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 $$14.$$MathematicsRS AgarwalStandard X

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