Question

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Answer 1

$sqrt{2x+8} = x, \ x geq 0, \ 2x+8 = x^2, \ x^2-2x-8 = 0, \ x_1=-2<0, x_2=4, \ x=4.$
$sqrt{5x+1} = 1-x, \ 1-x geq 0, x leq 1, \ 5x+1=(1-x)^2, \ 5x+1=1-2x+x^2, \ x^2-7x=0, \ x(x-7)=0, \ x_1=0, \ x-7=0, x_2=7>1, \ x=0.$

Answer 2

$sqrt{2x+8} =x \ left { {{2x+8=x^2} atop {x geq 0}} right. \ left { {{x^2-2x-8=0} atop {x geq 0}} right. \ left { {{(x-4)(x+2)=0} atop {x geq 0}} right. \ left { {{x_1=4; x_2=-2} atop {x geq 0}} right. \ x=4$
Ответ: 4

$sqrt{5x+1} =1-x \ left { {{5x+1=(1-x)^2} atop {1-x geq 0}} right. \ left { {{5x+1=1-2x+x^2} atop {x leq 1}} right. \ left { {{x^2-7x=0} atop {x leq 1}} right. \ left { {{x(x-7)=0} atop {x leq 1}} right. \ left { {{x_1=0; x_2=7} atop {x leq 1}} right. \ x=0$
Ответ: 0