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Question-88021




Question Number 88021 by naka3546 last updated on 07/Apr/20
Commented by MJS last updated on 07/Apr/20
I get 4∨5... not sure how I got there...
$$\mathrm{I}\:\mathrm{get}\:\mathrm{4}\vee\mathrm{5}…\:\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{I}\:\mathrm{got}\:\mathrm{there}… \\ $$
Commented by naka3546 last updated on 08/Apr/20
you′re  right , sir.
$${you}'{re}\:\:{right}\:,\:{sir}. \\ $$
Answered by $@ty@m123 last updated on 08/Apr/20
p+q+r+s=9 {where p=(a/b) and so on  pq+qr+rs+sp=20  ⇒(p+r)(q+s)=20  Equation whose roots are p+r, q+s is  x^2 −9x+20=0  x=4,5
$${p}+{q}+{r}+{s}=\mathrm{9}\:\left\{{where}\:{p}=\frac{{a}}{{b}}\:{and}\:{so}\:{on}\right. \\ $$$${pq}+{qr}+{rs}+{sp}=\mathrm{20} \\ $$$$\Rightarrow\left({p}+{r}\right)\left({q}+{s}\right)=\mathrm{20} \\ $$$${Equation}\:{whose}\:{roots}\:{are}\:{p}+{r},\:{q}+{s}\:{is} \\ $$$${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{20}=\mathrm{0} \\ $$$${x}=\mathrm{4},\mathrm{5} \\ $$
Commented by Prithwish Sen 1 last updated on 08/Apr/20
I think it will be  (p+r)(q+s)=20
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{p}+\mathrm{r}\right)\left(\mathrm{q}+\mathrm{s}\right)=\mathrm{20} \\ $$
Commented by $@ty@m123 last updated on 08/Apr/20
Thanks for correction.
$${Thanks}\:{for}\:{correction}. \\ $$

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