Question 1209228
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Let a and b be the roots of the quadratic 2x^2 - 8x + 7 = -3x^2 + 15x + 11.  
Compute 1/a^2 + 1/b^2.
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<pre>
This given equation is reduced to the standard form quadratic equation

    5x^2 - 23x - 4 = 0.    (1)


Therefore, according to Vieta's theorem,  

    a + b = {{{23/5}}},    (2)

    ab = {{{-4/5}}}.       (3)



Next,  

    {{{1/a^2}}} + {{{1/b^2}}} = {{{(a^2 + b^2)/(a^2*b^2)}}}.    (4)


The numerator in (4) is

    a^2 + b^2 = (a^2 + 2ab + b^2) - 2ab = (a+b)^2 - 2ab = 

                replace here a+b by {{{23/5}}}  and replace ab by  {{{-4/5}}}  based on (2),(3) and continue

              = {{{(23/5)^2}}} - {{{2*(-4/5)}}} = {{{529/25}}} + {{{8/5}}} = {{{(529+5*8)/25}}} = {{{569/25}}}.


Therefore

    {{{1/a^2}}} + {{{1/b^2}}} = {{{(a^2 + b^2)/(a^2*b^2)}}} = {{{((569/25))/((16/25))}}} = {{{569/16}}} = 35{{{9/16}}} = 35.5625.    <U>ANSWER</U>
</pre>

Solved.