Question-205092

Question Number 205092 by BaliramKumar last updated on 08/Mar/24

Answered by A5T last updated on 08/Mar/24

n^2 +19n+92−x^2 =0 n=((−19+_− (√(361−4(92−x^2 ))))/2)=((−19+_− (√(4x^2 −7)))/2) 4x^2 −7=p^2 ⇒(2x−p)(2x+p)=7=1×7=−1×−7 2x−p=1∧2x+p=7⇒x=2 2x−p=−1∧2x+p=−7⇒x=−2 n=((−19+_− 3)/2)=−8 or −11⇒sum=−19⇒(d)

$${n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${n}=\frac{−\mathrm{19}\underset{−} {+}\sqrt{\mathrm{361}−\mathrm{4}\left(\mathrm{92}−{x}^{\mathrm{2}} \right)}}{\mathrm{2}}=\frac{−\mathrm{19}\underset{−} {+}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{7}}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{7}={p}^{\mathrm{2}} \Rightarrow\left(\mathrm{2}{x}−{p}\right)\left(\mathrm{2}{x}+{p}\right)=\mathrm{7}=\mathrm{1}×\mathrm{7}=−\mathrm{1}×−\mathrm{7} \\ $$$$\mathrm{2}{x}−{p}=\mathrm{1}\wedge\mathrm{2}{x}+{p}=\mathrm{7}\Rightarrow{x}=\mathrm{2} \\ $$$$\mathrm{2}{x}−{p}=−\mathrm{1}\wedge\mathrm{2}{x}+{p}=−\mathrm{7}\Rightarrow{x}=−\mathrm{2} \\ $$$${n}=\frac{−\mathrm{19}\underset{−} {+}\mathrm{3}}{\mathrm{2}}=−\mathrm{8}\:{or}\:−\mathrm{11}\Rightarrow{sum}=−\mathrm{19}\Rightarrow\left({d}\right) \\ $$

Commented by BaliramKumar last updated on 08/Mar/24

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by A5T last updated on 08/Mar/24

(n+9)^2 <n^2 +19n+92<(n+10)^2 when n≥0, so n^2 +19n+92 can′t be a perfect square (n+10)^2 =n^2 +20n+100=n^2 +19n+92+n+8 when n=−8, then n^2 +19n+92=(−8+10)^2 =4 but when n<−8, (n+10)^2 <n^2 +19n+92 (n+9)^2 =n^2 +18n+81=n^2 +19n+92−n−11 when n=−11,n^2 +19n+92=(−11+9)^2 =4 but when n<−11; (n+9)^2 >n^2 +19n+92 ⇒when n<−11; (n+10)^2 <n^2 +19n+92<(n+9)^2 Equality holds at either −8 or −11 So,one can easily check n=−10,−9,−7,−6,...−1 to see that none satisfies.

$$\left({n}+\mathrm{9}\right)^{\mathrm{2}} <{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}<\left({n}+\mathrm{10}\right)^{\mathrm{2}} \\ $$$${when}\:{n}\geqslant\mathrm{0},\:{so}\:{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}\:{can}'{t}\:{be}\:{a}\:{perfect}\:{square} \\ $$$$\left({n}+\mathrm{10}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\mathrm{20}{n}+\mathrm{100}={n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}+{n}+\mathrm{8} \\ $$$${when}\:{n}=−\mathrm{8},\:{then}\:{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}=\left(−\mathrm{8}+\mathrm{10}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${but}\:{when}\:{n}<−\mathrm{8},\:\left({n}+\mathrm{10}\right)^{\mathrm{2}} <{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92} \\ $$$$\left({n}+\mathrm{9}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\mathrm{18}{n}+\mathrm{81}={n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}−{n}−\mathrm{11} \\ $$$${when}\:{n}=−\mathrm{11},{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}=\left(−\mathrm{11}+\mathrm{9}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${but}\:{when}\:{n}<−\mathrm{11};\:\left({n}+\mathrm{9}\right)^{\mathrm{2}} >{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92} \\ $$$$\Rightarrow{when}\:{n}<−\mathrm{11};\:\left({n}+\mathrm{10}\right)^{\mathrm{2}} <{n}^{\mathrm{2}} +\mathrm{19}{n}+\mathrm{92}<\left({n}+\mathrm{9}\right)^{\mathrm{2}} \\ $$$${Equality}\:{holds}\:{at}\:{either}\:−\mathrm{8}\:{or}\:−\mathrm{11} \\ $$$${So},{one}\:{can}\:{easily}\:{check}\:{n}=−\mathrm{10},−\mathrm{9},−\mathrm{7},−\mathrm{6},…−\mathrm{1} \\ $$$${to}\:{see}\:{that}\:{none}\:{satisfies}. \\ $$

Commented by BaliramKumar last updated on 08/Mar/24

difficult method example n^2 + 2n + 361 twelve value of n

$$\mathrm{difficult}\:\mathrm{method} \\ $$$$\boldsymbol{\mathrm{example}}\:\:\:\:\:\:\boldsymbol{\mathrm{n}}^{\mathrm{2}} \:+\:\mathrm{2}\boldsymbol{\mathrm{n}}\:+\:\mathrm{361}\:\:\:\:\:\:\:\:\mathrm{twelve}\:\mathrm{value}\:\mathrm{of}\:\:\:\boldsymbol{\mathrm{n}} \\ $$

Commented by A5T last updated on 08/Mar/24

This sort of method could be useful for questions like n^4 +n^3 +n^2 +n+1=x^2 (Q204397) where it would be difficult to make n the subject of formula in terms of x. So,methods have their disadvantages and advantages.

$${This}\:{sort}\:{of}\:{method}\:{could}\:{be}\:{useful}\:{for}\:{questions} \\ $$$${like}\:{n}^{\mathrm{4}} +{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}={x}^{\mathrm{2}} \left({Q}\mathrm{204397}\right)\:{where}\:{it}\: \\ $$$${would}\:{be}\:{difficult}\:{to}\:{make}\:{n}\:{the}\:{subject}\:{of}\: \\ $$$${formula}\:{in}\:{terms}\:{of}\:{x}. \\ $$$${So},{methods}\:{have}\:{their}\:{disadvantages}\:{and} \\ $$$${advantages}. \\ $$

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