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Question Number 37324 by behi83417@gmail.com last updated on 11/Jun/18

Commented by math khazana by abdo last updated on 12/Jun/18

$$b)\frac{\sqrt{1+x}}{1+\sqrt{x}}=x\Rightarrow \sqrt{1+x}=x(1+\sqrt{x})withx⩾0ifxreal$$$$1+x=x^2(1+2\sqrt{x}+x)\Rightarrow 1+x=x^3+x^2+2x^2\sqrt{x}$$$${(1+x-(x^3+x))}^2=4x^5\Rightarrow$$$${(1+x)}^2-2(1+x)(x^3+x)+{(x^3+x)}^2-4x^5=0\Rightarrow$$$$x^2+2x+1-2(x^3+x+x^4+x^2)+x^6+2x^4+x^2$$$$-4x^5=0\Rightarrow$$$$x^6-4x^5+2x^4+2x^2+2x+1-2x^3-2x-2x^4-2x^2=0$$$$\Rightarrow x^6-4x^5-2x^3+2x+1=0$$$$therootsoftbepolynom$$$$p\left(x\right)=x^6-4x^5-2x^3+2x+1are$$$$x_1\sim 4,1105\left(real\right)$$$$x_2\sim 0,8572\left(real\right)$$$$x_3\sim i\left(complex\right)$$$$x_4\sim -i\left(complex\right)$$$$x_5\sim -0,4839+0,2229i\left(complex\right)$$$$x_6\sim -0,4839-0,2229i\left(complex\right).$$

Commented by math khazana by abdo last updated on 12/Jun/18

$$wecanalsodividep\left(x\right)by{x}^2+1andweget$$$$apolynomwithdegre4inwichwefindtheroots...$$

Commented by behi83417@gmail.com last updated on 12/Jun/18

$$thankyouverymuch.$$

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