[02]
It is a quadratic equation
d^2-5d+6=0
d^2-3d-2d+6=0
d(d-3)-2(d-3)=0
(d-3)(d-2)=0
The product of two or more numbers,terms or expressions cannot be zero until and unless one of them is zero
Therefore,either d-3=0 or d-2=0
If d-3=0,then d=3
If d-2=0,then d=2
Therefore d=3 or 2
2007-10-16 03:07:53
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answer #1
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answered by alpha 7
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d^2 - 5d + 6 = 0
find factors of +6 that can add up to -5
+6&+1
or
-6&-1
or
+3&+2
or
-3&-2
in this case, -3&-2 are the only ones that could add up to -5
so (d - 3)(d - 2) = 0
d= 3, d= 2
2007-10-16 10:07:05
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answer #2
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answered by Anonymous
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d = 3
3(squared) - 5*3 + 6 = 0
9 - 15 + 6 = 0
2007-10-16 10:03:38
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answer #3
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answered by Bigfoot 3
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A quadratic equation.
d^2 - 5d + 6 = 0
Factored:
(d - 3)(d - 2) = 0
Solutions:
d - 3 = 0
d = 3
d - 2 = 0
d = 2.
2007-10-16 10:02:53
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answer #4
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answered by S. B. 6
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I believe I answered a question like this earlier - the same method works here.
What factor pairs of 6 (two numbers that multiply to 6) have a sum of -5?
The answer is -2 and -3!
So,
(d - 2) (d - 3) = 0
So the solutions to d are {2, 3}
Enjoy!
2007-10-16 10:03:00
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answer #5
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answered by zelljrc 2
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x-kate-x
Use FOIL again and factorize
d^2 -5d +6 =0 ; factorize
(d-3)(d-2) =0
d= 3 or 2.
Note when using FOIL (d*d, then -2d, -3d, then +6)
See it now?
Hope this helps.
2007-10-16 10:04:07
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answer #6
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answered by pyz01 7
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It's obvious that you knew what it was, so I don't understand why you're asking about it, unless, of course, you were just trying to get someone to do your homework for you.
2007-10-16 15:57:52
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answer #7
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answered by skaizun 6
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