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Question Number 59270 by pete last updated on 07/May/19
Solve the equation 3x^(1/2) +5−2x^(1/2) =0
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{3x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{5}−\mathrm{2x}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{0} \\ $$
Answered by Joel578 last updated on 07/May/19
3(√x) + 5 − 2(√x) = 0  ⇔ (√x) = −5  But for x ∈ R, (√x) must ≥ 0, so the equation has no real solution
$$\mathrm{3}\sqrt{{x}}\:+\:\mathrm{5}\:−\:\mathrm{2}\sqrt{{x}}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\sqrt{{x}}\:=\:−\mathrm{5} \\ $$$$\mathrm{But}\:\mathrm{for}\:{x}\:\in\:\mathbb{R},\:\sqrt{{x}}\:\mathrm{must}\:\geqslant\:\mathrm{0},\:\mathrm{so}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$
Commented by Joel578 last updated on 07/May/19
If x = 25i^4  , then (√x) = 5i^2  = −5  Hence the equation has complex solution  Am I wrong?
$$\mathrm{If}\:{x}\:=\:\mathrm{25}{i}^{\mathrm{4}} \:,\:\mathrm{then}\:\sqrt{{x}}\:=\:\mathrm{5}{i}^{\mathrm{2}} \:=\:−\mathrm{5} \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{complex}\:\mathrm{solution} \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{wrong}? \\ $$
Commented by pete last updated on 07/May/19
you are right
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$

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