Question 729484
given:

The sum of the squares of the solution {{{x^2+2kx-5=0}}} is {{{26}}}. 

Find: the value of {{{k}}}

first find the solution {{{x^2+2kx-5=0}}} using quadratic formula

{{{x^2+2kx-5=0}}}.....note that coefficient {{{a=1}}}, {{{b=2k}}} and {{{c=-5}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...plug in coefficients


{{{x = (-(2k) +- sqrt( (2k)^2-4*1*(-5) ))/(2*1) }}}


{{{x = (-2k +- sqrt( 4k^2+20 ))/2 }}}


{{{x = (-2k +- sqrt( 4(k^2+5) ))/2 }}}


{{{x = (-2k +- 2sqrt( (k^2+5) ))/2 }}}


{{{x = (-cross(2)k +- cross(2)sqrt( (k^2+5) ))/cross(2) }}}


{{{x = (-k +- sqrt( (k^2+5) )) }}}

solutions involving {{{k}}}:

{{{x = -k + sqrt(k^2+5) }}} or {{{x = -k - sqrt(k^2+5) }}}

since given the sum of the squares of the solution is {{{26}}}, we have


{{{ (-k + sqrt(k^2+5))^2+(-k -sqrt(k^2+5 ))^2=26}}}


{{{ (-k)^2 -2sqrt(k^2+5)+ (sqrt(k^2+5))^2+(-k)^2 +2sqrt(k^2+5)+ (sqrt( k^2+5))^2=26}}}

{{{ k^2 -cross(2sqrt(k^2+5))+ k^2+5+k^2 +cross(2sqrt(k^2+5))+k^2+5=26}}}


{{{ k^2 + k^2+5+k^2 +k^2+5=26}}}

{{{ 4k^2 +10=26}}}

{{{ 4k^2 =26-10}}}

{{{ 4k^2 =16}}}

{{{ k^2 =16/4}}}

{{{ k^2 =4}}}

{{{ k =sqrt(4)}}}

{{{ k =2}}} or {{{ k =-2}}}


check:

{{{x^2+2kx-5=0}}}...plug in value for {{{k}}}

if {{{ k =2}}}

{{{x^2+2*2x-5=0}}}


{{{x^2+4x-5=0}}}........find solutions

{{{x = (-4 +- sqrt( 4^2-4*1*(-5) ))/(2*1) }}}

{{{x = (-4 +- sqrt( 16+20 ))/2 }}}

{{{x = (-4 +- sqrt( 36 ))/2 }}}

{{{x = (-4 +- 6)/2 }}}

solutions:

{{{x = (-4 + 6)/2 }}}

{{{x = 2/2 }}}

{{{x =1 }}}

and

{{{x = (-4 - 6)/2 }}}

{{{x = -10/2 }}}

{{{x =-5 }}}

if we do second solution above we will get {{{x =-1 }}} and {{{x =5 }}}

so, solutions of the {{{x^2+2kx-5=0}}} are:

{{{x =1 }}} and {{{x =-5 }}}
or
{{{x =-1 }}} and {{{x =5 }}}

now check if the sum of the squares of the solution {{{x^2+2kx-5=0}}} is {{{26}}}:

{{{1^2+(-5)^2=26 }}}

{{{1+25=26 }}}

{{{26=26 }}}......true

or

{{{(-1)^2+5^2=26 }}}

{{{1+25=26 }}}

{{{26=26 }}}......true