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One positive integer is 6 less than twice another. The sum of their squares is 164. How do you find the integers?

Answer 1

Genesis Allen

The numbers are #8 and 10#

Let one of the integers be #x#

The other integer is then #2x-6#

The sum of their squares is #164#: Write an equation:

#x^2 + (2x-6)^2 =164#

#x^2 + 4x^2 -24x+36 = 164" "larr# make = 0#

#5x^2 -24x -128 =0" "larr# find factors

#(5x+16)(x-8=0#

Set each factor equal to #0#

#5x+16 = 0 " "rarr x = -16/5" "# reject as a solution

#x-8 = 0 " "rarr x =8#

Check: The numbers are #8 and 10#

#8^2 +102 = 64 +100 = 164#

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Answer 2

Ethan Adkins

The integers are 5 and 11.

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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.