Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 181676 by yaslm last updated on 28/Nov/22

Answered by floor(10²Eta[1]) last updated on 28/Nov/22

$$\mathrm{def}\mathrm{of}\mathrm{square}\mathrm{root}:$$$$\sqrt{\mathrm{b}}=\mathrm{a}\iff \mathrm{b}={\mathrm{a}}^2\wedge \mathrm{a}⩾0$$$$$$$$\sqrt{\mathrm{x}}-7=\sqrt{\mathrm{x}-7}\iff \mathrm{x}-7={(\sqrt{\mathrm{x}}-7)}^2\wedge \sqrt{\mathrm{x}}⩾7$$$$\iff \mathrm{x}-7=\mathrm{x}-14\sqrt{\mathrm{x}}+49\wedge \mathrm{x}⩾49$$$$\iff \sqrt{\mathrm{x}}=4\wedge \mathrm{x}⩾49$$$$\iff \mathrm{x}=16\wedge \mathrm{x}⩾49$$$$\Rightarrow \mathrm{dont}\mathrm{exist}\mathrm{x}\in \mathbb{R}\mathrm{s}.\mathrm{t}.\sqrt{\mathrm{x}}-7=\sqrt{\mathrm{x}-7}$$

Commented by yaslm last updated on 28/Nov/22

great sir

Answered by Rasheed.Sindhi last updated on 30/Nov/22

$$\sqrt{x}-\sqrt{x-7}=7........\left(i\right)$$$$(\sqrt{x}-\sqrt{x-7})(\sqrt{x}+\sqrt{x-7})=7(\sqrt{x}+\sqrt{x-7})$$$$7(\sqrt{x}+\sqrt{x-7})=x-(x-7)$$$$\sqrt{x}+\sqrt{x-7}=1.........\left(ii\right)$$$$\left(i\right)+\left(ii\right):$$$$2\sqrt{x}=8\Rightarrow x=16.........\left(iii\right)$$$$\left(i\right)-\left(ii\right):-2\sqrt{x-7}=6$$$$\sqrt{x-7}=-3......\left(iv\right)$$$$\left(iii\right)\mathrm{\&}\left(iv\right):$$$$x=16\wedge \sqrt{x-7}=-3\Rightarrow x\notin \mathbb{R}$$

Commented by Frix last updated on 29/Nov/22

$$\mathrm{you}\mathrm{must}\mathrm{test}\mathrm{your}\mathrm{solution}$$$$\sqrt{16}-\sqrt{16-7}=7$$$$\sqrt{16}-\sqrt{9}=7$$$$4-3=7\mathrm{wrong}$$

Commented by Rasheed.Sindhi last updated on 30/Nov/22

$$ForgotthatI{}^{\prime }vesquaredbothsides.$$$$...Ofcoursethesolutionisincorrect!$$$$\mathcal{T}han\mathcal{X}muchsir!$$

Commented by Rasheed.Sindhi last updated on 30/Nov/22

$$...HadIstudiedsir{floor}^{\prime }sanswer$$$$properly!$$

Commented by Rasheed.Sindhi last updated on 30/Nov/22

$$Frix\mathit{s}\mathit{i}\mathit{r},pleaserevisitmyanswers$$$$ofthisquestion.$$

Commented by Frix last updated on 30/Nov/22

$$\mathrm{Sorry}\mathrm{I}\mathrm{get}\mathrm{no}\mathrm{notifications}\mathrm{from}\mathrm{this}\mathrm{app},$$$$\mathrm{so}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{a}\mathrm{bit}\mathrm{late}...\mathrm{your}\mathrm{answers}\mathrm{are}\mathrm{now}$$$$\mathrm{correct}.$$

Commented by Rasheed.Sindhi last updated on 01/Dec/22

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{a}\mathbf{lot}\mathbf{sir}!$$

Answered by Rasheed.Sindhi last updated on 30/Nov/22

$$\underset{a}{\underbrace{\sqrt{x}}}-\underset{b}{\underbrace{\sqrt{x-7}}}=7\Rightarrow a-b=7....\left(i\right)$$$$a^2-b^2=x-(x-7)=7$$$$\frac{a^2-b^2}{a-b}=\frac{7}{7}=1$$$$a+b=1......\left(ii\right)$$$$\left(i\right)+\left(ii\right)\mathrm{\&}\left(i\right)-\left(ii\right)$$$$a=4\wedge b=-3$$$$\sqrt{x}=4\wedge \sqrt{x-7}=-3\Rightarrow x\notin \mathbb{R}$$

Answered by Frix last updated on 30/Nov/22

$$\mathrm{For}x,a\in \mathbb{R}$$$$\sqrt{x}-a=\sqrt{x-a}$$$$\mathrm{Squaring}\mathrm{both}\mathrm{sides}$$$$x-2a\sqrt{x}+a^2=x-a$$$$2a\sqrt{x}-a(a+1)=0$$$$a=0\mathrm{is}\mathrm{a}\mathrm{solution}$$$$2\sqrt{x}-(a+1)=0$$$$x=\frac{{(a+1)}^2}{4}\wedge a+1⩾0$$$$\mathrm{Inserting}\mathrm{this}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{gives}$$$$\frac{\mid a+1\mid }{2}-a=\frac{\mid a-1\mid }{2}$$$$\mathrm{This}\mathrm{is}\mathrm{true}\mathrm{for}-1⩽a⩽1$$$$\Rightarrow$$$$\sqrt{x}-a=\sqrt{x-a}\mathrm{can}\mathrm{only}\mathrm{be}\mathrm{solved}\mathrm{for}-1⩽a⩽1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com