Question

Respondam, é biquadrada

a)x²(x²-10)+9=(x+1)(x-1)
b)9x^4-13x²+4=0

Answer 1

Olá Alana,

$x^{2} ( x^{2} -10)+9=(x+1)(x-1)\\ x^4-10x^2+9=x^2-1\\ x^4-10x^2+9-x^2+1=0\\ x^4-10x^2-x^2+9+1=0\\ x^4-11x^2+10=0\\ (x^2)^2-11x^2+10=0\\\\ x^2=k\\\\ k^2-11k+10=0$

$\Delta=b^2-4ac\\ \Delta=(-11)^2-4*1*10\\ \Delta=121-40\\ \Delta=81\\\\ k= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-(-11)\pm \sqrt{81} }{2*1}= \dfrac{11\pm9}{2}\begin{cases}k'= \dfrac{11-9}{2}= \dfrac{2}{2}=1\\\\ k''= \dfrac{11+9}{2}= \dfrac{20}{2}=10 \end{cases}\\\\ x^{2} =k\\\\ x^{2} =1~~~~~~~~~~~~~~~~~~ x^{2} ~=10\\ x=\pm \sqrt{1}~~~~~~~~~~~~~~x=\pm \sqrt{10} \\ x=\pm1\\\\ \boxed{S=\{1,-1, \sqrt{10},- \sqrt{10}\}}$

________________________

$9x^4-13x^2+4=0\\ (3x^2)^2-13x^2+4=0\\\\ x^2=k\\\\ 3k^2-13k+4=0\\\\ \Delta=(-13)^2-4*3*4\\ \Delta=169-48\\ \Delta=121\\\\ k= \dfrac{-(-13)\pm \sqrt{121} }{2*3}= \dfrac{13\pm11}{6}\begin{cases}k'= \dfrac{13-11}{6}= \dfrac{2}{6}= \dfrac{1}{3}\\\\ k''= \dfrac{13+11}{6} = \dfrac{24}{6}=4 \end{cases}$

$x^{2} =k:\\\\ x^{2} = \dfrac{1}{3}~~~~~~~~~~~~~~~~~~~~~~ x^{2} =4\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x~=\pm \sqrt{4} \\ x=\pm \sqrt{ \dfrac{1}{3} }~~~~~~~~~~~~~~~~~~x=\pm2\\\\\\ x=\pm \dfrac{ \sqrt{1} }{ \sqrt{3} }=\pm \dfrac{1}{ \sqrt{3} }=\pm \dfrac{1* \sqrt{3} }{ \sqrt{3}* \sqrt{3} }~\to~x'~~e~~x''=\pm \dfrac{ \sqrt{3} }{3}$

$S=\left\{ \dfrac{ \sqrt{3} }{3},- \dfrac{ \sqrt{3} }{3},~2,-2 \right\}$

A outra é mais fácil, faça o mesmo.^^

Tenha ótimos estudos.