Sumeet D.

asked • 07/24/18

Maths Polynomials-2

Evaluate (997-w)^2 +(993-w)^2 if (997-w)(993-w)=21

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Paul M. answered • 07/24/18

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Write (997-w)(993-w) = 21 as (1000-w-3)(1000-w-7)=21
(1000-w)2 - 10(100-w) +21 = 21
Subtract 21 from each side and then divide out (1000-w)
1000 - w -10 = 0 and w=990
 
Then the initial pair of squares just becomes 72 + 32 = 58
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Sumeet D.

When I was on my fifth try I realised that a2 +b2 = (a-b)2 + 2ab. So, I substituted the values of a for 997-w and b for 993-w into the expression (a-b)2 + 2ab and got The answer 46. Isn’t this also correct?
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07/24/18

Paul M.

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If you set a=997-w & b = 993-w, you get a-b = 4. Then using the identity, you get (a^2) + (b^2) =
16 + 2(997-w)(993-w) = 16 + 42 = 58 as before. Your idea is a good one, but I think you made an error somewhere. Since you didn’t show your work, I can’t help you find an error. If you can’t find your error, show me your work so that I can help you find it. However: Good Thinking! Nice work!
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07/24/18

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