How to solve y2+y-12
Answers
Answered by
10
by factorisation method
=y^2+4y-3y-12
=y(y+4)-3(y+4)
=(y+4)(y-3)
y=-4 and y=3
=y^2+4y-3y-12
=y(y+4)-3(y+4)
=(y+4)(y-3)
y=-4 and y=3
Answered by
1
it can be solved by quadratic eq
y^2+y-12=0
by factorize method remove factors of -12 such as if we multiply then it should be -12 and if we add or subtract it should be 1
that is
y^2+4y-3y-12=0
take common
y(y+4)-3(y+4)=0
we get
y-3=0 or y+4=0
i.e
y=3 or y=-4
hope this helped u
y^2+y-12=0
by factorize method remove factors of -12 such as if we multiply then it should be -12 and if we add or subtract it should be 1
that is
y^2+4y-3y-12=0
take common
y(y+4)-3(y+4)=0
we get
y-3=0 or y+4=0
i.e
y=3 or y=-4
hope this helped u
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