+0  
 
0
72
6
avatar+73 

Three consecutive positive odd integers $a,b$ and $c$ satisfy $b^{2}-a^{2}=344$ and $c^{2}-b^{2}>0.$ What is the value of $c^{2}-b{^2}$?

 Nov 18, 2020
 #1
avatar+129 
-1

Using algebra, c^2 - b^2 = 348.

 Nov 18, 2020
 #2
avatar+73 
0

No explanation? Just "use algebra"? Thanks for trying.. but I need a good explanation I'm not here just for a quick answer.

MathzSolver111  Nov 18, 2020
 #3
avatar+9198 
0

b2 - a2 > 0   so  b2 > a2  and so since a, b are positive,  b > a

c2 - b2 > 0   so  c2 > b2  and so since c, b are positive,   c > b

 

So   a < b < c
 

And since a, b, and c are consecutive positive odd integers:

 

b  =  a + 2

c  =  a + 4

 

Now we can find  a  using this equation:

 

b2 - a2  =  344

(a + 2)2  -  a2  =  344

(a + 2)(a + 2) - a2  =  344

a2 + 4a + 4 - a2  =  344

4a + 4  =  344

4a  =  340

a  =  85

 

Since  a = 85,  b = 87  and  c = 89

 

And so   c2 - b2  =  892 - 872  =  352

 Nov 18, 2020
 #4
avatar+73 
+1

Thanks hectictar! Also, happy $21$st and $22$nd birthday (sorry I couldn't say this at your real birthday I joined two weeks ago).

MathzSolver111  Nov 18, 2020
edited by MathzSolver111  Nov 18, 2020
 #5
avatar+9198 
0

Thanks! xD

hectictar  Nov 18, 2020
 #6
avatar+73 
+1

The answer was correct and the exact explanation for how to solve after I put in $352$ was almost exactly what you said! Thank you a million!

MathzSolver111  Nov 18, 2020

24 Online Users

avatar