Question Number 181182 by depressiveshrek last updated on 22/Nov/22
$${For}\:{what}\:{values}\:{of}\:{a}\:{does}\:{the}\:{system} \\ $$$${of}\:{equations}\:{only}\:{have}\:{one}\:{solution}: \\ $$$$\begin{cases}{{a}\left({x}^{\mathrm{4}} +\mathrm{1}\right)={y}+\mathrm{2}−\mid{x}\mid}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$
Answered by mr W last updated on 22/Nov/22
$${y}={a}\left({x}^{\mathrm{4}} +\mathrm{1}\right)+\mid{x}\mid−\mathrm{2} \\ $$$${let}\:{t}=\mid{x}\mid>\mathrm{0} \\ $$$${y}={at}^{\mathrm{4}} +{t}+{a}−\mathrm{2}\:\:\:...\left({i}\right) \\ $$$${y}^{\mathrm{2}} +{t}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)\:{and}\:\left({ii}\right)\:{should}\:{touch}\:{at}\:{one}\:{and} \\ $$$${only}\:{one}\:{point}.\:{due}\:{to}\:{symmetry},\: \\ $$$${this}\:{point}\:{must}\:{be}\:{at}\:{t}=\mathrm{0},\:{i}.{e}.\:\left(\mathrm{0},\mathrm{2}\right) \\ $$$${a}\left(\mathrm{0}^{\mathrm{4}} +\mathrm{1}\right)+\mathrm{0}−\mathrm{2}=\mathrm{2} \\ $$$$\Rightarrow{a}=\mathrm{4}\:\checkmark \\ $$
Commented by mr W last updated on 22/Nov/22
Commented by depressiveshrek last updated on 22/Nov/22
$${Thank}\:{you}\:{sir},\:{you}\:{are}\:{helping}\:{me} \\ $$$${a}\:{lot}.\:{Can}\:{I}\:{contact}\:{you}\:{somehow}? \\ $$$${I}\:{need}\:{help}\:{to}\:{practice}\:{for}\:{my} \\ $$$${country}'{s}\:{national}\:{math}\:{olympiad}. \\ $$
Commented by mr W last updated on 22/Nov/22
$${here}\:{is}\:{a}\:{good}\:{place}.\:{all}\:{people}\:{can}\: \\ $$$${help}\:{you}. \\ $$