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Question Number 137817 by bramlexs22 last updated on 07/Apr/21

Let f(x)=x^2 −2x−3; x≥1 & g(x)=1+(√(x+4)) ; x≥−4 then the number of real solutions of equation f(x)=g(x) is ...

$${Let}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3};\:{x}\geqslant\mathrm{1}\:\& \\ $$ $${g}\left({x}\right)=\mathrm{1}+\sqrt{{x}+\mathrm{4}}\:;\:{x}\geqslant−\mathrm{4}\:{then} \\ $$ $${the}\:{number}\:{of}\:{real}\:{solutions} \\ $$ $${of}\:{equation}\:{f}\left({x}\right)={g}\left({x}\right)\:{is}\:... \\ $$

Answered by MJS_new last updated on 07/Apr/21

x^2 −2x−3=1+(√(x+4)) x^2 −2x−4=(√(x+4)) (x^2 −2x−4)^2 =x+4 x^4 −4x^3 −4x^2 +15x+12=0 I found no ♮niceε exact solution of the 4 solutions only 2 solve the given equation x_1 ≈−1.56155 x_2 ≈3.79129 but because of x≥1 the only solution is x_2

$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{1}+\sqrt{{x}+\mathrm{4}} \\ $$ $${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{4}=\sqrt{{x}+\mathrm{4}} \\ $$ $$\left({x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{4}\right)^{\mathrm{2}} ={x}+\mathrm{4} \\ $$ $${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{12}=\mathrm{0} \\ $$ $$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\natural\mathrm{nice}\varepsilon\:\mathrm{exact}\:\mathrm{solution} \\ $$ $$\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{only}\:\mathrm{2}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$ $${x}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{56155} \\ $$ $${x}_{\mathrm{2}} \approx\mathrm{3}.\mathrm{79129} \\ $$ $$\mathrm{but}\:\mathrm{because}\:\mathrm{of}\:{x}\geqslant\mathrm{1}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:{x}_{\mathrm{2}} \\ $$

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