Question 1209228
<br>
{{{2x^2 - 8x + 7 = -3x^2 + 15x + 11}}}<br>
{{{5x^2-23x-4=0}}}<br>
First a slow algebraic method for finding the answer, using the quadratic formula to find the two roots....<br>
{{{a=(23+sqrt(23^2-4(5)(-4)))/10=(23+sqrt(609))/10}}}
{{{b=(23-sqrt(23^2-4(5)(-4)))/10=(23-sqrt(609))/10}}}<br>
{{{1/a^2=100/(23^2+609+46sqrt(609))=100/(1138+46sqrt(609))}}}
{{{1/b^2=100/(23^2+609-46sqrt(609))=100/(1138-46sqrt(609))}}}<br>
{{{1/a^2+1/b^2=100/(1138+46sqrt(609))+100/(1138-46sqrt(609))}}}
{{{1/a^2+1/b^2=(100(1138-46sqrt(609))+100(1138+46sqrt(609)))/(1138^2-46^2*609)=(100(2276))/6400=2276/64=569/16}}}<br>
ANSWER: 569/16<br>
And now a MUCH easier solution, using Vieta's Theorem....<br>
Given the equation {{{5x^2-23x-4=0}}}, Vieta's Theorem tells us<br>
(a+b) = 23/5
(ab) = -4/5<br>
Rewrite the expression {{{1/a^2+1/b^2}}} in terms of (a+b) and (ab).<br>
{{{1/a^2+1/b^2=(a^2+b^2)/(a^2b^2)=((a^2+2ab+b^2)-2ab)/((ab)^2)=((a+b)^2-2ab)/((ab)^2)}}}<br>
So<br>
{{{1/a^2+1/b^2=((23/5)^2-2(-4/5))/((-4/5)^2)=(529/25+8/5)/(16/25)=((529+40)/25)/(16/25)=569/16}}}<br>
ANSWER (again, of course): 569/16<br>