Help quick! Don’t have that much left
2 answers:
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This is a polynomial of degree 2, and remember that complex roots always come in pairs; therefore, in a second degree polynomial you can only have 0 real roots and 2 complex roots, or 2 real roots and 0 complex roots. There is no way you can have 1 real root and 1 complex root because, as stated before, complex (non-real) roots always come in pairs; so the only way of having 1 real root is dumping the polynomial of degree 2 (
) by changing the 14 to 0. That way we'll left with 
which only has one real root:
3:5 ratio mean there are 8 "parts" (3 + 5 = 8)
The distance between the x-coordinates is |-3 - 5| = 8.
So each "part" is 8/8 = 1 unit long in the x-direction.
You want I to be 3(1) = 3 units from D, so the x-coordinate of I is -3 + 3 = 0.
Same deal for y.
|2 - 5| = 3 is the distance between D and E
Each part is 3/8.
3(3/8) = 9/8
2 + 9/8 = 25/8
So the point I is (0, 25/8)
The distance between the x-coordinates is |-3 - 5| = 8.
So each "part" is 8/8 = 1 unit long in the x-direction.
You want I to be 3(1) = 3 units from D, so the x-coordinate of I is -3 + 3 = 0.
Same deal for y.
|2 - 5| = 3 is the distance between D and E
Each part is 3/8.
3(3/8) = 9/8
2 + 9/8 = 25/8
So the point I is (0, 25/8)