Question 885979
Let m and n be the roots of 3x^2-10x+3=0
Let p and q be the roots of the new quadratic equation(ax^2+bx+c)
Then 3m=p, 3n=q

m+n={{{-b/2a}}}={{{-(-10)/2(3)}}}={{{10/6}}}={{{5/3}}}
mn= {{{c/a}}}={{{3/3}}}={{{1}}}

so,
p+q = 3m+3n = 3(m+n) = 3(5/3) = 5
pq = (3m)(3n) = 9mn = 9(1) = 9

Since the new {{{-b/2a}}} = 5 and the new {{{c/a}}} = 9
Then a=1 , b=-10 , c=9

so {{{ax^2+bx+c}}} = {{{1(x^2)+(-10)x+9}}} = {{{x^2-10x+9}}}