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Question Number 217326 by Rasheed.Sindhi last updated on 10/Mar/25

Find three consecutive integers such that the sum of their squares is 50.

$${Find}\:{three}\:{consecutive}\:{integers}\: \\ $$$${such}\:{that}\:{the}\:{sum}\:{of}\:{their}\:{squares} \\ $$$$\:{is}\:\mathrm{50}. \\ $$

Answered by Hanuda354 last updated on 10/Mar/25

Let n, n+1 and n+2 are required numbers. n^2 + (n+1)^2 + (n+2)^2 = 50 3n^2 + 6n + 5 = 50 3n^2 + 6n − 45 = 0 3(n^2 + 2n − 15) = 0 3(n+5)(n−3) = 0 n = −5 ∨ n = 3 n = −5 ⇒ −5, −4, −3 n = 3 ⇒ 3, 4, 5

$$\mathrm{Let}\:{n},\:{n}+\mathrm{1}\:\:\mathrm{and}\:\:{n}+\mathrm{2}\:\:\mathrm{are}\:\:\mathrm{required}\:\:\mathrm{numbers}. \\ $$$$ \\ $$$${n}^{\mathrm{2}} \:+\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:+\:\left({n}+\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{50} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{6}{n}\:+\:\mathrm{5}\:=\:\mathrm{50} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{6}{n}\:−\:\mathrm{45}\:\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left({n}^{\mathrm{2}} \:+\:\mathrm{2}{n}\:−\:\mathrm{15}\right)\:\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left({n}+\mathrm{5}\right)\left({n}−\mathrm{3}\right)\:\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{n}\:=\:−\mathrm{5}\:\:\vee\:\:{n}\:=\:\mathrm{3} \\ $$$$ \\ $$$${n}\:=\:−\mathrm{5}\:\:\Rightarrow\:−\mathrm{5},\:−\mathrm{4},\:−\mathrm{3} \\ $$$${n}\:=\:\mathrm{3}\:\:\:\:\:\:\:\Rightarrow\:\mathrm{3},\:\mathrm{4},\:\mathrm{5} \\ $$

Commented by Rasheed.Sindhi last updated on 10/Mar/25

ThanX sir!

$$\mathcal{T}{han}\mathcal{X}\:{sir}! \\ $$

Answered by Rasheed.Sindhi last updated on 10/Mar/25

b−a=1⇒a^2 +b^2 −2ab=1 c−b=1⇒b^2 +c^2 −2bc=1 c−a=2⇒c^2 +a^2 −2ca=4 Adding: 2((a^2 +b^2 +c^2 )−(ab+bc+ca))=6 50−(ab+bc+ca)=3 ab+bc+ca=47 (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca) (3b)^2 =50+2(47) 9b^2 =50+94=144 b^2 =16 b= 4,−4 b−1=3,−5 b+1=5, −3

$${b}−{a}=\mathrm{1}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{1} \\ $$$${c}−{b}=\mathrm{1}\Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}=\mathrm{1} \\ $$$${c}−{a}=\mathrm{2}\Rightarrow{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ca}=\mathrm{4} \\ $$$$\mathrm{Adding}: \\ $$$$\mathrm{2}\left(\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({ab}+{bc}+{ca}\right)\right)=\mathrm{6} \\ $$$$\:\:\:\mathrm{50}−\left({ab}+{bc}+{ca}\right)=\mathrm{3} \\ $$$${ab}+{bc}+{ca}=\mathrm{47} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\:\:\left(\mathrm{3}{b}\right)^{\mathrm{2}} =\mathrm{50}+\mathrm{2}\left(\mathrm{47}\right) \\ $$$$\mathrm{9}{b}^{\mathrm{2}} =\mathrm{50}+\mathrm{94}=\mathrm{144} \\ $$$${b}^{\mathrm{2}} =\mathrm{16} \\ $$$${b}=\:\:\:\:\:\:\:\mathrm{4},−\mathrm{4} \\ $$$${b}−\mathrm{1}=\mathrm{3},−\mathrm{5} \\ $$$${b}+\mathrm{1}=\mathrm{5},\:−\mathrm{3} \\ $$

Answered by mr W last updated on 10/Mar/25

(n−1)^2 +n^2 +(n+1)^2 =50 n^2 =16 n=±4 ⇒(−5, −4, −3), (3, 4, 5)

$$\left({n}−\mathrm{1}\right)^{\mathrm{2}} +{n}^{\mathrm{2}} +\left({n}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{50} \\ $$$${n}^{\mathrm{2}} =\mathrm{16} \\ $$$${n}=\pm\mathrm{4} \\ $$$$\Rightarrow\left(−\mathrm{5},\:−\mathrm{4},\:−\mathrm{3}\right),\:\left(\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 10/Mar/25

Thanks sir!

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{sir}! \\ $$

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