Question 1122815
In the quadratic formula, for repeated solutions
(1) {{{ b^2 - 4*a*c = 0 }}}
and
(2) {{{ x = -b/( 2a ) }}}
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{{{ k*x^2 + x + 100k = 0 }}}
{{{ b = 1 }}}
{{{ a = k }}}
{{{ c = 100k }}}
(1) {{{ 1^2 - 4*k*100k = 0 }}}
(1) {{{ 1 - 400k^2 = 0 }}}
(1) {{{ k^2 = 1/400 }}}
{{{ k = 1/20 }}}
or
{{{ k = -1/20 }}}
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The equation is
{{{ (1/20)*x^2 + x + 100*( 1/20) }}}
{{{ (1/20)*x^2 + x + 5 = 0 }}}
Check both solutions with quadratic equation