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If $a$ and $b$ are roots of $x^2-3x+p=0$ and $c$ and $d$ are roots of $x^2-12x+q=0$, where $a,b,c,d$ form a G.P. then prove that $(q+p):(q-p)=17:15$

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  • $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$ then $\alpha+\beta=-\large\frac{b}{a}$ and $\alpha.\beta=\large\frac{c}{a}$
Given that $a$ and $b$ are roots of $x^2-3x+p=0$
$\Rightarrow\:a+b=3$ and $ab=p$........(i)
It is given that $c$ and $d$ are roots of $x^2-12x+q=0$
$\Rightarrow\:c+d=12$ and $cd=q$........(ii)
Also given that $a,b,c,d$ are in G.P.
Let $a,b,c,d$ be the first four terms of a G.P.
$\Rightarrow\:$ $a=a,\:\:b=ar\:\:c=ar^2\:\:d=ar^3$
Step 2
$c+d=12$ $\Rightarrow\:ar^2+ar^3=12$
From (iii) and (iv) we get
$\Rightarrow\:r^2=4$ or $r=\pm 2$
Substituting the value of $r$ in (iii) we get $a=\large\frac{3}{1+2}$$=1$
$\therefore b=ar=2$
$c=ar^2=2^2=4$ and $d=ar^3=2^3=8$
$\Rightarrow\:ab=p=2$ and $cd=4\times 8=32$
$\Rightarrow\:q+p=32+2=34$ and $q-p=32-2=30$
Hence proved.
answered Mar 31, 2014 by rvidyagovindarajan_1

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