Question

1)Determine no conjuntos dos reais a solução de cada uma das equações biquadradas a seguir:

a)x⁴-x²-12= 0
b)x⁴-5x²+4= 0
c)x⁴-x²-2= 0
d)9x⁴-40x²+16= 0
e)x²(x²-9)= -20​

Answer 1

 Pura fatoração...

a -

$x^4-x^2-12\\x^4+3.x^2-4.x^2-12=0\\x^2.(x^2+3)-4.(x^2+3)=0\\(x^2+3).(x^2-4)=0\\\\x^2+3=0\\x^2=-3\\\\x^2-4=0\\x^2=4\\x=+-\sqrt{4} \\\\x'=2\\x''=-2$

 S = { -2, 2}

b -

$x^4-5.x^2+4=0\\x^4-x^2-4.x^2+4=0\\x^2.(x^2-1)-4.(x^2-1)=0\\(x^2-4).(x^2-1)=0\\\\x^2-4=0\\x^2=4\\x=+-\sqrt{4}\\\\x'=2\\x''=-2\\\\x^2-1=0\\x^2=1\\x=+-\sqrt{1}\\\\x'=1\\x''=-1$

 S = { -2, -1, 1, 2}

c -

$x^4-x^2-2=0\\x^4-2.x^2+x^2-2=0\\x^2.(x^2-2)+(x^2-2)=0\\(x^2+1).(x^2-2)=0\\\\x^2+1=0\\x^2=-1\\\\x^2-2=0x^2=2\\x=+-\sqrt{2}\\\\x'=-\sqrt{2}\\x'' =\sqrt{2}$

 S = { -√2, √2}

d -

$9.x^4-40.x^2+16=0\\9.x^4-36.x^2-4.x^2+16=0\\9.x^2.(x^2-4)-4.(x^2-4)=0\\(9.x^2-4).(x^2-4)=0\\\\9.x^2-4=0\\9.x^2=4\\x^2=\frac{4}{9}\\x=+-\sqrt{\frac{4}{9}}\\x=+-\frac{2}{3}\\\\x'=\frac{2}{3}\\x''=-\frac{2}{3}\\\\x^2-4=0\\x^2=4\\x=+-\sqrt{4}\\x=+-2\\\\x'=2\\x''=-2$

 S = { -2, -2/3, 2/3, 2}

e -

$x^2.(x^2-9)=-20\\x^4-9.x^2+20=0\\x^4-4x^2-5.x^2+20=0\\x^2.(x^2-4)-5.(x^2-4)=0\\(x^2-5).(x^2-4)=0\\\\x^2-5=0\\x^2=5\\x=+-\sqrt{5}\\x'=\sqrt{5}\\x''=-\sqrt{5}\\\\x^2-4=0\\x^2=4\\x=+-\sqrt{4}\\x=+-2\\x'=2\\x''=-2$

 S = { -√5, -2, 2, √5}

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