Question 539168
{{{ a + b = 10 }}}
{{{ a^2 + b^2 = 82 }}}
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{{{ b = 10 - a }}}
By substitution:
{{{ a^2 + ( 10 - a )^2 = 82 }}}
{{{ a^2 + 100 - 20a + a^2 = 82 }}}
{{{ 2a^2 - 20a + 100 = 82 }}}
{{{ 2a^2 - 20a + 18 = 0 }}}
{{{ a^2 - 10a + 9 = 0 }}}
complete the square
{{{ a^2 - 10a = -9 }}}
{{{a^2 - 10a + (-10/2)^2 = -9 + (-10/2)^2 }}}
{{{ a^2 - 10a + 25 = 16 }}}
{{{ ( a - 5 )^2 = 4^2 }}}
Take the square root of both sides
{{{ a - 5 = 4 }}}
{{{ a = 9 }}}
It is also true that
{{{ a - 5 = -4 }}}
{{{ a = 1 }}}
If I say that {{{ a = 9 }}}, then
{{{ a + b = 10 }}}
{{{ b = 1 }}}
The integers are 9 and 1
check:
{{{ 9^2 + 1^2 = 82 }}}
{{{ 81 + 1 = 82 }}}
{{{ 82 = 82 }}}
OK