Find two consecutive positive odd numbers whose squares have the sum 71 ( Quadratic Equations )
Answers
Answer:
Let one of the odd positive integer be x
then the other odd positive integer is x+2
their sum of squares
=x
2
+(x+2)
2
=x
2
+x
2
+4x+4
=2x
2
+4x+4
Given that their sum of squares = 290
2x
2
+4x+4=290
2x
2
+4x=286
2x
2
+4x−286=0
x
2
+2x−143=0
x
2
+13x−11x−143=0
x(x+13)−11(x+13)=0
(x−11)=0,(x+13)=0
Therfore ,x=11or−13
We always take positive value of x
So , x=11 and (x+2)=11+2=13
Therefore , the odd positive integers are 11 and 13
Step-by-step explanation:
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Answer:
Answer:
Answer:
Let one of the odd positive integer be x
then the other odd positive integer is x+2
their sum of squares
=x
2
+(x+2)
2
=x
2
+x
2
+4x+4
=2x
2
+4x+4
Given that their sum of squares = 290
2x
2
+4x+4=290
2x
2
+4x=286
2x
2
+4x−286=0
x
2
+2x−143=0
x
2
+13x−11x−143=0
x(x+13)−11(x+13)=0
(x−11)=0,(x+13)=0
Therfore ,x=11or−13
We always take positive value of x
So , x=11 and (x+2)=11+2=13
Therefore , the odd positive integers are 11 and 13
Step-by-step explanation:
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