A b c are intezer and a2+b2=45 and b²+c²=40 then find a, b, c
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If a,b,c are positive integers such that a2+b2=45 and b2+c2=40, find the values of a,b,c
Let ^ denote an exponent...
This is what I did.
a^2 + b^2 = 45
b^2 + c^2 = 40 <-- subtract the equations to eliminate b^2...since b^2 are the same...
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a^2 - c^2 = 5 <-- factor (a^2 - c^2) as a difference of squares..
A difference of squares is a special product in the form (a^2 - b^2) and are factored into the form (a-b)(a+b)...
(a - c)(a + c) = 5
Now we have an equation where (a - c) is multiplied by (a+c) to produce a result of 5.
Since a, b and c has to be integers...(a - c) and (a + c) must be integers...and the only positive integer factors that multiply to 5 are 1 and 5...so set one equation equal to 1 and one equation equal to 5...and solve for a and c..
highly recommend you set a - c = 1 and a + c = 5...
a - c = 1
a + c = 5 <-- add the equations to eliminate "c"... since c's are opposites..
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2a = 6 <-- divide both sides of this equation by 6 to isolate a..
2a/2 = 6/2 <-- 2/2 = 1 which isolates a...
a = 3
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Now plug 3 for a in the first original equation and solve for b...
a^2 + b^2 = 45
(3)^2 + b^2 = 45
9 + b^2 = 45 <-- subtract 9 to both sides of this equation..
9 + b^2 - 9 = 45 - 9
b^2 = 36 <-- now square root both sides of this equation..
√(b^2) = √(36)
b = 6 <-- this could also be -6, but it asks for positive integers...
============
Now plug 6 for b in the second original equation and solve for c...
b^2 + c^2 = 40
(6)^2 + c^2 = 40
36 + c^2 = 40 <-- subtract 36 both sides of the equation..
36 + c^2 - 36 = 40 - 36
c^2 = 4 <-- square root both sides of the equation..
√(c^2) = √(4)
c = 2 <---- this could also be -2, but question asks for positive integers..
===============
a = 3
b = 6
c = 2
This is the answer: value of a is 3, value of b is 6, and value of c is 2...
Let ^ denote an exponent...
This is what I did.
a^2 + b^2 = 45
b^2 + c^2 = 40 <-- subtract the equations to eliminate b^2...since b^2 are the same...
---------------------
a^2 - c^2 = 5 <-- factor (a^2 - c^2) as a difference of squares..
A difference of squares is a special product in the form (a^2 - b^2) and are factored into the form (a-b)(a+b)...
(a - c)(a + c) = 5
Now we have an equation where (a - c) is multiplied by (a+c) to produce a result of 5.
Since a, b and c has to be integers...(a - c) and (a + c) must be integers...and the only positive integer factors that multiply to 5 are 1 and 5...so set one equation equal to 1 and one equation equal to 5...and solve for a and c..
highly recommend you set a - c = 1 and a + c = 5...
a - c = 1
a + c = 5 <-- add the equations to eliminate "c"... since c's are opposites..
--------------
2a = 6 <-- divide both sides of this equation by 6 to isolate a..
2a/2 = 6/2 <-- 2/2 = 1 which isolates a...
a = 3
===========
Now plug 3 for a in the first original equation and solve for b...
a^2 + b^2 = 45
(3)^2 + b^2 = 45
9 + b^2 = 45 <-- subtract 9 to both sides of this equation..
9 + b^2 - 9 = 45 - 9
b^2 = 36 <-- now square root both sides of this equation..
√(b^2) = √(36)
b = 6 <-- this could also be -6, but it asks for positive integers...
============
Now plug 6 for b in the second original equation and solve for c...
b^2 + c^2 = 40
(6)^2 + c^2 = 40
36 + c^2 = 40 <-- subtract 36 both sides of the equation..
36 + c^2 - 36 = 40 - 36
c^2 = 4 <-- square root both sides of the equation..
√(c^2) = √(4)
c = 2 <---- this could also be -2, but question asks for positive integers..
===============
a = 3
b = 6
c = 2
This is the answer: value of a is 3, value of b is 6, and value of c is 2...
Answered by
1
here is the answer explained.
answer:
a- 3
b- 6
c- 2
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