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

Apr 14, 2017

The numbers are $8 \mathmr{and} 10$

#### Explanation:

Let one of the integers be $x$

The other integer is then $2 x - 6$

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

${x}^{2} + {\left(2 x - 6\right)}^{2} = 164$

${x}^{2} + 4 {x}^{2} - 24 x + 36 = 164 \text{ } \leftarrow$ make = 0

$5 {x}^{2} - 24 x - 128 = 0 \text{ } \leftarrow$ find factors

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

Set each factor equal to $0$

$5 x + 16 = 0 \text{ "rarr x = -16/5" }$ reject as a solution

$x - 8 = 0 \text{ } \rightarrow x = 8$

Check: The numbers are $8 \mathmr{and} 10$

${8}^{2} + 102 = 64 + 100 = 164$