Найдите увеличенную в 5 раз сумму корней уравнения

Решите задачу:

$\sqrt{1+2x \sqrt{1- x^{2} } } =2-x \\ \sqrt{(x+ \sqrt{1- x^{2} })^2 } =2-x \\ |x+ \sqrt{1- x^{2}}|=2-x \\ \\ 1)x+ \sqrt{1- x^{2}}=2-x \\ 2)x+ \sqrt{1- x^{2}}=x-2 \\ \\ \\ 1)x+ \sqrt{1- x^{2}}=2-x \\ \\ \sqrt{1- x^{2}}=2-2x\ \textless \ =\ \textgreater \ \left \{ {{2-2x \geq 0} \atop {1- x^{2} =4-8x+4 x^{2} }} \right. \\ \\ \left \{ {{2x \leq 2} \atop {5 x^{2} -8x+3=0}} \right.\ \ \left \{ {x \leq 1} \atop {5 x^{2} -8x+3=0}} \right. \\ \\ 5 x^{2} -8x+3=0 \\ \\ D=64-4*5*3=64-60=4=2^2 \\ \\$

$x_1= \frac{8-2}{2*5} =0.6 \\ \\ x_2= \frac{8+2}{2*5} =1$

$1) \left \{ {{x \leq 1} \atop {x_1=0.6;\ x_2=1}} \right. \ \textless \ =\ \textgreater \ x_1=0.6;\ x_2=1 \\ \\ \\2) x+ \sqrt{1- x^{2}}=x-2 \\ \sqrt{1- x^{2} } =-2\ \ =\ \textgreater \ KOPHEI\ HET \\ \\ 1)\ 5(x_1+x_2)= 5(0.6+1)=5*1.6=8 \\ \\ OTBET:\ 8$