Question

Разность квадратов корней уравнения х^2-8х+b=0 равна 32 найдите b

Answer 1

$x^2-8x+b=0 \\ x_1^2-x_2^2=32 \\ \\ a^2-b^2=(a-b)(a+b) \\ \\ x_1^2-x_2^2=(x_1-x_2)(x_1+x_2)=32 \\ x_1+x_2=8 \\ 8(x_1-x_2)=32 \\ x_1-x_2= 4 \\ \\ \left \{ {{ x_1^2-x_2^2=32} \atop {x_1-x_2= 4 }} \right. \\ \left \{ {{ x_1^2-x_2^2=32} \atop {x_1= 4 +x_2}} \right. \\ \left \{ {{ (4 +x_2)^2-x_2^2=32} \atop {x_1= 4 +x_2}} \right. \\$

$(4 +x_2)^2-x_2^2=32} \\ (4 +x)^2-x^2=32} \\ 16+8x+x^2-x^2=32 \\ 8x=32-16 \\ 8x=16 \\ x=2 \\ x_2=2 \\ \\ \\ \left \{ {{ x_2=2} \atop {x_1= 4 +2} \right. \\ \\ \left \{ {{ x_2=2} \atop {x_1= 6} \right. \\ \\ b=x_1x_2 \\ b=6*2=12 \\ \\ \\ x^2-8x+6=0$