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# Positive real number is four less than another. if the sum of the squares of the two numbers is 72 what are the numbers?

### 3 Answers

- AmyLv 71 month ago
Write statement as an equation

x^2 + (x+4)^2 = 72

Expand the squared term

2x^2 + 8x + 16 = 72

Combine like terms

2x^2 + 8x - 56 = 0

Divide all by 2

x^2 + 4x - 28 = 0

Plug into quadratic formula

x = [-4 ± √(4^2 + 4*1*28) ] / [2*1]

Simplify

x = -2 ± 4√2

Only the positive answer is allowed

x = -2 + 4√2

- lenpol7Lv 71 month ago
Let the numbers be 'n' & n - 4'

Squared

n = n^2 = n- 4 = ( n-4)^2

Hence the sum of their squares is

n^2 + ( n- 4)^2 = 72

Expand the brackets

n^2 + n^2 - 8n + 16 = 72

Collect 'like terms'

2n^2 - 8n = 56

Factor out '2'

n^2 - 4n = 28

This will not factor so 'Complete the Square'.

(n - 2)^2 - (-2)^2 = 28

( n- 2)^2 - 4 = 28

( n- 2)^2 = 24

Square root both sides

n - 2 = 2sqrt(6)

n = 2 +/-2sqrt(6)

n = 2 +/- 2.44948....

Hence the two numbers are

n = 4.44948...

&

n = 0.44948....

- Wayne DeguManLv 71 month ago
Letting the numbers be a and b we have:

a - 4 = b

i.e. a - b = 4...(1)

Also, a² + b² = 72...(2)

Then, (b + 4)² + b² = 72

so, 2b² + 8b - 56 = 0

or, b² + 4b - 28 = 0

Using the quadratic formula yields:

b = 3.66 or -7.66

Hence, a = 7.66 or -3.66

As the numbers are positive and real we have a = 7.66 and b = 3.66

Note: we see that 3² + 7² = 58

and 4² + 8² = 80

so, by inspection we see that a is between 7 and 8, therefore b is between 3 and 4.

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