Question 772758
If one of the roots of the quadratic equation {{{x^2-(2k-5)x+k^2-5k+56/9=0}}} is twice the other,show that one of the roots is {{{(2k-5)/3}}}.

 {{{((2k-5)/3)(2((2k-5)/3))=0}}}

 {{{(2k-5)(2(2k-5))=0}}}

 {{{(2k-5)(4k-10)=0}}}

{{{8k^2-20k-20k+50=0}}}

{{{8k^2-40k+50=0}}}

{{{k = (-(-40) +- sqrt( (-40)^2-4*8*50 ))/(2*8) }}}

{{{k = (40 +- sqrt(1600-1600 ))/16 }}}

{{{k = (40 +- sqrt(0 ))/16 }}}

{{{k = 40 /16 }}}

{{{k = 5 /2 }}}.... the values of the constant {{{k}}}

so, plug it in given equation

{{{x^2-(2k-5)x+k^2-5k+56/9=0}}}

=> {{{x^2-(0)x+(5/2)^2-5(5/2)+56/9=0}}}

=> {{{x^2+25/4-25/2+56/9=0}}}

=> {{{x^2+25/4-50/4+56/9=0}}}

=> {{{x^2-25/4+56/9=0}}}....{{{-(25*9)/(4*9)+(56*4)/(9*4)=-225/36+216/36=-1/36}}}

=> {{{x^2-1/36=0}}}

roots:

{{{(2k-5)/3}}}=>{{{(2(5 /2)-5)/3=(5-5)/3=0}}}

{{{2(2k-5)/3}}}=>{{{2(2(5 /2)-5)/3=2(5-5)/3=0}}}

{{{ graph(600, 600, -10, 10, -10, 10, x^2-1/36) }}}