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

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Answer 1 · 2017-04-14T11:16:34

The numbers are $8 and 10$ $8 and 10$

Let one of the integers be $x$ $x$

The other integer is then $2 x - 6$ $2x-6$

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

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

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

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

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

Set each factor equal to $0$ $0$

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

$x - 8 = 0 \to x = 8$ $x-8 = 0 " "rarr x =8$

Check: The numbers are $8 and 10$ $8 and 10$

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