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Question Number 124463 by mathocean1 last updated on 03/Dec/20
we are in N^3 . (S): { ((p^2 +q^2 =r^2 )),((q+p+r=24 and r<p+q)) :} 1) Show that (p;q;r) is solution of (S) if only r<12 and p ; q are solution of the equation: n^2 −(24−r)n+24(12−r)=0^ where n is unknown.
$${we}\:{are}\:{in}\:\mathbb{N}^{\mathrm{3}} . \\ $$$$\left({S}\right):\:\begin{cases}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={r}^{\mathrm{2}} }\\{{q}+{p}+{r}=\mathrm{24}\:{and}\:{r}<{p}+{q}}\end{cases} \\ $$$$\left.\mathrm{1}\right)\:{Show}\:{that}\:\left({p};{q};{r}\right)\:{is}\:{solution}\:{of}\:\left({S}\right) \\ $$$${if}\:{only}\:{r}<\mathrm{12}\:{and}\:{p}\:;\:{q}\:{are}\:{solution}\:{of} \\ $$$${the}\:{equation}:\:{n}^{\mathrm{2}} −\left(\mathrm{24}−{r}\right){n}+\mathrm{24}\left(\mathrm{12}−{r}\right)=\mathrm{0}^{} \\ $$$${where}\:{n}\:{is}\:{unknown}. \\ $$
Commented by mathocean1 last updated on 03/Dec/20
thanks sir
$${thanks}\:{sir} \\ $$
Answered by floor(10²Eta[1]) last updated on 03/Dec/20
(I)p+q=24−r, but r<p+q ⇒24−r>r⇒r<12 (II)p^2 +q^2 =r^2 p^2 +q^2 +2pq=r^2 +2pq (p+q)^2 =r^2 +2pq (24−r)^2 =r^2 +2pq 24^2 −48r+r^2 =r^2 +2pq 24(24−2r)=2pq pq=24(12−r) and also p+q=24−r let p and q be the solutions of the the quadratic equation p(n), since we know the sum and product of the roots∴ p(n)=n^2 −(24−r)n+24(12−r)
$$\left(\mathrm{I}\right)\mathrm{p}+\mathrm{q}=\mathrm{24}−\mathrm{r},\:\mathrm{but}\:\mathrm{r}<\mathrm{p}+\mathrm{q} \\ $$$$\Rightarrow\mathrm{24}−\mathrm{r}>\mathrm{r}\Rightarrow\mathrm{r}<\mathrm{12} \\ $$$$\left(\mathrm{II}\right)\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{2pq}=\mathrm{r}^{\mathrm{2}} +\mathrm{2pq} \\ $$$$\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} +\mathrm{2pq} \\ $$$$\left(\mathrm{24}−\mathrm{r}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} +\mathrm{2pq} \\ $$$$\mathrm{24}^{\mathrm{2}} −\mathrm{48r}+\mathrm{r}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} +\mathrm{2pq} \\ $$$$\mathrm{24}\left(\mathrm{24}−\mathrm{2r}\right)=\mathrm{2pq} \\ $$$$\mathrm{pq}=\mathrm{24}\left(\mathrm{12}−\mathrm{r}\right) \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{p}+\mathrm{q}=\mathrm{24}−\mathrm{r} \\ $$$$\mathrm{let}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{be}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{the}\: \\ $$$$\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{p}\left(\mathrm{n}\right),\:\mathrm{since}\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{and}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\therefore \\ $$$$\mathrm{p}\left(\mathrm{n}\right)=\mathrm{n}^{\mathrm{2}} −\left(\mathrm{24}−\mathrm{r}\right)\mathrm{n}+\mathrm{24}\left(\mathrm{12}−\mathrm{r}\right) \\ $$
Commented by mathocean1 last updated on 03/Dec/20
thanks sir... so we can solve then (S) ? it seem difficult for me solve it...
$${thanks}\:{sir}… \\ $$$${so}\:{we}\:{can}\:{solve}\:{then}\:\left({S}\right)\:?\: \\ $$$${it}\:{seem}\:{difficult}\:{for}\:{me}\:{solve}\:{it}… \\ $$
Commented by floor(10²Eta[1]) last updated on 07/Dec/20
you can′t solve S because you have 2 equations but you have 3 variables
$$\mathrm{you}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{S}\:\mathrm{because}\:\mathrm{you}\:\mathrm{have} \\ $$$$\mathrm{2}\:\mathrm{equations}\:\mathrm{but}\:\mathrm{you}\:\mathrm{have}\:\mathrm{3}\:\mathrm{variables} \\ $$

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