Question

the sum of the squares of 2 positive integers is 208. if the square of larger number is 18 times the smaller , find the numbers

Answer 1

Let the nos. be x and y

∴x²+y²=208 (i)

∴x²=18y (ii)

Now, putting the value of x² in (i)

18y+y²=208

⇒y²+18y-208=0

⇒y=-18+-(324+832)^(1/2)/2

⇒y=-18+-(1156)^(1/2)/2

⇒y=-18+-34/2

⇒y=-18+34/2 or -18-34/2

⇒y=16/2 or -52/2

⇒y=8 or -26

Putting the value of y=8 in (i)

x²+8²=208

⇒x²+64=208

⇒x²=208-64

⇒x²=144

⇒x=12

Answer 2

Sum of 2 positive integer= 208

Let smaller no= X

Let larger no= 18x

= X2+18x=208

= X2 +18x- 208=0

X2 +26x-6x-208=0

X( x+26)8( x+26)

(X+26) ( x+8)

X- 26= 0 , x+8= 0

Thus a square can't be negative

So x-26= 0(Ignored)

Thus x= 8 &

Larger no= 18(8)

✓ 144= 12

Thus the numbers are 8& 12

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