Question

Resolva as equações biquadradas e as equações literais em R: (Preciso dos cálculos)
a) x^4 - 10x^2 + 9 = 0

b) x^4 + 5x^2 - 36 = 0

c) x^2 - 6mx + 5m^2 = 0

d) x^2 - (m + 3)x + 3m = 0

Answer 1

$\begin{array}{l}\mathsf{Fixemos~x^2=z,~teremos~o~seguinte:}\\\\\mathsf{x^4-10x^2+9=0}\\\\\mathsf{z^2-10z+9=0}\\\\\mathsf{z^2-10z=-9}\\\\\mathsf{z^2-10z+5^2=-9+5^2}\\\\\mathsf{(z-5)^2=16}\\\\\mathsf{\sqrt{(z-5)^2}=\sqrt{16}}\\\\\mathsf{|z-5|=4}\\\\\mathsf{z-5=\pm4}\\\\\\\begin{Bmatrix}\mathsf{z=-4+5~\rightarrow~z=1}\\\\\mathsf{ou}\\\\\mathsf{z=4+5~\rightarrow~z=9}\end.\end{array}$

$\begin{array}{l}\mathsf{Mas~~z=x^2,~ent\~ao}\\\\\\\mathsf{x^2=1}\\\\\mathsf{\sqrt{x^2}=\sqrt{1}}\\\\\mathsf{|x|=1}\\\\\mathsf{x=\pm1}\\\\\\\mathsf{x^2=9}\\\\\mathsf{\sqrt{x^2}=\sqrt{9}}\\\\\mathsf{|x|=3}\\\\\mathsf{x=\pm3}\\\\\\\mathsf{S=\begin{Bmatrix}\mathsf{-3,-1,~1,~3}\end{Bmatrix}}\end{array}$

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$\begin{array}{l}\textsf{Repetindo o mesmo processo utilizado na equa\c{c}\~ao anterior:}\\\\\\\mathsf{x^4+5x^2-36=0}\\\\\mathsf{z^2+5z-36=0}\\\\\mathsf{z^2+5z=36}\\\\\mathsf{z^2+5z+(\frac{5}{2})^2=36+(\frac{5}{2})^2}\\\\\mathsf{(z+\frac{5}{2})^2=36+\frac{25}{4}}\\\\\mathsf{(z+\frac{5}{2})^2=\frac{144}{4}+\frac{25}{4}}\\\\\mathsf{(z+\frac{5}{2})^2=\frac{169}{4}}\\\\\mathsf{\sqrt{(z+\frac{5}{2})^2}=\sqrt{\frac{169}{4}}}\end{array}$
$\begin{array}{l}\mathsf{|z+\frac{5}{2}|=\frac{13}{2}}\\\\\mathsf{z+\frac{5}{2}=\pm\frac{13}{2}}\\\\\\\begin{Bmatrix}\mathsf{z=-\frac{13}{2}-\frac{5}{2}~\rightarrow~z=-\frac{18}{2}~\rightarrow~z=-9}\\\\\mathsf{ou}\\\\\mathsf{z=\frac{13}{2}-\frac{5}{2}~\rightarrow~z=\frac{8}{2}~\rightarrow~z=4}\end.\end{array}$

$\begin{array}{l}\mathsf{z=x^2}\\\\\mathsf{x^2=-9}\\\\\mathsf{\sqrt{x^2}=\sqrt{-9}~~~(indeterminado~no~conjunto~dos~reais)}\\\\\\\mathsf{x^2=4}\\\\\mathsf{\sqrt{x^2}=\sqrt{4}}\\\\\mathsf{|x|=2}\\\\\mathsf{x=\pm2}\\\\\\\mathsf{S=}~~\begin{Bmatrix}\mathsf{-2,~2}\end{Bmatrix}\end{array}$

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$\begin{array}{l}\mathsf{x^2-6mx+5m^2=0}\\\\\mathsf{\Delta=(-6m)^2-4\cdot1\cdot5m^2}\\\\\mathsf{\Delta=36m^2-20m^2}\\\\\mathsf{\Delta=16m^2}\\\\\\\mathsf{\dfrac{-(-6m)\pm\sqrt{16m^2}}{2}}\\\\~~~~~~~~~~~~\mathsf{\dfrac{6m\pm4m}{2}}\\\\~~~~~~~~\mathsf{x_1=m~~~~~~x_2=5m}\\\\\\\mathsf{S=}\begin{Bmatrix}\mathsf{m,~5m}\end{Bmatrix}\end{array}$

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$\begin{array}{l}\mathsf{x^2-(m+3)x+3m=0}\\\\\mathsf{\Delta=(-m-3)^2-4\cdot1\cdot3m}\\\\\mathsf{\Delta=m^2+6m+9-12m}\\\\\mathsf{\Delta=m^2-6m+9}\\\\\mathsf{\Delta=(m-3)^2}\\\\\\\mathsf{\dfrac{-(-m-3)\pm\sqrt{(m-3)^2}}{2}}\\\\\mathsf{~~~~~~\dfrac{(m+3)\pm(m-3)}{2}}\\\\\mathsf{~~~~~~~~~~~x_1=m~~~~x_2=3}\\\\\\\mathsf{S=\begin{Bmatrix}\mathsf{m,~~3}\end{Bmatrix}}\end{array}$