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The sum of the squares of two positive numbers is 53 and the difference of the squares of the numbers is 45. How do you find the number?

Answer 1

Hazel Anglin

#7,2#

Let the two numbers be #a and b#, repectively.

#=> a^2+b^2=53 ---- (1)# #=> a^2-b^2=45 ----(2)#

#(1)+(2) => 2a^2=98# #=> a^2=49# #=> a=+-7# #=> b=+-2#

Given that #a# and #b# are both positive, hence, #a=7, b=2#

Check :

#7^2+2^2=49+4=53# #7^2-2^2=49-4=45#

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Answer 2

Eli Assouline

Let (x) and (y) be the two positive numbers.

Given:

(x^2 + y^2 = 53)
(x^2 - y^2 = 45)

Adding equation 1 and equation 2, we get: [2x^2 = 98] [x^2 = 49] [x = 7]

Substituting (x = 7) into equation 1, we get: [7^2 + y^2 = 53] [49 + y^2 = 53] [y^2 = 4] [y = 2]

So, the two positive numbers are 7 and 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.