Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 153464 by roxceefocus last updated on 07/Sep/21

Commented by roxceefocus last updated on 07/Sep/21

pls i need answers on this question

$${pls}\:{i}\:{need}\:{answers}\:{on}\:{this}\:{question} \\ $$

Answered by Rasheed.Sindhi last updated on 07/Sep/21

(√(3x+1))−(√(2x−1))=1 ( (√(3x+1))−(√(2x−1))=1 )^2 3x+1+2x−1−2(√((3x+1)(2x−1))=1 ( 5x−1=2(√((3x+1)(2x−1))) )^2 (5x−1)^2 =4(3x+1)(2x−1) 25x^2 −10x+1=4(6x^2 −3x+2x−1) 25x^2 −10x+1=24x^2 −4x−4 x^2 −6x+5=0 (x−1)(x−5)=0 x=1 ∣ x=5 On verification we can see that both roots satisfy the original eqn

$$\sqrt{\mathrm{3}{x}+\mathrm{1}}−\sqrt{\mathrm{2}{x}−\mathrm{1}}=\mathrm{1} \\ $$$$\left(\:\sqrt{\mathrm{3}{x}+\mathrm{1}}−\sqrt{\mathrm{2}{x}−\mathrm{1}}=\mathrm{1}\:\right)^{\mathrm{2}} \\ $$$$\mathrm{3}{x}+\mathrm{1}+\mathrm{2}{x}−\mathrm{1}−\mathrm{2}\sqrt{\left(\mathrm{3}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}−\mathrm{1}\right.}=\mathrm{1} \\ $$$$\left(\:\mathrm{5}{x}−\mathrm{1}=\mathrm{2}\sqrt{\left(\mathrm{3}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)}\:\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{3}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$\mathrm{25}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{1}=\mathrm{4}\left(\mathrm{6}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$\mathrm{25}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{1}=\mathrm{24}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\mid\:{x}=\mathrm{5} \\ $$$${On}\:{verification}\:{we}\:{can}\:{see}\:{that} \\ $$$${both}\:{roots}\:{satisfy}\:{the}\:{original}\:{eqn} \\ $$

Answered by puissant last updated on 07/Sep/21

x≥−(1/3) and x≥(1/2) ⇒ x≥(1/2). we have (√(3x+1))=1+(√(2x−1)) ⇒ 3x+1=1+2(√(2x−1))+2x−1 ⇒ x+1=2(√(2x−1)) ⇒ x^2 +2x+1=8x−4 ⇒ x^2 −6x+5=0 x_1 =1 and x_2 =5 S_R ={1,5}.

$${x}\geqslant−\frac{\mathrm{1}}{\mathrm{3}}\:{and}\:{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$${we}\:{have}\:\sqrt{\mathrm{3}{x}+\mathrm{1}}=\mathrm{1}+\sqrt{\mathrm{2}{x}−\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{3}{x}+\mathrm{1}=\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}}+\mathrm{2}{x}−\mathrm{1} \\ $$$$\Rightarrow\:{x}+\mathrm{1}=\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{8}{x}−\mathrm{4} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\mathrm{5} \\ $$$${S}_{\mathbb{R}} =\left\{\mathrm{1},\mathrm{5}\right\}. \\ $$

Answered by Rasheed.Sindhi last updated on 08/Sep/21

≡An Alternative≡ (√(3x+1)) − (√(2x−1)) =1 ⇒ p−q=1 p^2 −q^2 =(3x+1)−(2x−1)=x+2 (p−q)(p+q)=x+2 p+q=x+2 p−q=1 2p=x+3⇒p=((x+3)/2) 2q=x+1⇒q=((x+1)/2) p=(√(3x+1))=((x+3)/2) 4(3x+1)=x^2 +6x+9 x^2 −6x+5=0 (x−1)(x−5)=0 x=1,5 On verification it can be verified that both roots are true. (Same result can be obtained by q=(√(2x−1)) =((x+1)/2) 4(2x−1)=x^2 +2x+1 x^2 −6x+5=0)

$$\equiv\mathbb{A}\mathrm{n}\:\mathbb{A}\mathrm{lternative}\equiv \\ $$$$\sqrt{\mathrm{3}{x}+\mathrm{1}}\:−\:\sqrt{\mathrm{2}{x}−\mathrm{1}}\:=\mathrm{1} \\ $$$$\:\Rightarrow\:{p}−{q}=\mathrm{1} \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\left(\mathrm{3}{x}+\mathrm{1}\right)−\left(\mathrm{2}{x}−\mathrm{1}\right)={x}+\mathrm{2} \\ $$$$\left({p}−{q}\right)\left({p}+{q}\right)={x}+\mathrm{2} \\ $$$${p}+{q}={x}+\mathrm{2} \\ $$$${p}−{q}=\mathrm{1} \\ $$$$\mathrm{2}{p}={x}+\mathrm{3}\Rightarrow{p}=\frac{{x}+\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}{q}={x}+\mathrm{1}\Rightarrow{q}=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$${p}=\sqrt{\mathrm{3}{x}+\mathrm{1}}=\frac{{x}+\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{4}\left(\mathrm{3}{x}+\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{1},\mathrm{5} \\ $$$${On}\:{verification}\:{it}\:{can}\:{be}\:{verified} \\ $$$${that}\:{both}\:{roots}\:{are}\:{true}. \\ $$$$ \\ $$$$\left({Same}\:{result}\:{can}\:{be}\:{obtained}\right. \\ $$$${by} \\ $$$${q}=\sqrt{\mathrm{2}{x}−\mathrm{1}}\:=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}\left(\mathrm{2}{x}−\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}=\mathrm{0}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com