Find two consecutive positive integers, the sum of whose squares is 365.
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Find two consecutive positive integers, the sum of whose squares is 365.
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Answer:
Let first integer =X
Second integer =X+1
Also given that
Sum of integers =365
(First number )^2 +(Second number )^2 =365
x^2+(X+1)^2=365
x^2+x^2+1^2+2 x X x1=365
2x^2+1+2x=365
2x^2+2x+1-365=0
2x^2+2x-365=0
2(x^2+x-182)=0
x^2+x-182=0/2
x^2+x-182=0
Now factorise by splitting middle term method
x^2+14x-13x-182=0
x(x+14)-13(x+14)=0
(x-13)(X+14)=0
x-13=0. | x+14=0
X=13. | X=-14
So the roots of equation are X=13 and X=14
Since we have to find consecutive positive numbers=x=13
First number =X=13
Second number=X+1=13+1=14
Hope it helps
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