Question

Find two consecutive odd positive integers whose sum of square is 202​

Answer 1

Answer:9,11

Step-by-step explanation:

Let x and x+2 be the two consecutive odd integers

x^2+(x+2)^2=202

x^2+x^2+4x+4=202

2x^2+4x+4-202=0

2x^2+4x-198=0

divide both sides by 2

x^2+2x-99=0

(x+11)(x-9)=0

x-9=0

x=9

x+2=11

Answer 2

LET THE NUMBERS BE x , x+2

AS PER THE ABOVE SITUATION:-

x2 + (x+2)2 = 202

x2 + x2 + 4x + 4 =202

2x2 + 4x = 198

DIVIDED BY 2

x2 + 2x - 99 = 0

BY FACTORISATION METHOD SPLITTING THE MIDDLE TERM

x2 +11x -9x -99 = 0

x(x + 11) -9 (x+11) =0

(x -9)(x+11) = 0

SO x can be either 9 or -11 (WE CANNOT USE -11 BECAUSE NUMBERS CANNOT BE NEGATIVE)

SO THE NUMBERS ARE : -

x = 9

x+2=9 +2=11

CROSS VERIFICATION:-

x2 + (x+2)2 = 202

9*2 + 11*2 =202

81 + 121 =202

202=202

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