Question

Resolva a equação literal. Por favor mim ajudem nessa​

Answer 1

Resposta:

$S=\{-2\sqrt{n},~2\sqrt{n},~-3\sqrt{n},~3\sqrt{n}\}$

Explicação passo-a-passo:

$x^{4}-13nx^{2}+36n^{2}=0\\\\x^{2}=y\\y^{2}-13ny+36n^{2}=0\\\\Aplicando~a~f\'{o}rmula~de~Bhaskara~para~y^{2}-13ny+36n^{2}=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-13n~e~c=36n^{2}\\\Delta=(b)^{2}-4(a)(c)=(-13n)^{2}-4(1)(36n^{2})=169n^{2}-144n^{2}=25n^{2}\\\sqrt{\Delta}=\sqrt{25n^{2}}=\sqrt{25} \sqrt{n^{2}} =5n\\\\y^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-13n)-5n}{2(1)} =\frac{13n-5n}{2} =\frac{8n}{2} =4n$

$y^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-13n)+5n}{2(1)} =\frac{13n+5n}{2} =\frac{18n}{2} =9n$

$x^{2}=y\\\\Para~y=y'=4n{:}\\\\x^{2}=y=y'=4n\\\\x=\pm\sqrt{4n} =\pm\sqrt{4} \sqrt{n}=\pm2\sqrt{n}\\\\Para~y=y'=9n{:}\\\\x^{2}=y=y'=9n\\\\x=\pm\sqrt{9n} =\pm\sqrt{9} \sqrt{n}=\pm3\sqrt{n}$