Answer :
[tex]\sqrt{x-1}+3=x;\ Domain:x-1\geq0\ \wedge\ x\geq3\to x\geq1\ \wedge\ x\geq3\\D:x\in[3;\ \infty)\\\\\sqrt{x-1}=x-3\ \ \ \ |square\ both\ sides\\\\x-1=(x-3)^2\ \ \ \ |use\ of\ the\ formula:(a-b)^2=a^2-2ab+b^2\\\\x-1=x^2-6x+9\\\\x^2-6x+9=x-1\\\\x^2-6x-x+9+1=0\\x^2-7x+10=0\ \ \ \ |use\ quadratic\ formula[/tex]
[tex]a=1;\ b=-7;\ c=10\\\\\Delta=b^2-4ac\to\Delta=(-7)^2-4\cdot1\cdot10=49-40=9\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt9=3\\\\x_1=\frac{7-3}{2\cdot1}=\frac{4}{2}=2\notin D\\\\x_2=\frac{7+3}{2\cdot1}=\frac{10}{2}=5\in D\\\\Solution:x=5[/tex]
[tex]a=1;\ b=-7;\ c=10\\\\\Delta=b^2-4ac\to\Delta=(-7)^2-4\cdot1\cdot10=49-40=9\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt9=3\\\\x_1=\frac{7-3}{2\cdot1}=\frac{4}{2}=2\notin D\\\\x_2=\frac{7+3}{2\cdot1}=\frac{10}{2}=5\in D\\\\Solution:x=5[/tex]
[tex]\sqrt{x-1}+3=x\\
\sqrt{x-1}=x-3\\
D:x-1\geq0 \wedge x-3\geq0\\
D:x\geq1 \wedge x\geq3\\
D:x\geq3\\
x-1=(x-3)^2\\
x-1=x^2-6x+9\\
x^2-7x+10=0\\
x^2-2x-5x+10=0\\
x(x-2)-5(x-2)=0\\
(x-5)(x-2)=0\\
x=5 \vee x=2\\
2\not \in D \Rightarrow x=5[/tex]
