Question 1196712
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I think you meant to say 3x^2-4x-7 = 0 instead of 3x^3-4x-7 = 0


Compare 3x^2-4x-7 = 0 to ax^2+bx+c=0
to find that
a = 3, b = -4, c = -7
These a,b values are NOT the roots mentioned
I wish your teacher used p,q for the roots (or anything else except for a,b)


Anyways this is what the steps look like for the quadratic formula.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-4)+-sqrt((-4)^2-4(3)(-7)))/(2(3))}}}


{{{x = (4+-sqrt(100))/(6)}}}


{{{x = (4+-  10)/(6)}}}


{{{x = (4+10)/(6)}}} or {{{x = (4-10)/(6)}}}


{{{x = 14/6}}} or  {{{x = -6/6}}}


{{{x = 7/3}}} or  {{{x = -1}}}
The two roots are p = 7/3 and q = -1
The order of the roots doesn't matter.


We're asked to compute {{{(p^2)/(q) + (q^2)/p}}} where p,q are the roots of 3x^2-4x-7 = 0


{{{(p^2)/(q) + (q^2)/p}}}


{{{((7/3)^2)/(-1) + ((-1)^2)/(7/3)}}}


{{{((49/9))/(-1) + (1)/(7/3)}}}


{{{-49/9 + 3/7}}}


{{{(-49*7)/(9*7) + (9*3)/(9*7)}}}


{{{(-343)/(63) + (27)/(63)}}}


{{{(-343+27)/(63)}}}


{{{-316/63}}}


Therefore, 
{{{(p^2)/(q) + (q^2)/p = -316/63}}}
when p = 7/3 and q = -1, which are the roots of 3x^2-4x-7 = 0
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