31. The sum of squares of two numbers is 80 and
the square of their difference is 36. The product of
the two numbers is<

2 thoughts on “31. The sum of squares of two numbers is 80 and<br />the square of their difference is 36. The product of<br />the two numbers is<”

1. Step-by-step explanation:

   The answer is 22.

   Let the two numbers be x, and y.

   The conditions given are:

   The sum of squares of two numbers is 80.

   x²+y²=80

   The square of difference between the two numbers is 36.

   (x-y)²=36

   x²-2xy+y²=36

   Take the second condition, and derive a value for x².

   x²-2xy+2xy+y²-y²=36+2xy-y²

   x²=-y²+2xy+36

   Replace x² in the first condition with the derived value.

   x²+y²=80

   (-y²+2xy+36)+y²=80

   y²-y²+2xy+36=80

   2xy+36–36=80–36

   2xy÷2=44÷2

   xy=22

   Thus the product of the two numbers (x,y) is 22.

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2. Answer:

   a)22

   Step-by-step explanation:

   we have,2 numbers be x and y

   first condition

   sum of squares of 2 numbers is 80

   [tex]x {}^{2} + y {}^{2} = 80[/tex]

   square difference of 2 numbers is 36

   [tex](x – y) {}^{2} = 36 \\ x { }^{2} -2 xy + y {}^{2} { = 36}[/tex]

   In Second condition, take derive a value for

   [tex]x {}^{2} [/tex]

   [tex]x { }^{2} { – 2xy} + 2xy + y {}^{2} – y {}^{2} = 36 + 2xy – y {}^{2} [/tex]

   Replace

   [tex]x {}^{2} [/tex]

   in first condition with derived value

   [tex]x {}^{2} + y {}^{2} = 80 \\ ( – y {}^{2} + 2xy + 36) + y {}^{2} = 80 \\ y {}^{2} – y {}^{2} + 2xy + 36 = 80 \\ 2xy + 36 – 36 = 80 – 36 \\ 2xy \div 2 = 44 \div 2 \\ xy = 22 \\ [/tex]

   PRODUCT OF 2 NUMBERS=

   [tex](x.y) = 22[/tex]

   Reply