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Question Number 16383 by aamaguestt last updated on 21/Jun/17

If P : Q : R = 2 : 3 : 4 and P^2 +Q^2 +R^2 =11600, then find (P+Q−R).

$$\mathrm{If}\:\mathrm{P}\::\:\mathrm{Q}\::\:\mathrm{R}\:=\:\mathrm{2}\::\:\mathrm{3}\::\:\mathrm{4}\:\mathrm{and}\:\mathrm{P}^{\mathrm{2}} +\mathrm{Q}^{\mathrm{2}} +\mathrm{R}^{\mathrm{2}} =\mathrm{11600}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\left(\mathrm{P}+\mathrm{Q}−\mathrm{R}\right). \\ $$

Answered by ajfour last updated on 21/Jun/17

p^2 [1+((q/p))^2 +((r/p))^2 ] = 29×400 p^2 [1+(9/4)+4] = 29×400 ⇒ p^2 =((29×400×4)/(29)) =1600 (p+q−r)^2 =p^2 +q^2 +r^2 +2(pq−qr−rp) ?^( 2) = 29×400+2p^2 [(q/p)−((q/p))((r/p))−(r/p)] = 29×400+2×1600[(3/2)−(3/2)×2−2] ?^( 2) =29×400+2×1600(− (7/2)) ?^( 2) =400(29−28) = 400 ? = ± 20 .

$$\:{p}^{\mathrm{2}} \left[\mathrm{1}+\left(\frac{{q}}{{p}}\right)^{\mathrm{2}} +\left(\frac{{r}}{{p}}\right)^{\mathrm{2}} \right]\:=\:\mathrm{29}×\mathrm{400} \\ $$$$\:{p}^{\mathrm{2}} \left[\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{4}\right]\:=\:\mathrm{29}×\mathrm{400} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} =\frac{\mathrm{29}×\mathrm{400}×\mathrm{4}}{\mathrm{29}}\:=\mathrm{1600} \\ $$$$\:\left({p}+{q}−{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\left({pq}−{qr}−{rp}\right) \\ $$$$\:\:?^{\:\mathrm{2}} \:=\:\mathrm{29}×\mathrm{400}+\mathrm{2}{p}^{\mathrm{2}} \left[\frac{{q}}{{p}}−\left(\frac{{q}}{{p}}\right)\left(\frac{{r}}{{p}}\right)−\frac{{r}}{{p}}\right] \\ $$$$\:\:\:=\:\mathrm{29}×\mathrm{400}+\mathrm{2}×\mathrm{1600}\left[\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{2}−\mathrm{2}\right] \\ $$$$\:?^{\:\mathrm{2}} =\mathrm{29}×\mathrm{400}+\mathrm{2}×\mathrm{1600}\left(−\:\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$$$\:\:?^{\:\mathrm{2}} \:=\mathrm{400}\left(\mathrm{29}−\mathrm{28}\right)\:=\:\mathrm{400} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:?\:=\:\pm\:\mathrm{20}\:. \\ $$

Commented by chux last updated on 21/Jun/17

mr Ajfour.... I love this style.

$$\mathrm{mr}\:\mathrm{Ajfour}....\:\mathrm{I}\:\mathrm{love}\:\mathrm{this}\:\mathrm{style}. \\ $$

Commented by ajfour last updated on 21/Jun/17

i call it convenience.., thanks.

$${i}\:{call}\:{it}\:{convenience}..,\:{thanks}. \\ $$

Commented by chux last updated on 21/Jun/17

exactly.... thats the word.

$$\mathrm{exactly}....\:\mathrm{thats}\:\mathrm{the}\:\mathrm{word}. \\ $$

Answered by RasheedSoomro last updated on 21/Jun/17

P : Q : R = 2 : 3 : 4 P^2 +Q^2 +R^2 =11600 P+Q−R=? −−−−−−−−−−−−−−−−−−−−−− Let P=2k,Q=3k and R=4k (2k)^2 +(3k)^2 +(4k)^2 =11600 29k^2 =11600 k^2 =((11600)/(29))=400 k=±20 P+Q−R=2k+3k−4k=k=±20

$$\mathrm{P}\::\:\mathrm{Q}\::\:\mathrm{R}\:=\:\mathrm{2}\::\:\mathrm{3}\::\:\mathrm{4} \\ $$$$\mathrm{P}^{\mathrm{2}} +\mathrm{Q}^{\mathrm{2}} +\mathrm{R}^{\mathrm{2}} =\mathrm{11600} \\ $$$$\mathrm{P}+\mathrm{Q}−\mathrm{R}=? \\ $$$$−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{Let}\:\mathrm{P}=\mathrm{2k},\mathrm{Q}=\mathrm{3k}\:\mathrm{and}\:\mathrm{R}=\mathrm{4k} \\ $$$$\left(\mathrm{2k}\right)^{\mathrm{2}} +\left(\mathrm{3k}\right)^{\mathrm{2}} +\left(\mathrm{4k}\right)^{\mathrm{2}} =\mathrm{11600} \\ $$$$\mathrm{29k}^{\mathrm{2}} =\mathrm{11600} \\ $$$$\mathrm{k}^{\mathrm{2}} =\frac{\mathrm{11600}}{\mathrm{29}}=\mathrm{400} \\ $$$$\mathrm{k}=\pm\mathrm{20} \\ $$$$\mathrm{P}+\mathrm{Q}−\mathrm{R}=\mathrm{2k}+\mathrm{3k}−\mathrm{4k}=\mathrm{k}=\pm\mathrm{20} \\ $$

Commented by ajfour last updated on 21/Jun/17

much better sir .

$${much}\:{better}\:{sir}\:. \\ $$

Commented by RasheedSoomro last updated on 21/Jun/17

Not better sir, only a different way.

$$\mathrm{Not}\:\mathrm{better}\:\mathrm{sir},\:\mathrm{only}\:\mathrm{a}\:\mathrm{different}\:\mathrm{way}. \\ $$

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