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If $x_1$ and $x_2$ are non-zero roots of the equations $ax^2+bx+c=0$ and $-ax^2+bx+c=0$

respectively. Prove that $\frac{a}{2}x^2+bx+c=0$ has a root between $x_1$ and $x_2$.

Please help me ..

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  • $\begingroup$ Over what? $ \Bbb R$? $\endgroup$
    – wesssg
    Dec 17, 2016 at 5:48
  • $\begingroup$ $x_1$and $x_2$ are non zero roots. $\endgroup$ Dec 17, 2016 at 5:51
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    $\begingroup$ Have you used the fact that $bx+c=0 $ so that $x=-c/b$ $\endgroup$
    – wesssg
    Dec 17, 2016 at 5:52
  • $\begingroup$ $ac^2/2b^2=0$ so then, $ac^2=0$ which implies that $a=0 or c=0$ $\endgroup$
    – wesssg
    Dec 17, 2016 at 5:56
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    $\begingroup$ I try $f(x_1)=\frac{a}{2}x_1^2+bx_1+c$ and $f(x_2)=\frac{a}{2}x_2^2+bx_2+c$ $\endgroup$ Dec 17, 2016 at 5:56

1 Answer 1

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Let $f(x)=\frac{a}{2}x^2+bx+c$, then $f(x_1)=-\frac{a}{2}x_1^2$ and $f(x_2)=\frac{3a}{2}x_2^2$. Consequently $f(x_1)f(x_2)<0$. By continuity it should have a root in $(x_1,x_2)$.

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  • $\begingroup$ $f(x_1)f(x_2) = 3a^2x_1^2x_2^2/4 > 0$. $\endgroup$ Dec 17, 2016 at 6:14
  • $\begingroup$ @martycohen there was a typo (cut paste error). I have fixed it. $\endgroup$
    – Anurag A
    Dec 17, 2016 at 6:17

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