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Question Number 23208 by Joel577 last updated on 27/Oct/17

Is it possible to find how many real roots   exist in the equation  x^4  + ∣x∣ = 3  without find all the value of  x?

$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{how}\:\mathrm{many}\:\mathrm{real}\:\mathrm{roots}\: \\ $$$$\mathrm{exist}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{4}} \:+\:\mid{x}\mid\:=\:\mathrm{3} \\ $$$$\mathrm{without}\:\mathrm{find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{x}? \\ $$

Answered by mrW1 last updated on 28/Oct/17

x^4 +∣x∣≥0  f(x)=x^4 +∣x∣ is symmetric, i.e.   f(−x)=f(x)  f(x) is strictly increasing for x>0  and strictly decreasing for x<0  therefore  x^4 +∣x∣=a (a<0) has no real root  x^4 +∣x∣=0 has one real root, x=0  x^4 +∣x∣=a (a>0)has two real roots

$$\mathrm{x}^{\mathrm{4}} +\mid\mathrm{x}\mid\geqslant\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mid\mathrm{x}\mid\:\mathrm{is}\:\mathrm{symmetric},\:\mathrm{i}.\mathrm{e}.\: \\ $$$$\mathrm{f}\left(−\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{increasing}\:\mathrm{for}\:\mathrm{x}>\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{strictly}\:\mathrm{decreasing}\:\mathrm{for}\:\mathrm{x}<\mathrm{0} \\ $$$$\mathrm{therefore} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mid\mathrm{x}\mid=\mathrm{a}\:\left(\mathrm{a}<\mathrm{0}\right)\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{root} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mid\mathrm{x}\mid=\mathrm{0}\:\mathrm{has}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root},\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mid\mathrm{x}\mid=\mathrm{a}\:\left(\mathrm{a}>\mathrm{0}\right)\mathrm{has}\:\mathrm{two}\:\mathrm{real}\:\mathrm{roots} \\ $$

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