Question

Помогите решить уравнение!

7x^2+20x-14=5√(x^4-20x^2+4)

Answer 1

$[tex]7x^2+20x-14=5sqrt{x^4-20x^2+4}\sqrt{x^4-20x^2+4}geq0\x^4-20x^2+4geq0\(x^2+4x-2)(x^2-4x-2)geq0\x^2+4x-2=0\x_1=-2+sqrt{6}; x_2=-2-sqrt{6}\x^2-4x^2-2=0\x_1=2+sqrt{6}; x_2=2-sqrt{6}$\ 5sqrt{x^4-20x^2+4}=7x^2+20x-14\25(x^4-20x^2+4)=49x^4+400x^2+196+280x^3-196^2-520x\25(x^4-20x^2+4)-49x^4-400x^2-196-280x^3+196^2+520x=0\25x^4-500x^2+100-49x^4-400x^2-196-280x^3+196^2+520x=0\-24x^4-704x^2-96-280x^3+560=0\-8(3x^4-6x^2+100x^2-6x^2+12+16x^3+20x^3-40x-30x)=0\3x^4-6x^2+100x^2-6x^2+12+16x^3+20x^3-40x-30x=0\3x^2(x^2+5x-2)+20x(x^2+5x-2)$</p><p>[tex]-6(x^2+5x-2)=0\(x^2+5x-2)(3x^2+20x-6)=0$

х∈[-∞;-2-√6]∪[-2-√6;-2+√6]∪[2+√6;+∞]

Теперь находим корни уровнения.

$(x^2+5x-2)(3x^2+20x-6)=0\a_1=x^2+5x-2=0\a_2=3x^2+20x-6=0\D(a_1)=+-sqrt{25+8}=+-sqrt{33}\x_1(a_1)neq frac{-5+sqrt{33}}{2}; x=frac{-5-sqrt{33}}{2}\D(a_2)=sqrt{400+72}=sqrt{472}=+-2sqrt{118}\x_1(a_2)=frac{-20+2sqrt{118}}{6}neqfrac{-10+sqrt{118}}{3};x_2(a_2)=frac{-10-sqrt{118}}{3}$