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Question Number 207897 by hardmath last updated on 29/May/24

Find: (√(12 ∙ 13 ∙ 14 ∙ 15 + 1)) = ?

$$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{12}\:\:\centerdot\:\:\mathrm{13}\:\:\centerdot\:\:\mathrm{14}\:\:\centerdot\:\:\mathrm{15}\:\:+\:\:\mathrm{1}}\:\:=\:\:? \\ $$

Answered by Frix last updated on 29/May/24

(√(x(x+1)(x+2)(x+3)+1))= =(√(x^4 +6x^3 +11x^2 +6x+1))= =(√((x^2 +3x+1)^2 ))= =x^2 +3x+1 =^(x=12) 181

$$\sqrt{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1}}= \\ $$$$=\sqrt{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{11}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{1}}= \\ $$$$=\sqrt{\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$={x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\:\overset{{x}=\mathrm{12}} {=}\:\mathrm{181} \\ $$

Answered by Rasheed.Sindhi last updated on 30/May/24

let x=13(1/2) =(√((x−(3/2))(x+(3/2))(x−(1/2))(x+(1/2))+1)) =(√((x^2 −(9/4))(x^2 −(1/4))+1)) =(√(x^4 −(x^2 /4)−((9x^2 )/4)+(9/(16))+1)) =(√(x^4 −((5x^2 )/2)+((25)/(16)))) =(√((x^2 )^2 −2(x^2 )((5/4))+((5/4))^2 )) =(√((x^2 −(5/4))^2 )) =∣x^2 −(5/4)∣ ⇒∣ (((27)/2))^2 −(5/4)∣ =((27^2 −5)/4)=181

$${let}\:{x}=\mathrm{13}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\sqrt{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}}\: \\ $$$$=\sqrt{\left({x}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{1}} \\ $$$$=\sqrt{{x}^{\mathrm{4}} −\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{9}}{\mathrm{16}}+\mathrm{1}}\: \\ $$$$=\sqrt{{x}^{\mathrm{4}} −\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{25}}{\mathrm{16}}}\: \\ $$$$=\sqrt{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({x}^{\mathrm{2}} \right)\left(\frac{\mathrm{5}}{\mathrm{4}}\right)+\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\left({x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$=\mid{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\mid \\ $$$$\Rightarrow\mid\:\left(\frac{\mathrm{27}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\mid \\ $$$$=\frac{\mathrm{27}^{\mathrm{2}} −\mathrm{5}}{\mathrm{4}}=\mathrm{181} \\ $$

Answered by BaliramKumar last updated on 31/May/24

(√(a∙(a+d)∙(a+2d)∙(a+3d) + d^4 )) = a(a+3d) + d^2 a = 12, d = 1 12(12 + 3×1) + 1^2 = 12^2 + 12×3 + 1 144 + 36 + 1 = 181

$$\sqrt{{a}\centerdot\left({a}+{d}\right)\centerdot\left({a}+\mathrm{2}{d}\right)\centerdot\left({a}+\mathrm{3}{d}\right)\:+\:{d}^{\mathrm{4}} }\:\:=\:{a}\left({a}+\mathrm{3}{d}\right)\:+\:{d}^{\mathrm{2}} \\ $$$${a}\:=\:\mathrm{12},\:\:\:\:\:\:\:\:\mathrm{d}\:=\:\mathrm{1} \\ $$$$\mathrm{12}\left(\mathrm{12}\:+\:\mathrm{3}×\mathrm{1}\right)\:+\:\mathrm{1}^{\mathrm{2}} \:=\:\mathrm{12}^{\mathrm{2}} \:+\:\mathrm{12}×\mathrm{3}\:+\:\mathrm{1}\: \\ $$$$\mathrm{144}\:+\:\mathrm{36}\:+\:\mathrm{1}\:=\:\mathrm{181} \\ $$

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