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

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lejandan | Student, Grade 10 | eNoter

Posted March 27, 2010 at 11:28 PM via web

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

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giorgiana1976 | College Teacher | Valedictorian

Posted March 28, 2010 at 12:10 AM (Answer #1)

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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.

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neela | High School Teacher | Valedictorian

Posted March 28, 2010 at 11:17 PM (Answer #2)

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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.
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isbeatbox | Student , Grade 11 | eNotes Newbie

Posted March 31, 2010 at 3:09 AM (Answer #3)

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(a+b-3)-6a=-a^2-9

-5a+b=2a-9

-7a+b=-9

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

a=b+1+2/7

then you do this

(a+b-3)-6a=-a^2-9

((b+1+2/7)+b-3)-6(b+1+2/7)=-(b+1+2/7)^2-9

2b+1+2/7-3-6b-7-5/7=2b+2+2/3-9

-4b+1+2/7-3-7-5/7=2b+2+2/3-9

-4b-6-3/8-3=2b+2+2/3-9

-4b-6-3=2b+2+2/3-9+3/8

-4b-6-3=2b+2+2-9

-6b-9=4-9

-6b-9=-5

-6b=4

b=-2/3

(a+(-2/3)-3)-6a=-a^2-9

-5a+(-2/3)-3=2a-9

-7a+(-2/3)-3=-9

-7a+(-2/3)=-6

-7a=-6+2/3

7a=6-2/3

a=6/7-2/21

a=16/21

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