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Help with quadratic equations please :D?
How would I solve these, I have to solve each by graphing./
m^2+3m=28
3 Answers
- TomVLv 71 decade agoFavorite Answer
Draw a graph of the function y = x² + 3x - 28
Where the graph crosses the x axis, the x-intercepts, are the zeros of the function, or the values which satisfy the relation m² +3m = 28
since m²+3m-28 can be factored to (m+7)(m-4), the intercepts, zeros, or solutions will be -7 and 4.
- Anonymous1 decade ago
First move over the 28 to the left side of the equation.
m^2+3m-28=0.
Then factor.
(m+7)(m-4)=0
Why did I pick 7 and -4?
Because 7 times -4 = -28.
7+(-4) = 3.
If you multiple (m+7)(m-4) this out, you will get m^2+3m-28 which is what we want b/c the two equations are equal.
Now just solve them independently.
m+7 = 0
m-4 = 0
m= -7, 4
:)
- Anonymous1 decade ago
m=3
