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Question Number 81771 by Power last updated on 15/Feb/20

Commented by Power last updated on 15/Feb/20

$$\mathrm{solution}$$

Commented by Power last updated on 15/Feb/20

$$\mathrm{What}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{parameter}\uparrow \mathrm{will}\mathrm{have}2\mathrm{roots}?$$

Commented by mr W last updated on 15/Feb/20

$$tworootswhen-4⩽a<2$$$$onerootwhena=2$$$$norootwhena<-4ora>2$$

Answered by mr W last updated on 15/Feb/20

$$letf\left(x\right)=\sqrt{4-\mid 2x\mid }$$$$letg\left(x\right)=x^2+a$$$$bothfunctionsareeven,i.e.$$$$f(-x)=f\left(x\right),g(-x)=g\left(x\right)$$$$weonlyneedtolookatx⩾0.$$$$f\left(x\right)=\sqrt{4-2x}:$$$$0⩽x⩽2$$$$f\left(0\right)=2=maximum$$$$f\left(2\right)=0=minimum$$$$g\left(x\right)=x^2+a:$$$$thisisaparabolaopeningupwards,$$$$itslowestpointis(0,a).$$$$$$$$suchthattheeqn.hastworoots,the$$$$curveg\left(x\right)mustintersectthecurveg\left(x\right)$$$$attwopoints.$$$$wehavefollowingcases:$$$$$$$$case1:a>2,g\left(x\right)istoohighand{can}^{\prime }t$$$$intersectthecurvef\left(x\right)\Rightarrow noroot.$$$$$$$$case2:a=2,thelowestpointofg\left(x\right)$$$$coincideswiththetopmostpointof$$$$f\left(x\right),oneintersectionpoint\Rightarrow one$$$$root.$$$$$$$$case3:-4⩽a<2,g\left(x\right)intersectswith$$$$f\left(x\right)attwopoints\Rightarrow tworoots.$$$$$$$$case4:a=-4,y=x^2-4,y=0\Rightarrow x=\pm 2.$$$$f\left(x\right)intersectswithg\left(x\right)at(-2,0)and$$$$(2,0)\Rightarrow tworoots.$$$$$$$$case5:a<-4,g\left(x\right)istoolowand{can}^{\prime }t$$$$intersectwithf\left(x\right)\Rightarrow noroot.$$

Commented by mr W last updated on 15/Feb/20

Commented by Power last updated on 15/Feb/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by mr W last updated on 15/Feb/20

$$isthisavalidproofaccordingtoyou?$$

Commented by Power last updated on 15/Feb/20

$$\mathrm{yes}\mathrm{sir}$$

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