Question 997519

a quadratic equation in {{{x}}} whose roots are the {{{squares}}} of the roots of 
{{{2x^2-8x+1=0}}}

find the roots of

{{{2x^2-8x+1=0}}}............use quadratic formula

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

{{{x = (-(-8) +- sqrt( (-8)^2-4*2*1 ))/(2*2) }}}

{{{x = (8 +- sqrt(64-8 ))/4 }}}

{{{x = (8 +- sqrt(56 ))/4 }}}

{{{x = (8 +- sqrt(4*14 ))/4 }}}

{{{x = (cross(8)4 +- cross(2)sqrt(14 ))/cross(4)2 }}}

{{{x = (4 +- sqrt(14 ))/2 }}}

so, exact roots are:

{{{x = (4 + sqrt(14 ))/2 }}}

{{{x = (4 - sqrt(14 ))/2 }}}

approximate:

{{{x[1] = (4 + 3.741657386773941)/2 =7.741657386773941/2=3.87}}}

{{{x[2] = (4 - 3.741657386773941)/2= 0.2583426132260586/2=0.13}}}


now find their squares:

{{{x[1]^2 = 14.98}}}

{{{x[2]^2 =0.0169}}}


now use zero product formula to find your equation:

{{{f(x)=(x-x[1])(x-x[2])}}}

{{{f(x)=(x-14.98)(x-0.0169)}}}

{{{f(x)=x^2-0.0169x-14.98x-14.98*(-0.0169)}}}

{{{f(x)=x^2-14.9969x+0.253162}}}