Asked by Tayler
Solve the equation by completing the square
3m^2 = -7m +40
2y^2 +51 = -17v
3m^2 = -7m +40
2y^2 +51 = -17v
Answers
Answered by
Reiny
3m^2 = -7m +40
I will do the first one, you do the second
I usually have the square term and the first degree term on the left and the constant on the right
3m^2 + 7m = 40
divide each term by the coefficient of the square term, in this case by 3
m^2 + (7/3)m = 40/3
Now take 1/2 of the coefficient of the linear term, square it, then add it to both sides
1/2 of 7/3 = 7/6, after squaring we have 49/36, now add it
m^2 + (7/3)m + 49/36 = 40/3 + 49/36
we now have created a perfect square on the left
(m + 7/6)^ = 529/36
take the square root of both sides
m + 7/6 = ± √529 /6 = ± 23/6
isolate m
m = -7/6 ± 23/6
= (-7 ± 23)/6 = 16/6 or -30/6
= 8/3 or -5
I will do the first one, you do the second
I usually have the square term and the first degree term on the left and the constant on the right
3m^2 + 7m = 40
divide each term by the coefficient of the square term, in this case by 3
m^2 + (7/3)m = 40/3
Now take 1/2 of the coefficient of the linear term, square it, then add it to both sides
1/2 of 7/3 = 7/6, after squaring we have 49/36, now add it
m^2 + (7/3)m + 49/36 = 40/3 + 49/36
we now have created a perfect square on the left
(m + 7/6)^ = 529/36
take the square root of both sides
m + 7/6 = ± √529 /6 = ± 23/6
isolate m
m = -7/6 ± 23/6
= (-7 ± 23)/6 = 16/6 or -30/6
= 8/3 or -5
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