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Compute the sum 101^2 - 97^2 + 93^2 - 89^2.

 Feb 11, 2021

Best Answer 

 #1
avatar+151 
+2

You can use difference of squares: a- b2 = (a+b)(a-b)

[1012 - 972] + [932 - 892]

(101-97)(101+97) + (93-89)(93+89)
(4)(198) + (4)(182)

(4)(198 + 182)    [using the distributive property: ab + ac = a(b+c)]
(4)(380) = 1520

 Feb 11, 2021
 #1
avatar+151 
+2
Best Answer

You can use difference of squares: a- b2 = (a+b)(a-b)

[1012 - 972] + [932 - 892]

(101-97)(101+97) + (93-89)(93+89)
(4)(198) + (4)(182)

(4)(198 + 182)    [using the distributive property: ab + ac = a(b+c)]
(4)(380) = 1520

ChowMein Feb 11, 2021
 #2
avatar+266 
+5

Good job ChowMein...! laugh

Here is another way to do it: 

 

\(101^2 - 97^2 + 93^2 - 89^2\)

 

\(=10201-97^2+93^2-89^2\)

 

\(=10201-9409+93^2-89^2\)


\(=10201-9409+8649-89^2\)

 

\(=10201-9409+8649-7921\)

 

\(=1520\)

(ChowMien's way was much easier though)

 

cheekycheekycheeky

 Feb 12, 2021

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