms together and solve:

#5y^{2}+2y^{2}+10y+8y-64 =0#

#7y^{2}+18y-64 =0#

Find two such numbers whose sum is the coefficient of middle term (18) and product is the product of coefficient of first term and last term (7 x -64 = -448)

Finding factors of -448 to get sum +18:
-448 and 1 = -447
-224 and 2 = -222
-112 and 4 = -108
-64 and 7 = -57
-56 and 8 = -48
-32 and 14 = -18
-28 and 16 = -12
-16 and 28 = 12
-14 and 32 = 18

Two such numbers are -14 and 32

#7y^{2}-14y+32y-64 =0#

pulling out common factors:

#7y(y-2)+32(y-2) =0#

#=> (7y+32)(y-2)=0#

#=> (7y+32=0 # or #y-2=0#

#therefore y= -32/7 # or #y=2#

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