Let a and b be two non-zero real numbers. If | vedclass

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(C) For the equation $x^{2}-8ax+2a=0$,the roots are $p$ and $r$. Thus,$p+r=8a$ and $pr=2a$. Then $\frac{1}{p}+\frac{1}{r} = \frac{p+r}{pr} = \frac{8a}

Vedclass · Accepted Answer

$38$

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(C) For the equation $x^{2}-8ax+2a=0$,the roots are $p$ and $r$. Thus,$p+r=8a$ and $pr=2a$. Then $\frac{1}{p}+\frac{1}{r} = \frac{p+r}{pr} = \frac{8a}{2a} = 4$. For the equation $x^{2}+12bx+6b=0$,the roots are $q$ and $s$. Thus,$q+s=-12b$ and $qs=6b$. Then $\frac{1}{q}+\frac{1}{s} = \frac{q+s}{qs} = \frac{-12b}{6b} = -2$. Let the $A$.$P$. be $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ with common difference $d$. Then $\frac{1}{q} = \frac{1}{p}+d$,$\frac{1}{r} = \frac{1}{p}+2d$,and $\frac{1}{s} = \frac{1}{p}+3d$. We have $\frac{1}{p}+\frac{1}{r} = \frac{2}{p}+2d = 4$,so $\frac{1}{p}+d = 2$. Thus $\frac{1}{q} = 2$. We have $\frac{1}{q}+\frac{1}{s} = \frac{2}{q}+2d = -2$,so $\frac{1}{q}+d = -1$. Since $\frac{1}{q}=2$,we have $2+d=-1$,so $d=-3$. Then $\frac{1}{p} = \frac{1}{q}-d = 2-(-3) = 5$,so $p = \frac{1}{5}$. Since $pr=2a$,$r = \frac{2a}{p} = 10a$. Also $\frac{1}{r} = \frac{1}{p}+2d = 5+2(-3) = -1$,so $r = -1$. Thus $10a = -1$,which gives $a = -\frac{1}{10}$,so $a^{-1} = -10$. Also $\frac{1}{s} = \frac{1}{q}+2d = 2+2(-3) = -4$,so $s = -\frac{1}{4}$. Since $qs=6b$,$q = \frac{1}{2}$,so $b = \frac{qs}{6} = \frac{(1/2)(-1/4)}{6} = -\frac{1}{48}$,so $b^{-1} = -48$. Finally,$a^{-1}-b^{-1} = -10 - (-48) = 38$.

Let $a$ and $b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2}-8ax+2a=0$ and $q$ and $s$ are the roots of the equation $x^{2}+12bx+6b=0$,such that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ are in $A$.$P$.,then $a^{-1}-b^{-1}$ is equal to $......$

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