Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Quadratic equations help

The question I have been asked to solve is:
y^2 - 5y = 3y^2 - 6y - 10

I have rearranged it to get -2y^2 + y + 10 = 0 but I don't understand how to factorise this as nothing multiplies to get 10 and adds to get 1

Original post by bulbs

The question I have been asked to solve is:
y^2 - 5y = 3y^2 - 6y - 10

I have rearranged it to get -2y^2 + y + 10 = 0 but I don't understand how to factorise this as nothing multiplies to get 10 and adds to get 1

factors of -20 that add to make 1

(edited 7 years ago)

Original post by bulbs

The question I have been asked to solve is:
y^2 - 5y = 3y^2 - 6y - 10

I have rearranged it to get -2y^2 + y + 10 = 0 but I don't understand how to factorise this as nothing multiplies to get 10 and adds to get 1

Try to get the

-2y^2

coefficient to be positive, by multiplying everything by -1. It should be easier from there.

OP

Original post by ValerieKR

factors of -20 that add to make 1

Thanks.
So -2y^2 - 4y - 5y + 10 = 0
-2y(y+2) + 5(y+2)
(-2y+5)(y+2) = 0
y = 5/2 or y = -2

Original post by bulbs

Thanks.
So -2y^2 - 4y - 5y + 10 = 0
-2y(y+2) + 5(y+2)
(-2y+5)(y+2) = 0
y = 5/2 or y = -2

You've mixed up some signs in your type up but it's right

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