Question

Calcule as 4 raízes do polinômio...

Answer 1

Explicação passo-a-passo:

1) $\sf 2x^4-10x^2+8=0$

$\sf x^2=y$

$\sf 2(x^2)^2-10x^2+8=0$

$\sf 2y^2-10y+8=0$

$\sf \Delta=b^2-4\cdot a\cdot c$

$\sf \Delta=(-10)^2-4\cdot2\cdot8$

$\sf \Delta=100-64$

$\sf \Delta=36$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-10)\pm\sqrt{36}}{2\cdot2}=\dfrac{10\pm6}{4}$

$\sf y'=\dfrac{10+6}{4}~\longrightarrow~y'=\dfrac{16}{4}~\longrightarrow~y'=4$

$\sf y"=\dfrac{10-6}{4}~\longrightarrow~y"=\dfrac{4}{4}~\longrightarrow~y"=1$

• Para $\sf y'=4$:

$\sf x^2=4$

$\sf x=\pm\sqrt{4}$

$\sf x=\pm2$

• Para $\sf y"=1$:

$\sf x^2=1$

$\sf x=\pm\sqrt{1}$

$\sf x=\pm1$

$\sf S=\{-2,-1,1,2\}$

2) $\sf x^4-13x^2+36=0$

$\sf x^2=y$

$\sf (x^2)^2-13x^2+36=0$

$\sf y^2-13y+36=0$

$\sf \Delta=b^2-4\cdot a\cdot c$

$\sf \Delta=(-13)^2-4\cdot1\cdot36$

$\sf \Delta=169-144$

$\sf \Delta=25$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-13)\pm\sqrt{25}}{2\cdot1}=\dfrac{13\pm5}{2}$

$\sf y'=\dfrac{13+5}{2}~\longrightarrow~y'=\dfrac{18}{2}~\longrightarrow~y'=9$

$\sf y"=\dfrac{13-5}{2}~\longrightarrow~y"=\dfrac{8}{2}~\longrightarrow~y"=4$

• Para $\sf y'=9$:

$\sf x^2=9$

$\sf x=\pm\sqrt{9}$

$\sf x=\pm3$

• Para $\sf y"=4$:

$\sf x^2=4$

$\sf x=\pm\sqrt{4}$

$\sf x=\pm2$

$\sf S=\{-3,-2,2,3\}$

3) $\sf x^4-26x^2+25=0$

$\sf x^2=y$

$\sf (x^2)^2-26x^2+25=0$

$\sf y^2-26y+25=0$

$\sf \Delta=b^2-4\cdot a\cdot c$

$\sf \Delta=(-26)^2-4\cdot1\cdot25$

$\sf \Delta=676-100$

$\sf \Delta=576$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-26)\pm\sqrt{576}}{2\cdot1}=\dfrac{26\pm24}{2}$

$\sf y'=\dfrac{26+24}{2}~\longrightarrow~y'=\dfrac{50}{2}~\longrightarrow~y'=25$

$\sf y"=\dfrac{26-24}{2}~\longrightarrow~y"=\dfrac{2}{2}~\longrightarrow~y"=1$

• Para $\sf y'=25$:

$\sf x^2=25$

$\sf x=\pm\sqrt{25}$

$\sf x=\pm5$

• Para $\sf y"=1$:

$\sf x^2=1$

$\sf x=\pm\sqrt{1}$

$\sf x=\pm1$

$\sf S=\{-5,-1,1,5\}$

4) $\sf x^4-17x^2+16=0$

$\sf x^2=y$

$\sf (x^2)^2-17x^2+16=0$

$\sf y^2-17y+16=0$

$\sf \Delta=b^2-4\cdot a\cdot c$

$\sf \Delta=(-17)^2-4\cdot1\cdot16$

$\sf \Delta=289-64$

$\sf \Delta=225$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-17)\pm\sqrt{225}}{2\cdot1}=\dfrac{17\pm15}{2}$

$\sf y'=\dfrac{17+15}{2}~\longrightarrow~y'=\dfrac{32}{2}~\longrightarrow~y'=16$

$\sf y"=\dfrac{17-15}{2}~\longrightarrow~y"=\dfrac{2}{2}~\longrightarrow~y"=1$

• Para $\sf y'=16$:

$\sf x^2=16$

$\sf x=\pm\sqrt{16}$

$\sf x=\pm4$

• Para $\sf y"=1$:

$\sf x^2=1$

$\sf x=\pm\sqrt{1}$

$\sf x=\pm1$

$\sf S=\{-4,-1,1,4\}$

5) $\sf x^4-10x^2+9=0$

$\sf x^2=y$

$\sf (x^2)^2-10x^2+9=0$

$\sf y^2-10y+9=0$

$\sf \Delta=b^2-4\cdot a\cdot c$

$\sf \Delta=(-10)^2-4\cdot1\cdot9$

$\sf \Delta=100-36$

$\sf \Delta=64$

$\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2\cdot a}$

$\sf y=\dfrac{-(-10)\pm\sqrt{64}}{2\cdot1}=\dfrac{10\pm8}{2}$

$\sf y'=\dfrac{10+8}{2}~\longrightarrow~y'=\dfrac{18}{2}~\longrightarrow~y'=9$

$\sf y"=\dfrac{10-8}{2}~\longrightarrow~y"=\dfrac{2}{2}~\longrightarrow~y"=1$

• Para $\sf y'=9$:

$\sf x^2=9$

$\sf x=\pm\sqrt{9}$

$\sf x=\pm3$

• Para $\sf y"=1$:

$\sf x^2=1$

$\sf x=\pm\sqrt{1}$

$\sf x=\pm2$

$\sf S=\{-3,-1,1,3\}$