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Question Number 114071 by bobhans last updated on 17/Sep/20

$$Findthesolutionset$$$$\sqrt{x^2-4x-5}⩾x$$

Commented by prakash jain last updated on 17/Sep/20

Commented by bemath last updated on 17/Sep/20

$$howdoyoumadethisgraphsir$$

Commented by prakash jain last updated on 17/Sep/20

$$\mathrm{From}\mathrm{this}\mathrm{app}\mathrm{only},\mathrm{i}\mathrm{was}\mathrm{just}$$$$\mathrm{checking}$$$$\mathrm{steps}:$$$$a.\mathrm{insert}\mathrm{drawing}$$$$\mathrm{b}.\mathrm{insert}\mathrm{equation}\mathrm{and}\mathrm{pasted}\mathrm{equation}1.$$$$\mathrm{c}.\mathrm{from}\mathrm{BUILD}\mathrm{menu}\mathrm{select}\mathrm{plot}$$$$\mathrm{d}.\mathrm{repeated}\mathrm{procedure}\mathrm{for}\mathrm{other}\mathrm{equation}.$$$$\mathrm{see}\mathrm{Q}113542.$$

Commented by bemath last updated on 17/Sep/20

$$thankyousir$$

Answered by bemath last updated on 17/Sep/20

$$\left(1\right)x^2-4x-5⩾0$$$$(x-5)(x+1)⩾0$$$$x⩽-1\cup x⩾5$$$$\left(2\right)checkforx=-1$$$$\sqrt{{(-1)}^2-4(-1)-5}=\sqrt{5-5}=0⩾-1\left(true\right)$$$$checkforx=5$$$$\sqrt{25-20-5}=0⩾5\left(false\right)$$$$thensolutionisx⩽-1$$

Commented by bemath last updated on 17/Sep/20

Commented by bobhans last updated on 17/Sep/20

$$santuyy$$

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