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Question

Find two consecutive positive integers, sum of whose squares is $$365$$. 


Solution

Let the first number is $$x,$$ then the next number will be $$x+1.$$
Then,
$${ x }^{ 2 }+({ x+1) }^{ 2 }=365$$     {Given}
$$\Rightarrow { x }^{ 2 }+{ x }^{ 2 }+2x+1=365$$
$$\Rightarrow 2{ x }^{ 2 }+2x-364=0$$
$$\Rightarrow { x }^{ 2 }+x-182=0$$
$$\Rightarrow { x }^{ 2 }+14x-13x-182=0$$
$$\Rightarrow x(x+14)-13(x+14)=0$$
$$\Rightarrow (x-13)(x+14)=0$$
$$\Rightarrow x=13,-14$$
Considering positive number, the numbers are $$13$$ and $$14.$$

Mathematics
RS Agarwal
Standard X

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