Find the number ab, when (a+b-3)-6a=-a^2-9.



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giorgiana1976's profile pic

Posted on (Answer #1)

If we are moving the terms from the right side to the left side of the equality and if we are grouping them with the term "-6a", the result will be:

   (a+b-3) + (-6a+a^2+9)=0

The new group (-6a+a^2+9) could be written as (a-3)^2.

   (a+b-3) + (a-3)^2=0

The condition of equivalence between 2 expressions is that the equivalent terms from both expressions to be equal.

   (a+b-3)=0 and (a-3)^2=0

From (a-3)^2=0 it results that a-3=0, so a=3

By substituting the value of a into the relation  (a+b-3)=0, this one will become:

3+b-3=0, so b=0.

From both relations it results that the number ab=30.

neela's profile pic

Posted on (Answer #2)

When (a+b-3)-6a= -a^2-9. To find ab. Solution: When a is given or known: b = =-a^2-9-a+3+6a = -a^2+5a+6. So , ab = a(-a^2+5a+6),if a is known we can get ab through a. When b is known: (a+b-3)-6a=-a^2-9.Or a^2-5a+6+b=0. So a = {5+sqrt[25-4(6+b)]^(1/2)}/2 Or a={5-sqrt[25-4(6+b)]^(1/2)}/2. Therefore, ab = b{5+sqrt[25-4(6+b)]6(1/2)}/2 = b{5+sqrt[1-b)]^(1/2)}/2.Or ab = b{5-sqrt[1-b)]}/2, when b is known.
isbeatbox's profile pic

Posted on (Answer #3)




-7a=-9-b or 7a=9+b


then you do this




















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