Question

найдите все значения а,при которых уравнение не имеет решений
a(x^2+x^-2)-(a+1)(x+x^-1)+5=0

Answer 1

$a(x^2+x^{-2})-(a+1)(x+x^{-1})+5=0, \ 1) a=0, \ -(x+x^{-1})+5=0, \ -x^2-1+5x=0, \ x^2-5x+1=0, \ D=21 textgreater 0, \ x_{1,2}in R; \ 2) a neq 0, \ a(x^2+2+x^{-2})-2a-(a+1)(x+x^{-1})+5=0, \ a(x+x^{-1})^2-(a+1)(x+x^{-1})-2a+5=0, \ x+x^{-1}=t, \ at^2-(a+1)t-2a+5=0, \ D=(a+1)^2-4a(5-2a)=a^2+2a+1-20a+8a^2=\=9a^2-18a+1;$
$2.1) D textless 0, 9a^2-18a+1 textless 0, \ 9a^2-18a+1=0, \ D_{/4}=72, \ a_{1,2}=frac{9pm6sqrt{2}}{9}=1pmfrac{2sqrt{2}}{3}, \ 9(a-1+frac{2sqrt{2}}{3})(a-1-frac{2sqrt{2}}{3}) textless 0, \ 1-frac{2sqrt{2}}{3} textless a textless 1+frac{2sqrt{2}}{3}, \ ain(1-frac{2sqrt{2}}{3};1+frac{2sqrt{2}}{3}) Rightarrow tinvarnothing, xinvarnothing;$
$2.2) D geq 0, 9a^2-18a+1 geq 0, \ ain(-infty;1-frac{2sqrt{2}}{3},)cup(1+frac{2sqrt{2}}{3},+infty), \ t_{1,2}=frac{a+1pmsqrt{9a^2-18a+1}}{2a}, \ left [ {{x+x^{-1}=frac{a+1-sqrt{9a^2-18a+1}}{2a},} atop {x+x^{-1}=frac{a+1+sqrt{9a^2-18a+1}}{2a};}} right. left [ {{x^2-frac{a+1-sqrt{9a^2-18a+1}}{2a}x+1=0,} atop {x^2-frac{a+1+sqrt{9a^2-18a+1}}{2a}x+1=0;}} right.$
$D=(frac{a+1pmsqrt{9a^2-18a+1}}{2a})^2-4=\=frac{(a+1)^2pm2(a+1)sqrt{9a^2-18a+1}+(9a^2-18a+1)}{4a^2}-4=\=frac{a^2+2a+1pm(a+1)sqrt{9a^2-18a+1}-9a^2+18a-1-16a^2}{4a^2}=\=frac{-24a^2+20apm(a+1)sqrt{9a^2-18a+1}}{4a^2}, \ D textless 0, -24a^2+20apm(a+1)sqrt{9a^2-18a+1} textless 0, \ a+1=0, a=-1, \ D=-24-20pm0=-44 textless 0, xinvarnothing;$
$a neq -1, left { {{sqrt{9a^2-18a+1} textless frac{24a^2-20a}{a+1},} atop {{sqrt{9a^2-18a+1} textgreater frac{20a-24a^2}{a+1};}} right.$