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 152764 by mnjuly1970 last updated on 01/Sep/21

Commented by prakash jain last updated on 01/Sep/21

1) x=n+f n≥0 n=(n+f)(n+f)=(n+f)^2 f^2 +2nf+n(n−1)=0 f=((−2n±(√(4n^2 +4n−4n^2 )))/2) f=−n±(√n) n=1,f=0 n=0,f=0 n<0 n=−(n+f)^2 f^2 +2nf+n(n+1)=0 f=−n±(√(−n)) f<1⇒no solution for n<0 (wrong) A mistake pointed by mr W. for n<0 n=−1⇒f=0 x=−1 n=−2⇒f=2−(√2) x=n+f=−(√2) −−−−−−−−−−− x=0 or x=1 x=−1, x=−(√2)

$$\left.\mathrm{1}\right)\:{x}={n}+{f} \\ $$$${n}\geqslant\mathrm{0} \\ $$$${n}=\left({n}+{f}\right)\left({n}+{f}\right)=\left({n}+{f}\right)^{\mathrm{2}} \\ $$$${f}^{\mathrm{2}} +\mathrm{2}{nf}+{n}\left({n}−\mathrm{1}\right)=\mathrm{0} \\ $$$${f}=\frac{−\mathrm{2}{n}\pm\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{4}{n}−\mathrm{4}{n}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${f}=−{n}\pm\sqrt{{n}} \\ $$$${n}=\mathrm{1},{f}=\mathrm{0} \\ $$$${n}=\mathrm{0},{f}=\mathrm{0} \\ $$$${n}<\mathrm{0} \\ $$$${n}=−\left({n}+{f}\right)^{\mathrm{2}} \\ $$$${f}^{\mathrm{2}} +\mathrm{2}{nf}+{n}\left({n}+\mathrm{1}\right)=\mathrm{0} \\ $$$${f}=−{n}\pm\sqrt{−{n}} \\ $$$${f}<\mathrm{1}\Rightarrow{no}\:{solution}\:{for}\:{n}<\mathrm{0}\:\left({wrong}\right) \\ $$$$\mathrm{A}\:\mathrm{mistake}\:\mathrm{pointed}\:\mathrm{by}\:\mathrm{mr}\:\mathrm{W}. \\ $$$$\mathrm{for}\:{n}<\mathrm{0}\: \\ $$$${n}=−\mathrm{1}\Rightarrow{f}=\mathrm{0} \\ $$$${x}=−\mathrm{1} \\ $$$${n}=−\mathrm{2}\Rightarrow{f}=\mathrm{2}−\sqrt{\mathrm{2}} \\ $$$${x}={n}+{f}=−\sqrt{\mathrm{2}} \\ $$$$−−−−−−−−−−− \\ $$$${x}=\mathrm{0}\:\mathrm{or}\:{x}=\mathrm{1} \\ $$$${x}=−\mathrm{1},\:{x}=−\sqrt{\mathrm{2}} \\ $$

Commented by mr W last updated on 01/Sep/21

i think x=−(√2) is also a solution. ⌊−(√2)⌋=−2 −(√2)×∣−(√2)∣=−2 ✓

$${i}\:{think}\:{x}=−\sqrt{\mathrm{2}}\:{is}\:{also}\:{a}\:{solution}. \\ $$$$\lfloor−\sqrt{\mathrm{2}}\rfloor=−\mathrm{2} \\ $$$$−\sqrt{\mathrm{2}}×\mid−\sqrt{\mathrm{2}}\mid=−\mathrm{2}\:\checkmark \\ $$

Answered by prakash jain last updated on 01/Sep/21

(2) is same as 1 put y=x−1

$$\left(\mathrm{2}\right)\:\mathrm{is}\:\mathrm{same}\:\mathrm{as}\:\mathrm{1}\:\mathrm{put}\:{y}={x}−\mathrm{1} \\ $$

Answered by MJS_new last updated on 02/Sep/21

(3) (a) obviously x=−2∨x=0∨x=1 (b) ∣⌊x⌋−1∣<∣x^2 −1∣ ∀ (x<−3)∨(x>1) ⇒ there could be a solution −3<x<−2 ⇒ ∣⌊x⌋−1∣=4∧∣x^2 −1∣=x^2 −1 ⇒ we have x^2 −1=4∧−3<x<−2 ⇒ x=−(√5)

$$\left(\mathrm{3}\right) \\ $$$$\left({a}\right)\:\mathrm{obviously}\:{x}=−\mathrm{2}\vee{x}=\mathrm{0}\vee{x}=\mathrm{1} \\ $$$$\left({b}\right) \\ $$$$\mid\lfloor{x}\rfloor−\mathrm{1}\mid<\mid{x}^{\mathrm{2}} −\mathrm{1}\mid\:\forall\:\left({x}<−\mathrm{3}\right)\vee\left({x}>\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{there}\:\mathrm{could}\:\mathrm{be}\:\mathrm{a}\:\mathrm{solution}\:−\mathrm{3}<{x}<−\mathrm{2} \\ $$$$\Rightarrow\:\mid\lfloor{x}\rfloor−\mathrm{1}\mid=\mathrm{4}\wedge\mid{x}^{\mathrm{2}} −\mathrm{1}\mid={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{4}\wedge−\mathrm{3}<{x}<−\mathrm{2}\:\Rightarrow\:{x}=−\sqrt{\mathrm{5}} \\ $$

Answered by mnjuly1970 last updated on 02/Sep/21

⌊ x ⌋=k x=(√(k )) if x ≥0 x= −(√(−k)) x <0 k≤ (√k) < k+1 ⇒ k=0 , 1 ∵ x =0 , 1 .......(★ ) k≤ −(√(−k)) <k+1 −k−1< (√(−k)) ≤ −k ⇒ k=0 ,−1, −2 ∵ x =0 , −1 , −(√2) ....(★ ) (★)∪ (★★) ⇒ x ∈ {−(√2), −1 , 0 , 1 }

$$\:\:\lfloor\:{x}\:\rfloor={k}\: \\ $$$$\:\:\:{x}=\sqrt{{k}\:}\:\:{if}\:\:{x}\:\geqslant\mathrm{0} \\ $$$$\:\:\:{x}=\:−\sqrt{−{k}}\:\:\:{x}\:<\mathrm{0} \\ $$$$\:\:\:\:\:\:{k}\leqslant\:\sqrt{{k}}\:<\:{k}+\mathrm{1}\:\:\Rightarrow\:\:{k}=\mathrm{0}\:,\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\because\:{x}\:=\mathrm{0}\:,\:\mathrm{1}\:.......\left(\bigstar\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:{k}\leqslant\:−\sqrt{−{k}}\:<{k}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{k}−\mathrm{1}<\:\sqrt{−{k}}\:\leqslant\:−{k}\:\Rightarrow\:{k}=\mathrm{0}\:,−\mathrm{1},\:−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\because\:{x}\:=\mathrm{0}\:,\:−\mathrm{1}\:,\:−\sqrt{\mathrm{2}}\:\:....\left(\bigstar\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\left(\bigstar\right)\cup\:\left(\bigstar\bigstar\right)\:\Rightarrow\:\:{x}\:\in\:\left\{−\sqrt{\mathrm{2}},\:−\mathrm{1}\:,\:\mathrm{0}\:,\:\mathrm{1}\:\right\}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com