Find two consecutive odd integers if the sum of their squares is 202
priyanshipaul:
11 is the answer
We have,
Let the two odd consecutive numbers xandx+2
So,
According to given question,
x
2
+(x+2)
2
=202
⇒x
2
+x
2
+4+4x=202
⇒2x
2
+4x=198
⇒x
2
+2x=99
⇒x
2
+2x−99=0
⇒x
2
+(11−9)x−99=0
⇒x
2
+11x−9x−99=0
⇒x(x+11)−9(x+11)=0
⇒(x+11)x−9=0
If,
x+11=0
⇒x=−11
If,
x−9=0
x=9
So,
x+2=9+2=11
Hence, this is the answer.
Let the two odd consecutive numbers xandx+2
So,
According to given question,
x
2
+(x+2)
2
=202
⇒x
2
+x
2
+4+4x=202
⇒2x
2
+4x=198
⇒x
2
+2x=99
⇒x
2
+2x−99=0
⇒x
2
+(11−9)x−99=0
⇒x
2
+11x−9x−99=0
⇒x(x+11)−9(x+11)=0
⇒(x+11)x−9=0
If,
x+11=0
⇒x=−11
If,
x−9=0
x=9
So,
x+2=9+2=11
Hence, this is the answer.
Answers
Answered by
3
Step-by-step explanation:
Find two consecutive odd integers if the sum of their squares is 202
Answered by
1
Answer:
We have,
Let the two odd consecutive numbers xandx+2
So,
According to given question,
x
2
+(x+2)
2
=202
⇒x
2
+x
2
+4+4x=202
⇒2x
2
+4x=198
⇒x
2
+2x=99
⇒x
2
+2x−99=0
⇒x
2
+(11−9)x−99=0
⇒x
2
+11x−9x−99=0
⇒x(x+11)−9(x+11)=0
⇒(x+11)x−9=0
If,
x+11=0
⇒x=−11
If,
x−9=0
x=9
So,
x+2=9+2=11
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