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balu736 [363]
2 years ago
11

How do you solve for (w-2)²-63=0 where w is a real number

Mathematics
1 answer:
Murrr4er [49]2 years ago
4 0
(w-2)^2 = 63

FOIL left hand side:
(w-2)^2 = w^2 - 4w + 4

So,
w^2 - 4w + 4 = 63
w^2 - 4w - 59 = 0

Then use quadratic equation to get values of w = 2 + 3*root(7) and w = 2 - 3*root(7).
±
3
√
7
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Possible cases for sum 3: {(1,2), (2,1)}  Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)}  Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)}  Count 4

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Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

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