Question

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

Answer 1

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

Answer 2

Step-by-step explanation:

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