Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 212219 by hardmath last updated on 06/Oct/24

Find: (√(21∙22∙23∙24 + 1)) = ?

$$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{21}\centerdot\mathrm{22}\centerdot\mathrm{23}\centerdot\mathrm{24}\:+\:\mathrm{1}}\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 06/Oct/24

(√((22.5−1.5)(22.5−.5)(22.5+.5)(22.5+1.5)+1)) (√((22.5^2 −.5^2 )(22.5^2 −1.5^2 )+1)) (√(506∙504+1)) (√((505+1)(505−1)+1)) (√((505^2 −1^2 )+1)) (√(505^2 −1+1)) (√(505^2 )) 505

$$\sqrt{\left(\mathrm{22}.\mathrm{5}−\mathrm{1}.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}−.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+\mathrm{1}.\mathrm{5}\right)+\mathrm{1}}\: \\ $$$$\sqrt{\left(\mathrm{22}.\mathrm{5}^{\mathrm{2}} −.\mathrm{5}^{\mathrm{2}} \right)\left(\mathrm{22}.\mathrm{5}^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} \right)+\mathrm{1}} \\ $$$$\sqrt{\mathrm{506}\centerdot\mathrm{504}+\mathrm{1}} \\ $$$$\sqrt{\left(\mathrm{505}+\mathrm{1}\right)\left(\mathrm{505}−\mathrm{1}\right)+\mathrm{1}} \\ $$$$\sqrt{\left(\mathrm{505}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)+\mathrm{1}} \\ $$$$\sqrt{\mathrm{505}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}} \\ $$$$\sqrt{\mathrm{505}^{\mathrm{2}} } \\ $$$$\mathrm{505} \\ $$

Commented by racer last updated on 13/Oct/24

hh

Answered by Frix last updated on 06/Oct/24

x(x+1)(x+2)(x+3)+1=(x^2 +3x+1)^2 x=21 ⇒ answer is 505

$${x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1}=\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}=\mathrm{21}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{505} \\ $$

Answered by Rasheed.Sindhi last updated on 07/Oct/24

(√(21∙22∙23∙24 + 1)) =(√(((42)/2)∙((44)/2)∙((46)/2)∙((48)/2)+1)) =(√(((42.44.46.48)/2^4 )+1)) =(√((((45−3)(45−1)(45+1)(45+3))/(16))+1)) =(√((((45^2 −1^2 )(45^2 −3^2 ))/(16))+1)) =(√(((2024∙2016)/(16))+1)) =(√((((2020+4)(2020−4))/(16))+1)) =(√(((2020^2 −4^2 )/4^2 )+1)) =(√(((2020^2 )/4^2 )−1+1)) =(√((2020)/4)) =505

$$\:\sqrt{\mathrm{21}\centerdot\mathrm{22}\centerdot\mathrm{23}\centerdot\mathrm{24}\:+\:\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{42}}{\mathrm{2}}\centerdot\frac{\mathrm{44}}{\mathrm{2}}\centerdot\frac{\mathrm{46}}{\mathrm{2}}\centerdot\frac{\mathrm{48}}{\mathrm{2}}+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{42}.\mathrm{44}.\mathrm{46}.\mathrm{48}}{\mathrm{2}^{\mathrm{4}} }+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{45}−\mathrm{3}\right)\left(\mathrm{45}−\mathrm{1}\right)\left(\mathrm{45}+\mathrm{1}\right)\left(\mathrm{45}+\mathrm{3}\right)}{\mathrm{16}}+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{45}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)\left(\mathrm{45}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)}{\mathrm{16}}+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{2024}\centerdot\mathrm{2016}}{\mathrm{16}}+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\left(\mathrm{2020}+\mathrm{4}\right)\left(\mathrm{2020}−\mathrm{4}\right)}{\mathrm{16}}+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{2020}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }+\mathrm{1}} \\ $$$$=\sqrt{\frac{\mathrm{2020}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }−\mathrm{1}+\mathrm{1}}\: \\ $$$$=\sqrt{\frac{\mathrm{2020}}{\mathrm{4}}} \\ $$$$=\mathrm{505} \\ $$

Commented by hardmath last updated on 08/Oct/24

thankyou dear professors

$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professors} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com